relations,functions, and graphs - northshore …functions, and graphs ... for more on this relation,...

Precalculus—

Relations, Functions,and GraphsCHAPTER OUTLINE

1.1 Rectangular Coordinates; Graphing Circles and Other Relations 2

1.2 Linear Equations and Rates of Change 19

1.3 Functions, Function Notation, and the Graph of a Function 33

1.4 Linear Functions, Special Forms, and More on Rates of Change 50

1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving 64

1.6 Linear Function Models and Real Data 79

CHAPTER CONNECTIONS

Viewing r elations and functions in ter ms of anequation, a table of values, and the r elatedgraph, often brings a clear er understanding ofthe relationships involved. For instance, whilemany business ar e aware that Inter net use isincreasing with time, they ar e ver y interestedin the rate of gr owth, in or der to pr epare anddevelop r elated goods and ser vices.

� This application appears as Exer cise 109 inSection 1.4.

1

A solid understanding of lines and their equations is fundamental to a study of differential calculus. TheConnections to Calculus feature for Chapter 1 reviews these essential concepts and skills, and provides anopportunity for practice in the context of a future calculus course.

Connectionsto Calculus

11

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1.1 Rectangular Coordinates; Graphing Circles and Other Relations

In everyday life, we encounter a large variety of relationships. For instance, the time ittakes us to get to work is related to our average speed; the monthly cost of heating ahome is related to the average outdoor temperature; and in many cases, the amount ofour charitable giving is related to changes in the cost of living. In each case we say thata relation exists between the two quantities.

A. Relations, Mapping Notation, and Ordered Pairs

In the most general sense, a relation is simply a corre-spondence between two sets. Relations can be repre-sented in many different ways and may even be very“unmathematical,” like the one shown in Figure 1.1between a set of people and the set of their correspond-ing birthdays. If P represents the set of people and Brepresents the set of birthdays, we say that elements ofP correspond to elements of B, or the birthday relationmaps elements of P to elements of B. Using what is calledmapping notation, we might simply write Froma purely practical standpoint, we note that while it is possible for two different peopleto share the same birthday, it is quite impossible for the same person to have two differentbirthdays. Later, this observation will help us mark the difference between a relationand special kind of relation called a function.

The bar graph in Figure 1.2 is also anexample of a relation. In the graph, eachyear is related to annual consumer spend-ing per person on cable and satellite televi-sion. As an alternative to mapping or a bargraph, this relation could also be repre-sented using ordered pairs. For example,the ordered pair (5, 234) would indicatethat in 2005, spending per person oncable and satellite TV in the UnitedStates averaged $234. When a relation isrepresented using ordered pairs, we saythe relation is pointwise-defined.

Over a long period of time, we couldcollect many ordered pairs of the form (t, s), where consumer spending s depends on thetime t. For this reason we often call the second coordinate of an ordered pair (in this case s)the dependent variable, with the first coordinate designated as the independent variable.The set of all first coordinates is called the domain of the relation. The set of all secondcoordinates is called the range.

P S B.

LEARNING OBJECTIVESIn Section 1.1 you will see

how we can:

A. Express a relation inmapping notation andordered pair form

B. Graph a relation

C. Graph relations on a calculator

D. Develop the equation andgraph of a circle usingthe distance and midpointformulas

2 1–2

Precalculus—

Con

sum

er s

pend

ing

(dol

lars

per

yea

r) 400

500

300

200

100

7 95

($192)

($234)

($281)

($375)($411-est)

3 11

Year (0 → 2000)

MissyJeff

AngieMegan

MackenzieMichaelMitchell

April 12

Nov 11

Sept 10

Nov 28

May 7

April 14

P BFigure 1.1

Figure 1.2

Source: 2009 Statistical Abstract of the United States,Table 1089 (some figures are estimates)

cob19537_ch01_001-018.qxd 1/31/11 5:14 PM

B. The Graph of a Relation

Relations can also be stated in equation form. The equation expresses arelation where each y-value is one less than the corresponding x-value (see Table 1.1).The equation expresses a relation where each x-value corresponds to theabsolute value of y (see Table 1.2). In each case, the relation is the set of all orderedpairs (x, y) that create a true statement when substituted, and a few ordered pair solu-tions are shown in the tables for each equation.

Relations can be expressed graphically using a rec-tangular coordinate system. It consists of a horizontalnumber line (the x-axis) and a vertical number line (they-axis) intersecting at their zero marks. The point of inter-section is called the origin. The x- and y-axes create aflat, two-dimensional surface called the xy-plane anddivide the plane into four regions called quadrants.These are labeled using a capital “Q” (for quadrant) andthe Roman numerals I through IV, beginning in theupper right and moving counterclockwise (Figure 1.3).The grid lines shown denote the integer values on eachaxis and further divide the plane into a coordinate grid,where every point in the plane corresponds to an orderedpair. Since a point at the origin has not moved along eitheraxis, it has coordinates (0, 0). To plot a point (x, y) meanswe place a dot at its location in the xy-plane. A few of theordered pairs from are plotted in Figure 1.4,where a noticeable pattern emerges—the points seem tolie along a straight line.

If a relation is pointwise-defined, the graph of therelation is simply the plotted points. The graph of a re-lation in equation form, such as , is the set ofall ordered pairs (x, y) that are solutions (make theequation true).

Solutions to an Equation in Two Variables

1. If substituting and results in a true equation, the ordered pair (a, b) is a solution and on the graph of the relation.

2. If the ordered pair (a, b) is on the graph of a relation, it is a solution(substituting and will result in a true equation).y � bx � a

y � bx � a

y � x � 1

y � x � 1

x � �y�

y � x � 1

1–3 Section 1.1 Rectangular Coordinates; Graphing Circles and Other Relations 3

Precalculus—

Table 1.1 y � x � 1

Table 1.2 x � �y �

EXAMPLE 1 � Expressing a Relation as a Mapping and as a P ointwise-Defined RelatioRepresent the relation from Figure 1.2 in mapping notation and as a pointwise-defined relation, then state its domain and range.

Solution � Let t represent the year and s represent consumer spending.The mapping gives the diagram shown. As a pointwise-defined relation we have (3, 192), (5, 234), (7, 281), (9, 375),and (11, 411). The domain is the set {3, 5, 7, 9, 11}; therange is {192, 234, 281, 375, 411}.

Now try Exercises 7 through 12 �

For more on this relation, see Exercise 93.

t S s

t s

192234281375411

357911

x

QIQII

QIVQIII

y

54321�5 �4 �3 �2 �1�1

�2

�3

�4

�5

2

3

4

5

1

x

y

5�5

�5

5

(�2, �3)

(�4, �5)

(0, �1)

(4, 3)

(2, 1)

Figure 1.3

Figure 1.4

A. You’ve just seen how

we can express a relation in

mapping notation and ordered

pair form

x y

0

2 1

4 3

�1

�3�2

�5�4

x y

2

1

0 0

1 1

2 2

�1

�2

cob19537_ch01_001-018.qxd 1/31/11 5:14 PM

4 CHAPTER 1 Relations, Functions, and Graphs 1–4

Precalculus—

(4, 3)

(2, 1)

(0, �1)(�2, �3)

(�4, �5)

y � x � 1

x

y

5�5

�5

5

(2, 2)

(0, 0)

(2, �2)

x � �y�

x

y

5�5

�5

5

Figure 1.5

Figure 1.7y � x � 1: selected integer values

Figure 1.9y � x � 1: selected real number values

Figure 1.10y � x � 1: all real number values

Figure 1.8y � x � 1: selected rational values

Figure 1.6

WORTHY OF NOTEAs the graphs in Example 2indicate, arrowheads are usedwhere appropriate to indicate theinfinite extension of a graph.


While we used only a few points to graph the relations in Example 2, they areactually made up of an infinite number of ordered pairs that satisfy each equation, includ-ing those that might be rational or irrational. This understanding is an important part ofreading and interpreting graphs, and is illustrated for you in Figures 1.7 through 1.10.

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

We generally use only a few select points to determine the shape of a graph, then draw astraight line or smooth curve through these points, as indicated by any patterns formed.

EXAMPLE 2 � Graphing RelationsGraph the relations and using the ordered pairs given in Tables 1.1and 1.2.

Solution � For , we plot the points then connect them with a straight line (Figure 1.5).For , the plotted points form a V-shaped graph made up of two half lines(Figure 1.6).

x � �y�y � x � 1

x � �y�y � x � 1

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Since there are an infinite number of ordered pairs forming the graph of , thedomain cannot be given in list form. Here we note x can be any real number and writeD: Likewise, y can be any real number and for the range we have R: Allof these points together make these graphs continuous, which for our purposes meansyou can draw the entire graph without lifting your pencil from the paper.

Actually, a majority of graphs cannot be drawn using only a straight line or directedline segments. In these cases, we rely on a “sufficient number” of points to outline thebasic shape of the graph, then connect the points with a smooth curve. As your experi-ence with graphing increases, this “sufficient number of points” tends to get smaller asyou learn to anticipate what the graph of a given relation should look like. In particular,for the linear graph in Figure 1.5 we notice that both the x- and y-variables have animplied exponent of 1. This is in fact a characteristic of linear equations and graphs. InExample 3 we’ll notice that if the exponent on one of the variables is 2 (either x or y issquared ) while the other exponent is 1, the result is a graph called a parabola. If thex-term is squared (Example 3a) the parabola is oriented vertically, as in Figure 1.11,and its highest or lowest point is called the vertex. If the y-term is squared (Example 3c),the parabola is oriented horizontally, as in Figure 1.13, and the leftmost or rightmostpoint is the vertex. The graphs and equations of other relations likewise have certainidentifying characteristics. See Exercises 85 through 92.

EXAMPLE 3 � Graphing RelationsGraph the following relations by completing the tables given. Then use the graphto state the domain and range of the relation.

a. b. c.

Solution � For each relation, we use each x-input in turn to determine the related y-output(s),if they exist. Results can be entered in a table and the ordered pairs used to assist in drawing a complete graph.

a. y � x2 � 2x

x � y2y � 29 � x2y � x2 � 2x

y � �.x � �.

y � x � 1


Precalculus—

Figure 1.11

(2, 0)

(3, 3)

(4, 8)(�2, 8)

(�1, 3)

(0, 0)

(1, �1)

y � x2 � 2x

x

y

5�5

�2

5

The resulting vertical parabola is shown in Figure 1.11. Although ( ) and( ) cannot be plotted here, the arrowheads indicate an infinite extension ofthe graph, which will include these points. This “infinite extension” in the upwarddirection shows there is no largest y-value (the graph becomes infinitely “tall”).Since the smallest possible y-value is [from the vertex ( )], the range is

. However, this extension also continues forever in the outward directionas well (the graph gets wider and wider). This means the x-value of all possibleordered pairs could vary from negative to positive infinity, and the domain is allreal numbers. We then have and .R: y � �1D: x � �

y � �11, �1�1

�3, 15�4, 24

(x, y) x y Ordered Pairs

24 ( )

15 ( )

8 ( )

3 ( )

0 0 (0, 0)

1 (1, )

2 0 (2, 0)

3 3 (3, 3)

4 8 (4, 8)

�1�1

�1, 3�1

�2, 8�2

�3, 15�3

�4, 24�4

cob19537_ch01_001-018.qxd 1/31/11 5:14 PM

b. y � 29 � x2

The result is the graph of a semicircle (Figure 1.12). The points with irrationalcoordinates were graphed by estimating their location. Note that when or , the relation does not represent a real number and nopoints can be graphed. Also note that no arrowheads are used since the graphterminates at ( ) and (3, 0). These observations and the graph itself showthat for this relation, , and .

c. Similar to the relation is defined only for since y2 isalways nonnegative ( has no real solutions). In addition, we reasonthat each positive x-value will correspond to two y-values. For example, given

, ( ) and (4, 2) are both solutions to .x � y24, �2x � 4

�1 � y2x � 0x � y2x � �y�,

R: 0 � x � 3D: �3 � x � 3�3, 0

y � 29 � x2x 7 3x 6 �3

This relation is a horizontal parabola, with a vertex at (0, 0) (Figure 1.13). Thegraph begins at and extends infinitely to the right, showing the domainis . Similar to Example 3a, this “infinite extension” also extends in boththe upward and downward directions and the y-value of all possible orderedpairs could vary from negative to positive infinity. We then have D: andR: .


y � �x � 0

x � 0x � 0


Precalculus—

Figure 1.12

Figure 1.13

(�3, 0) (3, 0)

(0, 3)

(�2, �5) (2, �5)(1, 2�2)(�1, 2�2)

y � �9 � x2

x

y

5�5

�5

5

(4, �2)(2, ��2)

(2, �2)

(4, 2)

x � y2

x

y

5�5

�5

5

(0, 0)

x � y2


not real —

0 ( )

( )

( )

0 3 (0, 3)

1 (1, )

2 (2, )

3 0 (3, 0)

4 not real —

1515

212212

�1, 212212�1

�2, 1515�2

�3, 0�3

�4

B. You’ve just seen how

we can graph relations

C. Graphing Relations on a Calculator

For relations given in equation form, the TABLE feature of a graphing calculator canbe used to compute ordered pairs, and the feature to draw the related graph. To usethese features, we first solve the equation for the variable y (write y in terms of x),then enter the right-hand expression on the calculator’s (equation editor) screen.Y=

GRAPH


not real —

not real —

0 0 (0, 0)

1 , 1 (1, ) and (1, 1)

2 (2, ) and (2, )

3 (3, ) and (3, )

4 , 2 (4, ) and (4, 2)�2�2

13�13�13, 13

12�12�12, 12

�1�1

�1

�2

cob19537_ch01_001-018.qxd 1/31/11 5:14 PM

We can then select either the feature, or set-up, create, and use the TABLEfeature. We’ll illustrate here using the relation .

1. Solve for y in terms of x.given equation

add 2x to each side

2. Enter the equation.Press the key to access the equation editor,then enter as Y1 (see Figure 1.14). Thecalculator automatically highlights the equal sign,showing that equation Y1 is now active. If thereare other equations on the screen, you can either

them or deactivate them by moving the cur-sor to overlay the equal sign and pressing .

3. Use the TABLE or . To set up the table, we use the keystrokes

(TBLSET). For this exercise, we’ll put thetable in the “Indpnt: Ask” mode, whichwill have the calculator automatically generate theinput and output values. In this mode, we can tellthe calculator where to start the inputs (we choseTblStart � �3), and have the calculator pro-duce the input values using any increment desired(we choose �Tbl � 1). See Figure 1.15A. Accessthe table using (TABLE), and the tableresulting from this setup is shown in Figure 1.15B.Notice that all ordered pairs satisfy the equation

, or “y is twice x increased by 3.”

Since much of our graphical work is centered at (0, 0)on the coordinate grid, the calculator’s default set-tings for the standard viewing are [ ] forboth x and y (Figure 1.16). The Xscl and Yscl valuesgive the scale used on each axis, and indicate here that each “tick mark” will be 1 unitapart. To graph the line in this window, we can use the key and select 6:ZStan-dard (Figure 1.17), which resets the window to these default settings and automati-cally graphs the line (Figure 1.18).

ZOOM

�10, 10WINDOW

y � 2x � 3

GRAPH2nd

AutoWINDOW

2nd

GRAPH

ENTER

CLEAR

2x � 3Y=

y � 2x � 3�2x � y � 3

�2x � y � 3GRAPH


Precalculus—

Figure 1.15A

Figure 1.15B

Figure 1.16 Figure 1.17

10

10

�10

�10

Figure 1.18

Figure 1.14

In addition to using the calculator’s TABLE feature to find ordered pairs for agiven graph, we can also use the calculator’s feature. As the name implies, thisfeature allows us to “trace” along the graph by moving a cursor to the left and right

using the arrow keys. The calculator displays the coordinates of the cursor’s loca-tion each time it moves. After pressing the key, the marker appears automaticallyand as you move it to the left or right, the current coordinates are shown at the bottom

TRACE

TRACE

cob19537_ch01_001-018.qxd 2/1/11 10:30 AM

of the screen (Figure 1.19). While not very“pretty,” (�0.8510638, 1.2978723) is a pointon this line (rounded to seven decimal places)and satisfies its equation. The calculator isdisplaying these decimal values because theviewing screen is exactly 95 pixels wide, 47pixels to the left of the y-axis, and 47 pixelsto the right. This means that each time youpress the left or right arrow, the x-valuechanges by 1/47th—which is not a nice roundnumber. To have the calculator through“friendlier” values, we can use the 4:ZDecimal feature, which sets Xminand Xmax , or 8:ZInteger, whichsets Xmin and Xmax . Let’suse the 4:ZDecimal option here, notingthe calculator automatically regraphs the line.Pressing the key once again and movingthe marker shows that more “friendly” or-dered pair solutions are displayed (Fig-ure 1.20). Other methods for finding a friendlywindow are discussed later in this section.

TRACE

ZOOM

� 47� �47ZOOM� 4.7

� �4.7ZOOM

TRACE


Precalculus—

EXAMPLE 4 � Graphing a Relation Using TechnologyUse a calculator to graph . Then use the TABLE feature to determinethe value of y when , and the value of x when . Write each result inordered pair form.

Solution � We begin by solving the equation for y, so we can enter it on the screen.given equation

subtract 2x (isolate the y-term)

divide by 3

Entering on the screen and using 6:ZStandard produces

the graph shown. Using the TABLE and scrolling as needed, shows that when ,, and when , . As ordered pairs we have and .1�3, 0210, �22x � �3y � 0y � �2

x � 0

ZOOMY=y ��2

3 x � 2

y ��2

3 x � 2

3y � �2x � 6 2x � 3y � �6

Y=

y � 0x � 02x � 3y � �6


D. The Equation and Graph of a Circle

Using the midpoint and distance formulas, we can develop the equation of anotherimportant relation, that of a circle. As the name suggests, the midpoint of a line segmentis located halfway between the endpoints. On a standard number line, the midpointof the line segment with endpoints 1 and 5 is 3, but more important, note that 3 is the

C. You’ve just seen how

we can graph a relation using

a calculator

10

10

�10

�10

4.7

3.1

�3.1

�4.7

10

10

�10

�10

Figure 1.19

Figure 1.20

cob19537_ch01_001-018.qxd 2/1/11 10:31 AM

The Midpoint Formula

Given any line segment with endpoints and , the midpoint M is given by

The midpoint formula can be used in many different ways. Here we’ll use it to findthe coordinates of the center of a circle.

M: ax1 � x2

2,

y1 � y2

2b

P2 � 1x2, y22P1 � 1x1, y12


Precalculus—

EXAMPLE 5 � Using the Midpoint FormulaThe diameter of a circle has endpoints at and . Use the midpoint formula to findthe coordinates of the center, then plot this point.

Solution � Midpoint:

The center is at (1, 1), which we graph directly on the diameter as shown.

M: a2

2,

2

2b � 11, 12

M: a�3 � 5

2,

�2 � 4

2b

ax1 � x2

2,

y1 � y2

2b

P2 � 15, 42P1� 1�3, �22


The Distance Formula

In addition to a line segment’s midpoint, we are often interested in the length of thesegment. For any two points (x1, y1) and (x2, y2) not lying on a horizontal or verticalline, a right triangle can be formed as in Figure 1.21. Regardless of the triangle’s ori-entation, the length of side a (the horizontal segment or base of the triangle) will havelength units, with side b (the vertical segment or height) having length

units. From the Pythagorean theorem (Appendix A.6), we see that corresponds to . By taking the square root of both sideswe obtain the length of the hypotenuse, which is identical to the distance between these

two points: . The result is called the distance formula, although it’s most often written using d for distance, rather than c. Note the absolutevalue bars are dropped from the formula, since the square of any quantity is alwaysnonnegative. This also means that either point can be used as the initial point in thecomputation.

c � 21x2 � x122 � 1y2 � y12

2

c2 � 1�x2 � x1�22 � 1�y2 � y1�22c2 � a2 � b2�y2 � y1�

�x2 � x1�

(x2, y2)

(x1, y1) (x2, y1)

c

x

y

b

a

Figure 1.21

average distance (from zero) of 1 unit and 5 units: This observation

can be extended to find the midpoint between any two points (x1, y1) and (x2, y2) in thexy-plane. We simply find the average distance between the x-coordinates and the aver-age distance between the y-coordinates.

1 � 5

2�

6

2� 3.

P2

P1

(1, 1)x

y

5�5

5

�5

The Distance Formula

Given any two points and the straight line distance dbetween them is

d � 21x2 � x122 � 1y2 � y12

2

P2 � 1x2, y22,P1 � 1x1, y12d

P2

b � �y

2 � y

1 �

P1 a � �x2� x1�

cob19537_ch01_001-018.qxd 1/31/11 5:15 PM


EXAMPLE 6 � Using the Distance FormulaUse the distance formula to find the diameter of the circle from Example 5.

Solution � For and the distance formula gives

The diameter of the circle is 10 units long.


A circle can be defined as the set of all points in a plane that are a fixed distance calledthe radius, from a fixed point called the center. Since the definition involves distance,we can construct the general equation of a circle using the distance formula. Assumethe center has coordinates (h, k), and let (x, y) represent any point on the graph. Thedistance between these points is equal to the radius r, and the distance formula yields:

Squaring both sides gives the equation of a circle in

standard form:

The Equation of a Circle

A circle of radius r with center at (h, k) has the equation

If and the circle is centeredat (0, 0) and the graph is a central circle withequation At other values for hor k, the center is at (h, k) with no change inthe radius. Note that an open dot is used forthe center, as it’s actually a point of referenceand not a part of the graph.

EXAMPLE 7 � Finding the Equation of a Circle in Standard F ormFind the equation of a circle with center ) and radius 4.

Solution � Since the center is at (0, ) we have and Using thestandard form we obtain

substitute 0 for h, for k, and 4 for r

simplifyx2 � 1y � 122 � 16�11x � 022 � 3y � 1�12 42 � 42

1x � h22 � 1y � k22 � r2r � 4.h � 0, k � �1,�1

10, �1

x2 � y2 � r2.

k � 0,h � 0

1x � h22 � 1y � k22 � r2

1x � h22 � 1y � k22 � r2.

21x � h22 � 1y � k22 � r.

� 1100 � 10

� 282 � 62

� 2 35 � 1�32 42 � 34 � 1�22 42 d � 21x2 � x12

2 � 1y2 � y122

1x2, y22 � 15, 42,1x1, y12 � 1�3, �22

r

r

h

k

(x, y)

x2 � y2 � r2

(x � h)2 � (y � k)2 � r2

(h, k)

(0, 0)

(x, y)

x

y

Centralcircle

Circle with centerat (h, k)

Precalculus—

cob19537_ch01_001-018.qxd 1/31/11 5:15 PM


The graph of is shown in the figure.


The graph of a circle can be obtained by first identifying the coordinates of the cen-ter and the length of the radius from the equation in standard form. After plotting thecenter point, we count a distance of r units left and right of center in the horizontaldirection, and up and down from center in the vertical direction, obtaining fourpoints on the circle. Neatly graph a circle containing these four points.

EXAMPLE 8 � Graphing a CircleGraph the circle represented by . Clearly label the centerand radius.

Solution � Comparing the given equation with the standard form, we find the center is atand the radius is

standard form

given equation

radius must be positive

Plot the center and count approximately 3.5 units in the horizontal andvertical directions. Complete the circle by freehand drawing or using a compass.The graph shown is obtained.


(2, �3)

(5.5, �3)

(2, 0.5)Some coordinatesare approximate

(2, �6.5)

(�1.5, �3)

y

xr 3.5~~

Circle

Center: (2, �3)Radius: r � 2�3

Endpoints of horizontal diameter(2 � 2�3, �3) and (2 � 2�3, �3)

Endpoints of vertical diameter(2, �3 � 2�3) and (2, �3 � 2�3)

12, �32

� 3.5r � 112 � 213k � �3h � 2

r2 � 12�k � 3�h � �21x � 222 � 1y � 322 � 12

1x � h22 � 1y � k22 � r2

r � 213 � 3.5.12, �32

1x � 222 � 1y � 322 � 12

(0, �1)

(0, 3)

Circle

Center: (0, �1)Radius: r � 4

Diameter: 2r � 8(4, �1)

(0, �5)

(�4, �1)x

y

r � 4

x2 � 1y � 122 � 16

Precalculus—

S S S

cob19537_ch01_001-018.qxd 1/31/11 5:15 PM

In Example 8, note the equation is composed of binomial squares in both x and y.By expanding the binomials and collecting like terms, we can write the equation of thecircle in general form:

standard form

expand binomials

combine like terms—general form

For future reference, observe the general form contains a sum of second-degreeterms in x and y, and that both terms have the same coefficient (in this case, “1”).

Since this form of the equation was derived by squaring binomials, it seems rea-sonable to assume we can go back to the standard form by creating binomial squaresin x and y. This is accomplished by completing the square.

EXAMPLE 9 � Finding the Center and Radius of a CircleFind the center and radius of the circle with equation Then sketch its graph and label the center and radius.

Solution � To find the center and radius, we complete the square in both x and y.

given equation

group x-terms and y-terms; add 4

complete each binomial square

adds 1 to left side adds 4 to left side add to right side

factor and simplify

The center is at and the radius is


EXAMPLE 10 � Applying the Equation of a CircleTo aid in a study of nocturnal animals, somenaturalists install a motion detector near apopular watering hole. The device has arange of 10 m in any direction. Assume thewater hole has coordinates (0, 0) and thedevice is placed at .

a. Write the equation of the circle thatmodels the maximum effective rangeof the device.

b. Use the distance formula to determineif the device will detect a badger thatis approaching the water and is now at coordinates .111, �52

12, �12

(�1, 2)

(�1, �1)

(2, 2)

(�1, 5)

(�4, 2)

y

x

r � 3

Circle

Center: (�1, 2)Radius: r � 3

r � 19 � 3.1�1, 22

1x � 122 � 1y � 222 � 91 � 4

1x2 � 2x � 12 � 1y2 � 4y � 42 � 4 � 1 � 4 1x2 � 2x � __2 � 1y2 � 4y � __2 � 4

x2 � y2 � 2x � 4y � 4 � 0

x2 � y2 � 2x � 4y � 4 � 0.

x2 � y2 � 4x � 6y � 1 � 0x2 � 4x � 4 � y2 � 6y � 9 � 12

1x � 222 � 1y � 322 � 12


Precalculus—

WORTHY OF NOTEAfter writing the equation instandard form, it is possible to endup with a constant that is zero ornegative. In the first case, thegraph is a single point. In thesecond case, no graph is possiblesince roots of the equation will becomplex numbers. These arecalled degenerate cases. SeeExercise 105.

x

y

10

�5

5

cob19537_ch01_001-018.qxd 1/31/11 5:15 PM

Solution � a. Since the device is at ( ) and the radius (or reach) of detection is 10 m, anymovement in the interior of the circle defined by will be detected.

b. Using the points ( ) and ( ) in the distance formula yields:

distance formula

substitute given values

simplify

compute squares

result

Since , the badger is within range of the device and will be detected.


When using a graphing calculator to study circles, it’s important to note that most standardviewing windows have the x- and y-values preset at [ ] even though the calculatorscreen is not square. This tends to compress the y-values and give a skewed image of thegraph. If the circle appears oval in shape, use 5:ZSquare to obtain the correct per-spective. Graphing calculators can produce the graph of a circle in various ways, and thechoice of method simply depends on what you’d like to accomplish. To simply view thegraph or compare two circular graphs, the DRAW command is used. From the homescreen press: (DRAW) 9:Circle(. This generates the “Circle(” command, withthe left parentheses indicating we need to supply three inputs, separated by commas.These inputs are the x-coordinate of the center, the y-coordinate of the center, and theradius of the circle. For the circle defined by the equation , weknow the center is at and the radius is 7 units. The resulting command and graphare shown in Figures 1.22 and 1.23.

While the DRAW command will graph any circle, we are unable to use the or CALC commands to interact with the graph. To make these features available, wemust first solve for x in terms of y, as we did previously (the 1:ClrDraw command isused to clear the graph). Consider the relation , which we know is theequation of a circle centered at (0, 0) with radius

original equation

isolate y 2

solve for y

Note that we can separate this result into two parts, enabling the calculator to graph(giving the “upper half” of the circle), and (giving

the “lower half”). Enter these on the screen (note that can be usedinstead of reentering the entire expression: ). If we graph Y1 and Y2 onthe standard screen, the result appears more oval than circular (Figure 1.24). Using the

5:ZSquare option, the tick marks become equally spaced on both axes (Figure 1.25).ZOOM

ENTERVARS

Y2 � �Y1Y=

Y2 � �225 � x2Y1 � 225 � x2

y � �225 � x2 y2 � 25 � x2

x2 � y2 � 25

r � 5.x2 � y2 � 25

TRACE

13, �221x � 322 � 1y � 222 � 49

PRGM2nd

ZOOM

�10, 10

9.85 6 10

� 197 � 9.85 � 181 � 16

� 292 � 1�422 � 2111 � 222 � 3�5 � 1�12 42

d � 21x2 � x122 � 1y2 � y12

2

11, �52, �1

1x � 222 � 1y � 122 � 1022, �1


Precalculus—

15.2

10

�10

�15.2 10

10

�10

�10

Figure 1.24

Figure 1.22, 1.23

15.2

10

�10

�15.2

Figure 1.25

cob19537_ch01_001-018.qxd 2/1/11 10:31 AM


Precalculus—

4. For , the center of the circle is at________ and the length of the radius is ________units. The graph is called a ________ circle.

5. Discuss/Explain how to find the center and radius ofthe circle defined by the equation .How would this circle differ from the one defined by

?

6. In Example 3(b) we graphed the semicircle defined by . Discuss how you would obtain theequation of the full circle from this equation, andhow the two equations are related.

y � 29 � x2

x2 � y2 � 6y � 7

x2 � y2 � 6x � 7

x2 � y2 � 25

� CONCEPTS AND VOCABULAR Y

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1.1 EXERCISES

1. If a relation is defined by a set of ordered pairs, thedomain is the set of all ________ components, therange is the set of all ________ components.

2. For the equation and the ordered pair (x, y), x is referred to as the input or ________variable, while y is called the ________ ordependent variable.

3. A circle is defined as the set of all points that arean equal distance, called the ________, from agiven point, called the ________.

y � x � 5

Although it is a much improved graph, the circle does not appear “closed” as the cal-culator lacks sufficient pixels to show the proper curvature. A second alternative is tomanually set a “friendly” window. Using , , ,and will generate a better graph due to the number of pixels available.Note that we can jump between the upper and lower halves of the circle using the up or down arrows. See Exercises 101 and 102.

Ymax � 6.2Ymin � �6.2Xmax � 9.4Xmin � �9.4

� DEVELOPING YOUR SKILLS

Represent each relation in mapping notation, then statethe domain and range.

7.

8.

Eff

icie

ncy

ratin

g

90

85

95

80

75

70

65

60

55

0

Month

4 5 6321

GPA

3.75

3.50

4.00

3.25

3.00

2.75

2.50

2.25

2.00

01 32 54

Year in college

State the domain and range of each pointwise-definedrelation.

9. {(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)}

10. {( ), ( ), ( ), ( ), (2, )}

11. {(4, 0), ( ), (2, 4), (4, 2), ( )}

12. {( ), (0, 4), (2, ), ( ), (2, 3)}

Complete each table using the given equation. ForExercises 15, 16, 21, and 22, each input may correspondto two outputs (be sure to find both if they exist). Usethese points to graph the relation. For Exercises 17through 24, also state the domain and range.

13. 14. y � �5

4x � 3y � �

2

3x � 1

�3, 4�5�1, 1

�3, 3�1, 5

�34, �5�1, 3�3, �5�2, 4

x y

0

3

6

8

�3

�6

x y

0

4

8

10

�4

�8

D. You’ve just seen howwe can develop the equationand graph of a circle using thedistance and midpointformulas

cob19537_ch01_001-018.qxd 1/31/11 5:16 PM


Precalculus—

15. 16. �y � 1� � xx � 2 � �y� Use a graphing calculator to graph the followingrelations. Then use the TABLE feature to determine thevalue of y when , and the value(s) of x when ,and write the results in ordered pair form.

25.

26.

27.

28.

Find the midpoint of each segment with the givenendpoints.

29. (1, 8), (5, )

30. (5, 6), (6, )

31. ( , 9.2), (3.1, )

32. (5.2, 7.1), (6.3, )

33.

34. a�3

4, �

1

3b, a

3

8,

5

6b

a1

5, �

2

3b, a�

1

10,

3

4b

�7.1

�9.8�4.5

�8

�6

y � x2 � 2x � 3

y � x2 � 4x

x � 2y � 6

�2x � 5y � 10

y � 0x � 0x y

0

1

3

6

7

�2

x y

0

1

3

5

6

7

x y

0

2

3

4

�2

�3

x y

0

1

2

3

�1

�2

x y

0

2

3

4

�3

�4

x y

0

3

5

12

�5

�12

x y

10

5

4

2

1.25

1

x y

0

1

2

7

�1

�2

x y

0

4

7

�1

�2

�9

x y

0

1

2

3

�1

�2

17. 18.

19. 20.

21. 22.

23. 24. y � 1x � 123y � 23 x � 1

y2 � 2 � xx � 1 � y2

y � 2169 � x2y � 225 � x2

y � �x2 � 3y � x2 � 1

Find the midpoint of each segment.

35. 36.

Find the center of each circle with the diameter shown.

37. 38.

39. Use the distance formula to find the length of theline segment in Exercise 35.

40. Use the distance formula to find the length of theline segment in Exercise 36.

41. Use the distance formula to find the length of thediameter of the circle in Exercise 37.

42. Use the distance formula to find the length of thediameter of the circle in Exercise 38.

x

y

54321�5�4�3�2�1�1�2�3�4�5

2345

1

x

y

54321�5�4�3�2�1�1�2�3�4�5

2345

1

x

y

54321�5�4�3�2�1�1�2�3�4�5

2345

1

x

y

54321�5�4�3�2�1�1�2�3�4�5

2345

1

cob19537_ch01_001-018.qxd 1/31/11 5:16 PM

In Exercises 43 to 48, three points that form thevertices of a triangle are given. Use the distanceformula to determine if any of the triangles are righttriangles (the three sides satisfy the PythagoreanTheorem ).

43. ( , 7), (2, 2), (5, 5)

44. (7, 0), ( , 0), (7, 4)

45. ( , 3), ( ), (3, )

46. (5, 2), (0, ), (4, )

47. ( , 2), ( , 5), ( , 4)

48. (0, 0), ( , 2), (2, )

Find the equation of a circle satisfying the conditionsgiven, then sketch its graph.

49. center (0, 0), radius 3

50. center (0, 0), radius 6

51. center (5, 0), radius

52. center (0, 4), radius

53. center (4, ), radius 2

54. center (3, ), radius 9

55. center ( ), radius

56. center ( ), radius

57. center (1, ), diameter 6

58. center ( , 3), diameter 10

59. center (4, 5), diameter

60. center (5, 1), diameter

61. center at (7, 1), graph contains the point (1, )

62. center at ( , 3), graph contains the point ( , 15)

63. center at (3, 4), graph contains the point (7, 9)

64. center at ( , 2), graph contains the point ( , 3)

65. diameter has endpoints (5, 1) and (5, 7)

66. diameter has endpoints (2, 3) and (8, 3)

Identify the center and radius of each circle, then graph.Also state the domain and range of the relation.

67.

68.

69.

70.

71.

72. x2 � 1y � 322 � 49

1x � 422 � y2 � 81

1x � 722 � 1y � 422 � 20

1x � 122 � 1y � 222 � 12

1x � 522 � 1y � 122 � 9

1x � 222 � 1y � 322 � 4

�1�5

�3�8

�7

415

413

�2

�2

16�2, �5

17�7, �4

�8

�3

15

13

�5�5

�6�1�3

�4�3

�2�7, �1�4

�1

�3

a2 � b2 � c2


Precalculus—

Write each equation in standard form to find the centerand radius of the circle. Then sketch the graph.

73.

74.

75.

76.

77.

78.

79.

80.

81.

82.

83.

84.

In this section we looked at characteristics of equationsthat generated linear graphs, and graphs of parabolasand circles. Use this information and ordered pairs ofyour choosing to match the eight graphs given with theircorresponding equation (two of the equations given haveno matching graph).

a. b.

c. d.

e. f.

g. h.

i. j.

85. 86.

87. 88.

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

6x � y � x2 � 94x � 3y � 12

1x � 122 � 1y � 222 � 91x � 322 � y2 � 16

1x � 122 � 1y � 222 � 49y ��3

2x � 4

3x � 4y � 12x2 � y � 9

x2 � 1y � 322 � 36y � x2 � 6x

3x2 � 3y2 � 24x � 18y � 3 � 0

2x2 � 2y2 � 12x � 20y � 4 � 0

x2 � y2 � 22y � 5 � 0

x2 � y2 � 14x � 12 � 0

x2 � y2 � 8x � 14y � 47 � 0

x2 � y2 � 4x � 10y � 18 � 0

x2 � y2 � 8x � 12 � 0

x2 � y2 � 6y � 5 � 0

x2 � y2 � 6x � 4y � 12 � 0

x2 � y2 � 10x � 4y � 4 � 0

x2 � y2 � 6x � 8y � 6 � 0

x2 � y2 � 10x � 12y � 4 � 0

cob19537_ch01_001-018.qxd 1/31/11 5:16 PM

89. 90.

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2


Precalculus—

91. 92.

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

x

y

108642�10�8�6�4�2�2�4�6�8

�10

468

10

2

� WORKING WITH FORMULAS

93. Spending on Cable and Satellite TV:

The data from Example 1 is closely modeled by theformula shown, where t represents the year (corresponds to the year 2000) and s represents theaverage amount spent per person, per year in theUnited States. (a) List five ordered pairs for thisrelation using . Does the modelgive a good approximation of the actual data? (b) According to the model, what will be theaverage amount spent on cable and satellite TV inthe year 2013? (c) According to the model, in whatyear will annual spending surpass $500? (d) Usethe table to graph this relation by hand.

t � 3, 5, 7, 9, 11

t � 0

s � 29t � 96 94. Radius of a circumscribed circle:

The radius r of a circlecircumscribed around asquare is found by usingthe formula given, whereA is the area of thesquare. Solve the formulafor A and use the result tofind the area of the squareshown.

r �B

A2

y

(�5, 0)

x

� APPLICATIONS

95. Radar detection: A luxury liner is located at mapcoordinates (5, 12) and has a radar system with arange of 25 nautical miles in any direction. (a) Writethe equation of the circle that models the range ofthe ship’s radar, and (b) use the distance formula todetermine if the radar can pick up the liner’s sistership located at coordinates (15, 36).

99. Radio broadcast range: Two radio stations maynot use the same frequency if their broadcast areasoverlap. Suppose station KXRQ has a broadcast areabounded by and WLRT hasa broadcast area bounded by Graph the circle representing each broadcast areaon the same grid to determine if both stations maybroadcast on the same frequency.

x2 � y2 � 10x � 4y � 0.x2 � y2 � 8x � 6y � 0

97. Inscribed circle: Find theequation for both the red andblue circles, then find the areaof the region shaded in blue.

98. Inscribed triangle: The areaof an equilateral triangleinscribed in a circle is given

by the formula ,

where r is the radius of thecircle. Find the area of theequilateral triangle shown.

A �313

4r2

y

(�2, 0)

x x

y

(3, 4)

96. Earthquake range: The epicenter (point of origin)of a large earthquake was located at mapcoordinates (3, 7), with the quake being felt up to 12 mi away. (a) Write the equation of the circlethat models the range of the earthquake’s effect. (b) Use the distance formula to determine if aperson living at coordinates (13, 1) would have feltthe quake.

cob19537_ch01_001-018.qxd 1/31/11 5:17 PM


Precalculus—

� EXTENDING THE CONCEPT

103. Although we use the word “domain” extensively inmathematics, it is also commonly seen in literatureand heard in everyday conversation. Using a college-level dictionary, look up and write out the variousmeanings of the word, noting how closely thedefinitions given are related to its mathematical use.

104. Consider the following statement, then determinewhether it is true or false and discuss why. A graphwill exhibit some form of symmetry if, given a pointthat is h units from the x-axis, k units from the y-axis,and d units from the origin, there is a second pointon the graph that is a like distance from the originand each axis.

105. When completing the square to find the center andradius of a circle, we sometimes encounter a valuefor r2 that is negative or zero. These are calleddegenerate cases. If , no circle is possible,while if , the “graph” of the circle is simplythe point (h, k). Find the center and radius of thefollowing circles (if possible).

a.b.c. x2 � y2 � 6x � 10y � 35 � 0

x2 � y2 � 2x � 8y � 8 � 0

x2 � y2 � 12x � 4y � 40 � 0

r2 � 0r2 6 0

� MAINTAINING YOUR SKILLS

106. (Appendix A.2) Evaluate/Simplify the followingexpressions.

a.

b.

c.

d.

e. (2m3n)2

f. 15x20 � 5x0

2723

125�13

33 � 32 � 31 � 30 � 3�1

x2x5

x3

107. (Appendix A.3) Solve the following equation.

108. (Appendix A.4) Solve by factoring.

109. (Appendix A.6) Solve and checksolutions by substitution. If a solution is extraneous,so state.

1 � 1n � 3 � �n

x2 � 27 � 6x

x

3�

1

4�

5

6

100. Radio broadcast range: The emergency radiobroadcast system is designed to alert thepopulation by relaying an emergency signal to allpoints of the country. A signal is sent from a station whose broadcast area is bounded by

(x and y in miles) and the signal ispicked up and relayed by a transmitter with range

. Graph the circlerepresenting each broadcast area on the same gridto determine the greatest distance from the originalstation that this signal can be received. Be sure toscale the axes appropriately.

1x � 2022 � 1y � 3022 � 900

x2 � y2 � 2500

101. Graph the circle defined by using afriendly window, then use the feature to findthe value of y when . Now find the value ofy when . Explain why the values seem“interchangeable.”

102. Graph the circle defined by using a friendly window, then use the featureto find the value of the y-intercepts. Show you getthe same intercept by computation.

TRACE

1x � 322 � y2 � 16

x � 4.8x � 3.6

TRACE

x2 � y2 � 36

cob19537_ch01_001-018.qxd 2/15/11 9:34 PM

Precalculus—

1.2 Linear Equations and Rates of Change

In preparation for sketching graphs of other equations, we’ll first look more closely at thecharacteristics of linear graphs. While linear graphs are fairly simple models, they havemany substantive and meaningful applications. For instance, most of us are aware thatsatellite and cable TV have been increas-ing in popularity since they were firstintroduced. A close look at Figure 1.2 fromSection 1.1 reveals that spending on theseforms of entertainment increased from$192 per person per year in 2003 to $281in 2007 (Figure 1.26). From an investor’sor a producer’s point of view, there is avery high interest in the questions, “Howfast are sales increasing? Can this relation-ship be modeled mathematically to helppredict sales in future years?” Answers tothese and other questions are preciselywhat our study in this section is all about.

A. The Graph of a Linear Equation

A linear equation can be identified using these three tests:

1. the exponent on any variable is one,2. no variable occurs in a denominator, and3. no two variables are multiplied together.

The equation 3y � 9 is a linear equation in one variable, while 2x � 3y � 12 andare linear equations in two variables. In general, we have the following

definition:

Linear Equations

A linear equation is one that can be written in the form

where a, b, and c are real numbers, with a and b not simultaneously equal to zero.

As in Section 1.1, the most basic method for graphing a line is to simply plot a fewpoints, then draw a straight line through the points.

ax � by � c

y � �23 x � 4


how we can:

A. Graph linear equationsusing the interceptmethod

B. Find the slope of a lineand interpret it as a rateof change

C. Graph horizontal andvertical lines

D. Identify parallel andperpendicular lines

E. Apply linear equations incontext Source: 2009 Statistical Abstract of the United States,

Table 1089 (some figures are estimates)

Con

sum

er s

pend

ing

(dol

lars

per

yea

r) 400

500

300

200

100

7 95

($192)

($234)

($281)

($375)($411-est)

3 11

Year (0 → 2000)

EXAMPLE 1 � Graphing a Linear Equation in Two VariablesGraph the equation by plotting points.

Solution � Selecting and as inputs,we compute the related outputs and enter the orderedpairs in a table. The result is

x � 4x � 1,x � 0,x � �2,

3x � 2y � 4

(0, 2)

(1, q)

(4, �4)

(�2, 5)

x

y

5�5

�5

5


WORTHY OF NOTEIf you cannot draw a straight linethrough the plotted points, acomputational error has beenmade. All points satisfying a linearequation lie on a straight line.

1–19 19

Figure 1.26

x input y output (x, y) ordered pairs

5

0 2

1

4 14, �42�4

11, 12 212

10, 22

1�2, 52�2

cob19537_ch01_019-032.qxd 1/28/11 8:17 PM

Precalculus—

Notice that the line in Example 1 crosses the y-axis at (0, 2), and this point is calledthe y-intercept of the line. In general, y-intercepts have the form (0, y). Although dif-ficult to see graphically, substituting 0 for y and solving for x shows this line crossesthe x-axis at ( ) and this point is called the x-intercept. In general, x-intercepts havethe form (x, 0). The x- and y-intercepts are usually easier to calculate than other points(since or , respectively) and we often graph linear equations using onlythese two points. This is called the intercept method for graphing linear equations.

The Intercept Method

1. Substitute 0 for x and solve for y. This will give the y-intercept (0, y).2. Substitute 0 for y and solve for x. This will give the x-intercept (x, 0).3. Plot the intercepts and use them to graph a straight line.

EXAMPLE 2 � Graphing Lines Using the Intercept MethodGraph using the intercept method.

Solution � Substitute 0 for x (y-intercept) Substitute 0 for y (x-intercept)

(3, 0)

a0, 9

2b

x � 3 y �

9

2

3x � 9 2y � 93x � 2102 � 9 3102 � 2y � 9

3x � 2y � 9

x � 0y � 0

43, 0

Now try Exercises 13 through 32 �A. You’ve just seen how

we can graph linear equations

using the intercept method

(3, 0)

x

y

5�5

�5

5

�0, t �3x � 2y � 9

B. The Slope of a Line and Rates of Change

After the x- and y-intercepts, we next consider the slope of a line. We see applicationsof this concept in many diverse areas, including the grade of a highway (trucking), thepitch of a roof (carpentry), the climb of an airplane (flying), the drainage of a field(landscaping), and the slope of a mountain (parksand recreation). While the general concept is anintuitive one, we seek to quantify the concept (as-sign it a numeric value) for purposes of compari-son and decision making. In each of the precedingexamples (grade, pitch, climb, etc.), slope is ameasure of “steepness,” as defined by the ratio

. Using a line segment through arbitrary points and ,we can create the right triangle shown in Fig-ure 1.27 to help us quantify this relationship. Thefigure illustrates that the vertical change or the

P2 � 1x2, y22P1 � 1x1, y12

vertical change

horizontal change

x2x1

y2

y1

(x2, y2)

(x1, y1)

x2 � x1run

y2 � y1rise

x

yFigure 1.27


cob19537_ch01_019-032.qxd 1/28/11 8:17 PM

1–21 Section 1.2 Linear Equations and Rates of Change 21

Precalculus—

change in y (also called the rise) is simply the difference in y-coordinates: .The horizontal change or change in x (also called the run) is the difference in x-coordinates: . In algebra, we typically use the letter “m” to represent slope, giving as the . The result is called the slope formula.change in y

change in xm �y2 � y1

x2 � x1

x2 � x1

y2 � y1

The Slope Formula

Given two points and , the slope of the nonvertical linethrough P1 and P2 is

Actually, the slope value does much more than quantify the slope of a line, itexpresses a rate of change between the quantities measured along each axis. Inapplications of slope, the ratio is symbolized as . The symbol is theGreek letter delta and has come to represent a change in some quantity, and thenotation is read, “slope is equal to the change in y over the change in x.”Interpreting slope as a rate of change has many significant applications in collegealgebra and beyond.

m �¢y¢x

¢¢y¢x

change in ychange in x

where x2 � x1.

m �y2 � y1

x2 � x1

P2 � 1x2, y22P1 � 1x1, y12

WORTHY OF NOTEWhile the original reason that “m”was chosen for slope is uncertain,some have speculated that it wasbecause in French, the verb for “to climb” is monter. Others say it could be due to the “modulus of slope,” the word modulusmeaning a numeric measure of agiven property, in this case theinclination of a line.

EXAMPLE 3 � Using the Slope FormulaFind the slope of the line through the given points, then use to find anadditional point on the line.

a. (2, 1) and (8, 4) b. ( ) and (4, 2)

Solution � a. For and , b. For and

The slope of this line is . The slope of this line is .�23

12

��4

6�

�2

3 �

3

6�

1

2

�2 � 6

4 � 1�22 �

4 � 1

8 � 2

m �y2 � y1

x2 � x1 m �

y2 � y1

x2 � x1

P2 � 14, 22,P1 � 1�2, 62P2 � 18, 42P1 � 12, 12

�2, 6

m �¢y

¢x


Using , we note that y

increases 1 unit (the y-value ispositive), as x increases 2 units. Since(8, 4) is known to be on the line, thepoint mustalso be on the line.

18 � 2, 4 � 12 � 110, 52

¢y

¢x�

1

2Using we note that y

decreases 2 units (the y-value isnegative), as x increases 3 units. Since(4, 2) is known to be on the line, thepoint mustalso be on the line.

14 � 3, 2 � 22 � 17, 02

¢y

¢x�

�2

3,

cob19537_ch01_019-032.qxd 1/28/11 8:17 PM


Precalculus—

CAUTION � When using the slope formula, try to avoid these common errors.

1. The order that the x- and y-coordinates are subtracted must be consistent, since .

2. The vertical change (involving the y-values) always occurs in the numerator:.

3. When x1 or y1 is negative, use parentheses when substituting into the formula toprevent confusing the negative sign with the subtraction operation.

y2 � y1

x2 � x1�

x2 � x1

y2 � y1

y2 � y1

x2 � x1�

y2 � y1

x1 � x2


Positive and Negative Slope

If you’ve ever traveled by air, you’ve likely heard the announcement, “Ladies and gen-tlemen, please return to your seats and fasten your seat belts as we begin our descent.”For a time, the descent of the airplane follows a linear path, but the slope of the line isnegative since the altitude of the plane is decreasing. Positive and negative slopes, aswell as the rate of change they represent, are important characteristics of linear graphs.In Example 3(a), the slope was a positive number ( ) and the line will slope up-ward from left to right since the y-values are increasing. If as in Example 3(b),the slope of the line is negative and the line slopes downward as you move left to rightsince y-values are decreasing.

m � 0, positive slopey-values increase from left to right

m � 0, negative slopey-values decrease from left to right

m 6 0m 7 0

EXAMPLE 4 � Interpreting the Slope Formula as a Rate of ChangeJimmy works on the assembly line for an auto parts remanufacturing company. By 9:00 A.M. his group has assembled 29 carburetors. By 12:00 noon, they havecompleted 87 carburetors. Assuming the relationship is linear, find the slope of theline and discuss its meaning in this context.

Solution � First write the information as ordered pairs using c to represent the carburetors assem-bled and t to represent time. This gives and Theslope formula then gives:

Here the slope ratio measures , and we see that Jimmy’s group canassemble 58 carburetors every 3 hr, or about carburetors per hour.191

3

carburetors assembledhours

�58

3 or 19.3

¢c

¢t�

c2 � c1

t2 � t1�

87 � 29

12 � 9

1t2, c22 � 112, 872.1t1, c12 � 19, 292WORTHY OF NOTEActually, the assignment of (t1, c1) to(9, 29) and (t2, c2) to (12, 87) wasarbitrary. The slope ratio will be thesame as long as the order ofsubtraction is the same. In otherwords, if we reverse this assignmentand use and

, we have

.m � 29 � 879 � 12 � �58

�3 � 583

1t2, c22 � 19, 2921t1, c12 � 112, 872

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Precalculus—

EXAMPLE 5 � Applying Slope as a Rate of Change in AltitudeAt a horizontal distance of 10 mi after take-off, an airline pilot receives instructions todecrease altitude from their current level of 20,000 ft. A short time later, they are17.5 mi from the airport at an altitude of 10,000 ft. Find the slope ratio for the descentof the plane and discuss its meaning in this context. Recall that

Solution � Let a represent the altitude of the plane and d its horizontal distance from theairport. Converting all measures to feet, we have and

, giving

Since this slope ratio measures , we note the plane is decreasing 25 ft inaltitude for every 99 ft it travels horizontally.


¢altitude¢distance

��10,000

39,600�

�25

99

¢a

¢d�

a2 � a1

d2 � d1�

10,000 � 20,000

92,400 � 52,800

1d2, a22 � 192,400, 10,00021d1, a12 � 152,800, 20,0002

1 mi � 5280 ft.

C. Horizontal Lines and Vertical Lines

Horizontal and vertical lines have a number of important applications, from finding theboundaries of a given graph (the domain and range), to performing certain tests on nonlin-ear graphs. To better understand them, consider that in one dimension, the graph of is a single point (Figure 1.28), indicating alocation on the number line 2 units from zeroin the positive direction. In two dimensions, theequation represents all points with an x-coordinate of 2. A few of these are graphed in Figure 1.29, but since there are an infinitenumber, we end up with a solid vertical line whose equation is (Figure 1.30).x � 2

x � 2

x � 2


we can find the slope of a line

and interpret it as a rate of

change

2�1�2�3�4�5 3 4

x � 2

510

Figure 1.28

(2, 5)

x

y

5�5

�5

5

(2, 3)

(2, 1)

(2, �1)

(2, �3)

x � 2

x

y

5�5

�5

5


The same idea can be applied to horizontal lines. In two dimensions, the equationrepresents all points with a y-coordinate of positive 4, and there are an infinite

number of these as well. The result is a solid horizontal line whose equation is .See Exercises 49 through 54.

y � 4y � 4

WORTHY OF NOTEIf we write the equation in the form , the equationbecomes , since theoriginal equation has no y-variable.Notice that regardless of the valuechosen for y, x will always be 2 andwe end up with the set of orderedpairs (2, y), which gives us a vertical line.

x � 0y � 2ax � by � c

x � 2

Horizontal Lines Vertical Lines

The equation of a horizontal line is The equation of a vertical line is

where (0, k) is the y-intercept. where (h, 0) is the x-intercept.

x � hy � k

cob19537_ch01_019-032.qxd 1/28/11 8:17 PM


Precalculus—

So far, the slope formula has only been applied to lines that were nonhorizontal ornonvertical. So what is the slope of a horizontal line? On an intuitive level, we expectthat a perfectly level highway would have an incline or slope of zero. In general, forany two points on a horizontal line, and , giving a slope of

. For any two points on a vertical line, and ,

making the slope ratio undefined: (see Figures 1.31 and 1.32).

For any horizontal line, For any vertical line,

The Slope of a Horizontal Line The Slope of a Vertical Line

The slope of any horizontal line The slope of any vertical line is zero. is undefined.

EXAMPLE 6 � Calculating SlopesThe federal minimum wage remained constant from 1997 through 2006. However, thebuying power (in 1996 dollars) of these wage earners fell each year due to inflation(see Table 1.3). This decrease in buying power is approximated by the red line shown.

a. Using the data or graph, find the slope of the line segment representing theminimum wage.

b. Select two points on the line representing buying power to approximate theslope of the line segment, and explain what it means in this context.

x

y

�y y2 � y1

�x x2 � x1�(x2, y2)

(x1, y1)

y2 � y1x1 � x1

�

y2 � y1

0�

undefined�

x

y

�y y2 � y1

�x x2 � x1�

(x2, y2)(x1, y1) y1 � y1x2 � x1

�

0x2 � x1

�

� 0

x2 � x1y2 � y1

m �y

2� y

1

0

x2 � x1 � 0x2 � x1m � 0x2 � x1

� 0y2 � y1 � 0y2 � y1

Time in years

Wag

es/B

uyin

g po

wer

1998

1999

2000

2001

2002

2003

2004

2005

2006

1997

4.05

4.15

4.25

4.35

4.45

4.55

4.65

4.75

4.85

4.95

5.05

5.15Time t Minimum Buying(years) wage w power p

1997 5.15 5.03

1998 5.15 4.96

1999 5.15 4.85

2000 5.15 4.69

2001 5.15 4.56

2002 5.15 4.49

2003 5.15 4.39

2004 5.15 4.28

2005 5.15 4.14

2006 5.15 4.04

Table 1.3


cob19537_ch01_019-032.qxd 1/28/11 8:17 PM


Precalculus—

Solution � a. Since the minimum wage did not increase or decrease from 1997 to 2006, theline segment has slope .

b. The points (1997, 5.03) and (2006, 4.04) from the table appear to be on or closeto the line drawn. For buying power p and time t, the slope formula yields:

The buying power of a minimum wage worker decreased by 11 per yearduring this time period.

Now try Exercises 55 and 56 �

D. Parallel and Perpendicular Lines

Two lines in the same plane that never intersect are called parallel lines. When we placethese lines on the coordinate grid, we find that “never intersect” is equivalent to saying“the lines have equal slopes but different y-intercepts.” In Figure 1.33, notice the riseand run of each line is identical, and that by counting both lines have slope .m � 3

4¢y¢x

¢

��0.99

9�

�0.11

1

�4.04 � 5.03

2006 � 1997

¢p

¢t�

p2 � p1

t2 � t1

m � 0

WORTHY OF NOTEIn the context of lines, try to avoidsaying that a horizontal line has “noslope,” since it’s unclear whether aslope of zero or an undefined slopeis intended.

C. You’ve just seen howwe can graph horizontal andvertical lines

L2

L1

L2

L1Generic plane

Rise

Run

Rise

Run

Coordinate plane

x

y

5�5

�5

5Figure 1.33

Parallel Lines

Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively.1. If , then L1 is parallel to L2.2. If L1 is parallel to L2, then .

In symbols, we write L1||L2.Any two vertical lines (undefined slope) are parallel.

EXAMPLE 7A � Determining Whether Two Lines Are ParallelTeladango Park has been mapped out on a rectangular coordinate system, with aranger station at (0, 0). Brendan and Kapi are at coordinates and haveset a direct course for the pond at (11, 10). Caden and Kymani are at ( , 1) andare heading straight to the lookout tower at ( , 21). Are they hiking on parallel ornonparallel courses?

�2�27

1�24, �182

m1 � m2

m1 � m2

cob19537_ch01_019-032.qxd 1/28/11 8:17 PM

Solution � To respond, we compute the slope of each trek across the park.

For Brendan and Kapi: For Caden and Kymani:

Since the slopes are equal, the two groups are hiking on parallel courses.

Two lines in the same plane that intersect at right angles are called perpendicularlines. Using the coordinate grid, we note that intersect at right angles suggests that theirslopes are negative reciprocals. While certainly not a proof, notice in Figure 1.34, the

for L1 is and the for L2 is . Alternatively, we can say their slopes have a product of , since .m1

# m2 � �1 implies m1 � � 1m2

�1

�34ratio rise

run43ratio rise

run

�20

25�

4

5 �

28

35�

4

5

�21 � 1

�2 � 1�272 �

10 � 1�182

11 � 1�242

m �y2 � y1

x2 � x1 m �

y2 � y1

x2 � x1


Precalculus—

L1

L2

Generic plane

RiseRun

Rise

Run

Coordinate plane

x

y

5�5

�5

5L1

L2

Figure 1.34

Perpendicular Lines

Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively.1. If , then L1 is perpendicular to L2.2. If L1 is perpendicular to L2, then .

In symbols we write . Any vertical line (undefined slope) is perpendicular

to any horizontal line (slope ).

We can easily find the slope of a line perpendicular to a second line whose slope isknown or can be found—just find the reciprocal and make it negative. For a line withslope , any line perpendicular to it will have a slope of . For the slope of any line perpendicular would be .

EXAMPLE 7B � Determining Whether Two Lines Are PerpendicularThe three points , and form the vertices of a triangle. Use these points to draw the triangle, then use the slope formula todetermine if they form a right triangle.

P3 � 1�3, 22P1 � 15, 12, P2 � 13, �22

m2 � 15

m1 � �5,m2 � 73m1 � �3

7

m � 0

L1 � L2

m1# m2 � �1

m1# m2 � �1

cob19537_ch01_019-032.qxd 1/28/11 8:17 PM


Precalculus—

Solution � For a right triangle to be formed, two of the linesthrough these points must be perpendicular (forminga right angle). From Figure 1.35, it appears a righttriangle is formed, but we must verify that two of thesides are actually perpendicular. Using the slopeformula, we have:

For P1 and P2 For P1 and P3

For P2 and P3

Since , the triangle has a right angleand must be a right triangle.


E. Applications of Linear Equations

The graph of a linear equation can be used to help solve many applied problems. If thenumbers you’re working with are either very small or very large, scale the axes appro-priately. This can be done by letting each tic mark represent a smaller or larger unit sothe data points given will fit on the grid. Also, many applications use only nonnegativevalues and although points with negative coordinates may be used to graph a line, onlyordered pairs in QI can be meaningfully interpreted.

EXAMPLE 8 � Applying a Linear Equation Model — Commission SalesUse the information given to create a linear equation model in two variables, thengraph the line and answer the question posed:

A salesperson gets a daily $20 meal allowance plus $7.50 for every item she sells.How many sales are needed for a daily income of $125?

Verify your answer by graphing the line on a calculator and using the feature.

Algebraic Solution � Let x represent sales and y represent income. This gives

verbal model: Daily income (y) equals equation model:

Using and , we find (0, 20) and (10, 95)are points on this line and these are used to sketchthe graph. From the graph, it appears that 14 salesare needed to generate a daily income of $125.00.

Since daily income is given as $125, we substitute125 for y and solve for x.

substitute 125 for y

subtract 20

divide by 7.5 14 � x 105 � 7.5x 125 � 7.5x � 20

x � 10x � 0

y � 7.5x � 20$7.5 per sale 1x2 � $20 for meals

TRACE

�4

�6�

2

�3

m1# m3 � �1 m3 �

2 � 1�22

�3 � 3

�1

�8 �

�3

�2�

3

2

m2 �2 � 1

�3 � 5 m1 �

�2 � 1

3 � 5

P3P1

P2

x

y

5�5

�5

5

Figure 1.35

D. You’ve just seen howwe can identify parallel andperpendicular lines

y � 7.5x � 20

Inco

me

Sales

(0, 20)

(10, 95)

16 x1284 6 10 1420

100

50

150

y

cob19537_ch01_019-032.qxd 1/28/11 8:17 PM

Graphical Solution � Begin by entering the equation on the screen, recognizing that in thiscontext, both the input and output values mustbe positive. Reasoning the 10 sales will net$95 (less than $125) and 20 sales will net $170 (more than $125), we set the viewing

as shown in Figure 1.36. We can thenthe equation and use the feature

to estimate the number of sales needed. Theresult shows that income is close to $125when x is close to 14 (Figure 1.37). Inaddition to letting us trace along a graph, the

option enables us to evaluate theequation at specific points. Simply enteringthe number “14” causes the calculator toaccept 14 as the desired input (Figure 1.38),and after pressing , it verifies that (14, 125)is indeed a point on the graph (Figure 1.39).


ENTER

TRACE

TRACEGRAPH

WINDOW

Y=

y � 7.5x � 20


Precalculus—

20

200

0

0

E. You’ve just seen howwe can apply linear equationsin context

4. The slope of a horizontal line is _______, the slopeof a vertical line is _______, and the slopes of twoparallel lines are ______.

5. Discuss/Explain If and , willthe lines intersect? If and , are thelines perpendicular?

6. Discuss/Explain the relationship between the slopeformula, the Pythagorean theorem, and the distanceformula. Include several illustrations.

m2 � � 23m1 � 23

m2 � 2.01m1 � 2.1



1.2 EXERCISES

1. To find the x-intercept of a line, substitute ______for y and solve for x. To find the y-intercept,substitute _________ for x and solve for y.

2. The slope formula is ______ ______,and indicates a rate of change between the x- andy-variables.

3. If the slope of the line is ______ and theline slopes _______ from left to right.

m 6 0,

�m �

Figure 1.36

Figure 1.37


cob19537_ch01_019-032.qxd 1/28/11 8:17 PM


Precalculus—


Create a table of values for each equation and sketch thegraph.

7. 8.

9. 10.

11. If you completed Exercise 9, verify that ( )and ( ) also satisfy the equation given. Do thesepoints appear to be on the graph you sketched?

12. If you completed Exercise 10, verify that( ) and also satisfy the equationgiven. Do these points appear to be on the graphyou sketched?

Graph the following equations using the interceptmethod. Plot a third point as a check.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

Compute the slope of the line through the given points,then graph the line and use to find twoadditional points on the line. Answers may vary.

33. (3, 5), (4, 6) 34. ( ), (5, 8)�2, 3

m �¢y¢x

�2x � 5y � 83y � 4x � 12

y � 3x � 02y � 3x � 0

y �3

4 x � 2y � �

2

5x � 2

y � 30 � 60xy � 25 � 50x

y �2

3 xy � �

1

2 x

�3x � 2y � 62x � 3y � �12

�6x � 4y � 82x � 5y � 4

3y � 4x � 9�5x � 2y � 6

�4y � x � 85y � x � 5

�2x � y � 123x � y � 6

1112 , 37

6 2�1.5, �5.5

12, 19

4

�3, �0.5

y �5

3x � 3y �

3

2x � 4

�3x � 5y � 102x � 3y � 6

35. (10, 3), (4, �5) 36. ( ), (0, 7)

37. (1, ), ( , 7) 38. ( , 5), (0, )

39. ( , 6), (4, 2) 40. ( ), ( )

41. The graph shown models the relationship betweenthe cost of a new home and the size of the home insquare feet. (a) Determine the slope of the line andinterpret what the slope ratio means in this contextand (b) estimate the cost of a 3000 ft2 home.

Exercise 41 Exercise 42

42. The graph shown models the relationship betweenthe volume of garbage that is dumped in a landfilland the number of commercial garbage trucks thatenter the site. (a) Determine the slope of the lineand interpret what the slope ratio means in thiscontext and (b) estimate the number of trucksentering the site daily if 1000 m3 of garbage isdumped per day.

43. The graph shown models the relationshipbetween the distance of an aircraft carrier fromits home port and the number of hours sincedeparture. (a) Determine the slope of the line and interpret what the slope ratio means in thiscontext and (b) estimate the distance from portafter 8.25 hours.

Exercise 43 Exercise 44

44. The graph shown models the relationship between thenumber of circuit boards that have been assembled ata factory and the number of hours since starting time.(a) Determine the slope of the line and interpret whatthe slope ratio means in this context and (b) estimatehow many hours the factory has been running if 225circuit boards have been assembled.

Cir

cuit

boar

ds

Hours1050

250

500

Dis

tanc

e (m

i)

Hours2010

150

300

0

Vol

ume

(m3 )

Trucks10050

480

240

1200

960

720

0

Cos

t ($1

000s

)

ft2 (1000s)

54310 2

250

500

�3, �1�2, �4�3

�5�5�3�8

�3, �1

x yx y

x y x y

cob19537_ch01_019-032.qxd 2/11/11 4:44 PM


Precalculus—

45. Height and weight: While there are manyexceptions, numerous studies have shown a closerelationship between an average height and averageweight. Suppose a person 70 in. tall weighs 165 lb,while a person 64 in. tall weighs 142 lb. Assuming therelationship is linear, (a) find the slope of the line anddiscuss its meaning in this context and (b) determinehow many pounds are added for each inch of height.

46. Rate of climb: Shortly after takeoff, a planeincreases altitude at a constant (linear) rate. In 5 min the altitude is 10,000 ft. Fifteen minutes aftertakeoff, the plane has reached its cruising altitude of32,000 ft. (a) Find the slope of the line and discussits meaning in this context and (b) determine how long it takes the plane to climb from 12,200 ft to 25,400 ft.

47. Sewer line slope: Fascinated at how quickly theplumber was working, Ryan watched with greatinterest as the new sewer line was laid from thehouse to the main line, a distance of 48 ft. At theedge of the house, the sewer line was 6 in. underground. If the plumber tied in to the main line at adepth of 18 in., what is the slope of the (sewer)line? What does this slope indicate?

48. Slope (pitch) of a roof: A contractor goes to alumber yard to purchase some trusses (thetriangular frames) for the roof of a house. Manysizes are available, so the contractor takes somemeasurements to ensure the roof will have thedesired slope. In one case, the height of the truss(base to ridge) was 4 ft, with a width of 24 ft (eaveto eave). Find the slope of the roof if these trussesare used. What does this slope indicate?

Graph each line using two or three ordered pairs thatsatisfy the equation.

49. 50.

51. 52.

Write the equation for each line L1 and L2 shown.Specifically state their point of intersection.

53. 54.

x

y

42�4 �2�1

�2

�3

�4

�5

2

3

4

5

1

L1

L2

x

y

42�4 �2

�2

�4

2

4

L1

L2

y � �2x � 2

y � 4x � �3

55. Supreme Court justices: The table given showsthe total number of justices j sitting on theSupreme Court of the United States for selectedtime periods t (in decades), along with the numberof nonmale, nonwhite justices n for the same years.(a) Use the data to graph the linear relationshipbetween t and j, then determine the slope of theline and discuss its meaning in this context. (b) Usethe data to graph the linear relationship between tand n, then determine the slope of the line anddiscuss its meaning.

Time t Justices Nonwhite,( ) j nonmale n

0 9 0

10 9 1

20 9 2

30 9 3

40 9 4

50 9 5

1960 S 0

Exercise 55

56. Boiling temperature: The table shown gives the boiling temperature t of water as related to the altitude h. Use the data to graph the linearrelationship between h and t, then determine theslope of the line and discuss its meaning in thiscontext.

Altitude h Boiling Temperature t (ft) ( )

0 212.0

1000 210.2

2000 208.4

3000 206.6

4000 204.8

5000 203.0

6000 201.2

�F

Exercise 56

Two points on L1 and two points on L2 are given. Use the slope formula to determine if lines L1 and L2 areparallel, perpendicular, or neither.

57. L1: ( , 0) and (0, 6) 58. L1: (1, 10) and ( , 7)L2: (1, 8) and (0, 5) L2: (0, 3) and (1, 5)

59. L1: ( ) and (0, 1) 60. L1: (6, 2) and (8, )L2: (0, 0) and ( , 4) L2: (5, 1) and (3, 0)

61. L1: (6, 3) and (8, 7) 62. L1: ( ) and (4, 4)L2: (7, 2) and (6, 0) L2: (4, ) and (8, 10)�7

�5, �1

�4�2�3, �4

�1�2

cob19537_ch01_019-032.qxd 1/28/11 8:18 PM


Precalculus—

In Exercises 63 to 68, three points that form the vertices of a triangle are given. Use the points to draw the triangle,then use the slope formula to determine if any of the triangles are right triangles. Also see Exercises 43–48 inSection 1.1.

66. (5, 2), (0, ), (4, )

67. ( , 2), ( , 5), ( , 4)

68. (0, 0), ( , 2), (2, )�5�5

�6�1�3

�4�3


69. Human life expectancy:

In the United States, the average life expectancyhas been steadily increasing over the years due tobetter living conditions and improved medical care.This relationship is modeled by the formula shown,where L is the average life expectancy and T isnumber of years since 1980. (a) What was the lifeexpectancy in the year 2010? (b) In what year willaverage life expectancy reach 79 yr?

L � 0.15T � 73.7 70. Interest earnings:

If $5000 dollars is invested in an account paying7% simple interest, the amount of interest earned isgiven by the formula shown, where I is the interestand T is the time in years. Begin by solving theformula for I. (a) How much interest is earned in 5 yr? (b) How much is earned in 10 yr? (c) Use thetwo points (5 yr, interest) and (10 yr, interest) tocalculate the slope of this line. What do younotice?

100I � 35,000T

� APPLICATIONS

Use the information given to build a linear equationmodel, then use the equation to respond. For exercises71 to 74, develop both an algebraic and a graphicalsolution.

71. Business depreciation: A business purchases acopier for $8500 and anticipates it will depreciatein value $1250 per year.

a. What is the copier’s value after 4 yr of use?

b. How many years will it take for this copier’svalue to decrease to $2250?

72. Baseball card value: After purchasing anautographed baseball card for $85, its valueincreases by $1.50 per year.

a. What is the card’s value 7 yr after purchase?

b. How many years will it take for this card’svalue to reach $100?

73. Water level: During a long drought, the water levelin a local lake decreased at a rate of 3 in. per month.The water level before the drought was 300 in.

a. What was the water level after 9 months ofdrought?

b. How many months will it take for the waterlevel to decrease to 20 ft?

74. Gas mileage: When empty, a large dump-truckgets about 15 mi per gallon. It is estimated that foreach 3 tons of cargo it hauls, gas mileage decreasesby mi per gallon.

a. If 10 tons of cargo is being carried, what is thetruck’s mileage?

b. If the truck’s mileage is down to 10 mi pergallon, how much weight is it carrying?

75. Parallel/nonparallel roads: Aberville is 38 minorth and 12 mi west of Boschertown, with astraight “farm and machinery” road (FM 1960)connecting the two cities. In the next county,Crownsburg is 30 mi north and 9.5 mi west ofDower, and these cities are likewise connected by astraight road (FM 830). If the two roads continuedindefinitely in both directions, would they intersectat some point?

76. Perpendicular/nonperpendicular courseheadings: Two shrimp trawlers depart CharlestonHarbor at the same time. One heads for theshrimping grounds located 12 mi north and 3 mieast of the harbor. The other heads for a point 2 misouth and 8 mi east of the harbor. Assuming theharbor is at (0, 0), are the routes of the trawlersperpendicular? If so, how far apart are the boatswhen they reach their destinations (to the nearestone-tenth mi)?

34

63. ( , 7), (2, 2), (5, 5)

64. (7, 0), ( , 0), (7, 4)

65. ( , 3), ( ), (3, )�2�7, �1�4

�1

�3

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Precalculus—

77. Cost of college: For the years 2000 to 2008, the cost of tuition and fees per semester (inconstant dollars) at a public 4-yr college can beapproximated by the equation ,where y represents the cost in dollars and represents the year 2000. Use the equation to find:(a) the cost of tuition and fees in 2010 and (b) theyear this cost will exceed $9000.Source: The College Board

78. Female physicians: In 1960 only about 7% ofphysicians were female. Soon after, this percentagebegan to grow dramatically. For the years 1990 to2000, the percentage of physicians that werefemale can be approximated by the equation

, where y represents the percentage(as a whole number) and represents the year1990. Use the equation to find: (a) the percentageof physicians that were female in 2000 and (b) theprojected year this percentage would haveexceeded 30%.Source: American Journal of Public Health

x � 0y � 0.6x � 18.1

x � 0y � 386x � 3500

79. Decrease in smokers: For the years 1990 to 2000, the percentage of the U.S. adult populationwho were smokers can be approximated by theequation , where y represents thepercentage of smokers (as a whole number) and

represents 1990. Use the equation to find: (a) the percentage of adults who smoked in the year2005 and (b) the year the percentage of smokers isprojected to fall below 15%.Source: WebMD

80. Temperature and cricket chirps: Biologists havefound a strong relationship between temperatureand the number of times a cricket chirps. This ismodeled by the equation , where N isthe number of times the cricket chirps per minuteand T is the temperature in Fahrenheit. Use theequation to find: (a) the outdoor temperature if thecricket is chirping 48 times per minute and (b) thenumber of times a cricket chirps if the temperatureis .70°

T � 14N � 40

x � 0

y � �1325 x � 28.7


81. If the lines and areperpendicular, what is the value of a?

82. Let m1, m2, m3, and m4 be the slopes of lines L1,L2, L3, and L4, respectively. Which of the followingstatements is true?

a.b.c.d.e. m1 6 m4 6 m3 6 m2

m1 6 m3 6 m4 6 m2

m3 6 m4 6 m2 6 m1

m3 6 m2 6 m4 6 m1

m4 6 m1 6 m3 6 m2

3y � ax � �24y � 2x � �5 83. An arithmetic sequence is a sequence of numberswhere each successive term is found by adding afixed constant, called the common difference d, tothe preceding term. For instance 3, 7, 11, 15, . . . isan arithmetic sequence with . The formulafor the “nth term” tn of an arithmetic sequence is alinear equation of the form ,where d is the common difference and t1 is the firstterm of the sequence. Use the equation to find theterm specified for each sequence.

a. 2, 9, 16, 23, 30, . . . ; 21st term

b. 7, 4, 1, , , . . . ; 31st term

c. 5.10, 5.25, 5.40, 5.55, . . . ; 27th term

d. , . . . ; 17th term32, 94, 3, 15

4 , 92

�5�2

tn � t1 � 1n � 12d

d � 4

L1L2

L3

L4 x

y


84. (1.1) Name the center and radius of the circledefined by

85. (Appendix A.6) Compute the sum and productindicated:

a.b.

86. (Appendix A.4) Solve the equation by factoring,then check the result(s) using substitution:

12x2 � 44x � 45 � 0

13 � 152 13 � 252

120 � 3145 � 15

1x � 322 � 1y � 422 � 16987. (Appendix A.5) Factor the following polynomials

completely:

a.b.c. x3 � 125

x2 � 23x � 24

x3 � 3x2 � 4x � 12

cob19537_ch01_019-032.qxd 1/28/11 8:18 PM

Precalculus—

In this section we introduce one of the most central ideas in mathematics—the conceptof a function. Functions can model the cause-and-effect relationship that is so impor-tant to using mathematics as a decision-making tool. In addition, the study will help tounify and expand on many ideas that are already familiar.

A. Functions and Relations

There is a special type of relation that merits further attention. A function is a relationwhere each element of the domain corresponds to exactly one element of the range. Inother words, for each first coordinate or input value, there is only one possible secondcoordinate or output.

Functions

A function is a relation that pairs each element from the domainwith exactly one element from the range.

If the relation is defined by a mapping, we need only check that each element ofthe domain is mapped to exactly one element of the range. This is indeed the case forthe mapping from Figure 1.1 (page 2), where we saw that each person corre-sponded to only one birthday, and that it was impossible for one person to be born ontwo different days. For the relation shown in Figure 1.6 (page 4), each elementof the domain except zero is paired with more than one element of the range. The rela-tion is not a function.x � �y�

x � �y�

P S B

LEARNING OBJECTIVES In Section 1.3 you will see

how we can:

A. Distinguish the graph of afunction from that of arelation

B. Determine the domainand range of a function

C. Use function notation andevaluate functions

D. Read and interpretinformation givengraphically

EXAMPLE 1 � Determining Whether a Relation is a FunctionThree different relations are given in mapping notation below. Determine whethereach relation is a function.

a. b. c.

Solution � Relation (a) is a function, since each person corresponds to exactly one room. Thisrelation pairs math professors with their respective office numbers. Notice thatwhile two people can be in one office, it is impossible for one person to physicallybe in two different offices.

Relation (b) is not a function, since we cannot tell whether Polly the Parrot weighs2 lb or 3 lb (one element of the domain is mapped to two elements of the range).

Relation (c) is a function, where each major war is paired with the year it began.


MariePesky

BoJohnnyRick

AnnieReece

270

268

274

276

272

282

Person Room

Fido

Bossy

Silver

Frisky

Polly

45055024083

Pet Weight (lb) War Year

Civil War

World War I

World War II

Korean War

Vietnam War

1963

1950

1939

1917

1861

1.3 Functions, Function Notation, and the Graph of a Function

1–33 33

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Precalculus—

If the relation is pointwise-defined or given as a set of individual and distinct plot-ted points, we need only check that no two points have the same first coordinate with adifferent second coordinate. This gives rise to an alternative definition for a function.

Functions (Alternate Definition

A function is a set of ordered pairs (x, y), in which each first component is paired with only one second component.

EXAMPLE 2 � Identifying FunctionsTwo relations named f and g are given; f is pointwise-defined (stated as a set ofordered pairs), while g is given as a set of plotted points. Determine whether eachis a function.

( ), and (6, 1)

Solution � The relation f is not a function, since is pairedwith two different outputs: and .

The relation g shown in the figure is a function.Each input corresponds to exactly one output,otherwise one point would be directly above theother and have the same first coordinate.

1�3, �221�3, 02�3

4, �510, �12,f: 1�3, 02, 11, 42, 12, �52, 14, 22, 1�3, �22, 13, 62,


(�4, 2)

(�2, 1)

(�1, �3)

(0, 5)

(4, �1)

(3, 1)

x

y

5�5

�5

5g

The graphs of and from Section 1.1 offer additional insight intothe definition of a function. Figure 1.40 shows the line with emphasis on theplotted points (4, 3) and The vertical movement shown from the x-axis to apoint on the graph illustrates the pairing of a given x-value with one related y-value.Note the vertical line shows only one related y-value ( is paired with only ).Figure 1.41 gives the graph of highlighting the points (4, 4) and (4, ). Thevertical movement shown here branches in two directions, associating one x-value withmore than one y-value. This shows the relation is also a function, while therelation is not.x � �y�

y � x � 1

�4x � �y�,y � 3x � 4

1�3, �42.y � x � 1

x � �y�y � x � 1

This “vertical connection” of a location on the x-axis to a point on the graph canbe generalized into a vertical line test for functions.

(4, 3)

y � x � 1

(�3, �4)

x

y

5�5

�5

5

(2, 2)

(4, 4)

(0, 0)

(2, �2)

(4, �4)

x � �y�

x

y

5�5

�5

5



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Precalculus—

Vertical Line Test

A given graph is the graph of a function, if and only if every vertical line intersects the graph in at most one point.

Applying the test to the graph in Figure 1.40 helps to illustrate that the graph ofany nonvertical line must be the graph of a function, as is the graph of any pointwise-defined relation where no x-coordinate is repeated. Compare the relations f and g fromExample 2.

(2, 0)

(3, 3)(�1, 3)

(0, 0)

(1, �1)

y � x2 � 2x

x

y

5�5

�2

5

(4, 8)(�2, 8)

(3, 0)

y � �9 � x2

x

y

5�5

�5

5

(�3, 0)

(0, 3)

(4, �2)(2, ��2)

(2, �2)

(4, 2)

x � y2

x

y

5�5

�5

5

(0, 0)

Figure 1.42 Figure 1.43 Figure 1.44

EXAMPLE 3 � Using the Vertical Line TestUse the vertical line test to determine if any of the relations shown (from Section 1.1)are functions.

Solution � Visualize a vertical line on each coordinate grid (shown in solid blue), then mentallyshift the line to the left and right as shown in Figures 1.42, 1.43, and 1.44 (dashedlines). In Figures 1.42 and 1.43, every vertical line intersects the graph only once, indicating both and are functions. In Figure 1.44, avertical line intersects the graph twice for any [for instance, both (4, 2) and

are on the graph]. The relation is not a function.x � y214, �22x 7 0

y � 29 � x2y � x2 � 2x

EXAMPLE 4 � Using the Vertical Line TestUse a table of values to graph the relations defined by

a. b. ,then use the vertical line test to determine whether each relation is a function.

Solution � a. For , using input values from to produces the followingtable and graph (Figure 1.45). Note the result is a V-shaped graph that “opensupward.” The point (0, 0) of this absolute value graph is called the vertex.Since any vertical line will intersect the graph in at most one point, this is thegraph of a function.

x � 4x � �4y � �x�

y � 1xy � �x�


WORTHY OF NOTEFor relations and functions, a goodway to view the distinction is toconsider a mail carrier. It is possiblefor the carrier to put more than oneletter into the same mailbox (morethan one x going to the same y), butquite impossible for the carrier toplace the same letter in two differentboxes (one x going to two y’s).

1–35 Section 1.3 Functions, Function Notation, and the Graph of a Function 35

cob19537_ch01_033-049.qxd 1/31/11 5:53 PM

Precalculus—

b. For , values less than zero do not produce a real number, so our graphactually begins at (0, 0) (see Figure 1.46). Completing the table for nonnegativevalues produces the graph shown, which appears to rise to the right and remainsin the first quadrant. Since any vertical line will intersect this graph in at mostone place, is also a function.y � 1x

y � 1x

y � �x�

x

y

5�5

�5

5

Figure 1.45

A. You’ve just seen howwe can distinguish the graphof a function from that of arelation

y � 1x

x

0 0

1 1

2

3

4 2

13 � 1.7

12 � 1.4

y � 1x

x

y

5�5

�5

5

Figure 1.46


B. The Domain and Range of a Function

Vertical Boundary Lines and the Domain

In addition to its use as a graphical test for functions, a vertical line can help determinethe domain of a function from its graph. For the graph of (Figure 1.46), a ver-tical line will not intersect the graph until , and then will intersect the graph forall values (showing the function is defined for these values). These verticalboundary lines indicate the domain is .

Instead of using a simple inequality to write the domain and range, we will oftenuse (1) a form of set notation, (2) a number line graph, or (3) interval notation. Inter-val notation is a symbolic way of indicating a selected interval of the real numbers.When a number acts as the boundary point for an interval (also called an endpoint),we use a left bracket “[” or a right bracket “]” to indicate inclusion of the endpoint. Ifthe boundary point is not included, we use a left parenthesis “(” or right parenthesis “).”

x � 0x � 0

x � 0y � 1x

WORTHY OF NOTEOn a number line, some texts willuse an open dot “º” to mark thelocation of an endpoint that is notincluded, and a closed dot “•” foran included endpoint.


x

4

3

2

1

0 0

1 1

2 2

3 3

4 4

�1

�2

�3

�4

y � �x�

cob19537_ch01_033-049.qxd 1/31/11 5:53 PM

Precalculus—

EXAMPLE 5 � Using Notation to State the Domain and RangeModel the given phrase using the correct inequality symbol. Then state the result inset notation, graphically, and in interval notation: “The set of real numbers greaterthan or equal to 1.”

Solution � Let n represent the number:

• Set notation: • Graph:

• Interval notation:


The “ ” symbol says the number n is an element of the set or interval given. The“ ” symbol represents positive infinity and indicates the interval continues forever tothe right. Note that the endpoints of an interval must occur in the same order as on thenumber line (smaller value on the left; larger value on the right).

A short summary of other possibilities is given here for any real number x. Manyvariations are possible.

q�

n � 31, q 2

5n |n � 16

n � 1.

WORTHY OF NOTESince infinity is really a concept andnot a number, it is never included(using a bracket) as an endpoint foran interval.

[1�1�2 2 3 540

)k

[k

))a b

[ )a b

))a b

Conditions ( ) Set Notation Number Line Interval Notation

x is greater than k

x is less than or equal to k

x is less than band greater than a

x is less than b and greater than or equal to a

x is less than a orx is greater than b x 7 b6

x � 1�q, a2 ´ 1b, q 25x |x 6 a or

x � 3a, b25x |a � x 6 b6

x � 1a, b25x |a 6 x 6 b6

x � 1�q, k 45x |x � k6

x � 1k, q 25x |x 7 k6

a � b

For the graph of (Figure 1.45), a vertical line will intersect the graph (or itsinfinite extension) for all values of x, and the domain is . Using verticallines in this way also affirms the domain of (Section 1.1, Figure 1.5) is

while the domain of the relation (Section 1.1, Figure 1.6) is .

Range and Horizontal Boundary Lines

The range of a relation can be found using a horizontal “boundary line,” since it willassociate a value on the y-axis with a point on the graph (if it exists). Simply visualize ahorizontal line and move the line up or down until you determine the graph will alwaysintersect the line, or will no longer intersect the line. This will give you the boundaries ofthe range. Mentally applying this idea to the graph of (Figure 1.46) shows therange is Although shaped very differently, a horizontal boundary line showsthe range of (Figure 1.45) is also

EXAMPLE 6 � Determining the Domain and Range of a FunctionUse a table of values to graph the functions defined by

a. b.

Then use boundary lines to determine the domain and range of each.

y � 13 xy � x2

y � 30, q 2.y � �x�y � 30, q 2.

y � 1x

x � 30, q 2x � �y�x � 1�q, q 2

y � x � 1x � 1�q, q 2

y � �x�


cob19537_ch01_033-049.qxd 1/31/11 5:53 PM

Solution � a. For , it seems convenient to use inputs from to ,producing the following table and graph. Note the result is a basic parabolathat “opens upward” (both ends point in the positive y direction), with a vertexat (0, 0). Figure 1.47 shows a vertical line will intersect the graph or itsextension anywhere it is placed. The domain is . Figure 1.48shows a horizontal line will intersect the graph only for values of y that aregreater than or equal to 0. The range is .y � 30, q 2

x � 1�q, q 2

x � 3x � �3y � x2

Precalculus—

x

y

5�5

�5

5y � x2

x

y

5�5

�5

5y � x2


b. For we select points that are perfect cubes where possible, then a fewothers to round out the graph. The resulting table and graph are shown. Noticethere is a “pivot point” at (0, 0) called a point of inflection, and the ends of the graph point in opposite directions. Figure 1.49 shows a vertical line willintersect the graph or its extension anywhere it is placed. Figure 1.50 shows ahorizontal line will likewise always intersect the graph. The domain is

, and the range is .y � 1�q, q 2x � 1�q, q 2

y � 13 x,

x

y

10�10

�5

5y � �x

3

x

y

�5

5y � �x

3

�10 10



Implied Domains

When stated in equation form, the domain of a function is implicitly given by theexpression used to define it, since the expression will dictate what input values areallowed. The implied domain is the set of all real numbers for which the function rep-resents a real number. If the function involves a rational expression, the domain willexclude any input that causes a denominator of zero, since division by zero is undefined.If the function involves a square root expression, the domain will exclude inputs thatcreate a negative radicand, since represents a real number only when .A � 01A


Squaring Function

x

9

4

1

0 0

1 1

2 4

3 9

�1

�2

�3

y � x2

Cube Root Function

x

0 0

1 1

4

8 2

� 1.6

�1�1

� �1.6�4

�2�8

y � 13 x

cob19537_ch01_033-049.qxd 1/31/11 5:53 PM

Precalculus—

EXAMPLE 7 � Determining Implied DomainsState the domain of each function using interval notation.

a. b.

c. d.

Solution � a. By inspection, we note an x-value of results in a zero denominator andmust be excluded. The domain is

b. Since the radicand must be nonnegative, we solve the inequality giving The domain is

c. To prevent division by zero, inputs of and 3 must be excluded (set and solve by factoring). The domain is

. Note that is in the domain

since is defined. See Figure 1.51, where .

d. Since squaring a number and multiplying a number by a constant are definedfor all real numbers, the domain is


EXAMPLE 8 � Determining Implied DomainsDetermine the domain of each function:

a. b.

Solution � a. For , we must have (for the radicand) and

(for the denominator). Since the numerator is always positive, we need, which gives . The domain is .

b. For , we must have and . This shows

we need , so . The domain is .


C. Function Notation

In our study of functions, you’ve likely noticed that the relationship between input andoutput values is an important one. To highlight this fact, think of a function as a simplemachine, which can process inputs using a stated sequence of operations, then delivera single output. The inputs are x-values, a program we’ll name f performs the opera-tions on x, and y is the resulting output (see Figure 1.52). Once again we see that “thevalue of y depends on the value of x,” or simply “y is a function of x.” Notationally, wewrite “y is a function of x” as using function notation. You are already famil-iar with letting a variable represent a number. Here we do something quite different, asthe letter f is used to represent a sequence of operations to be performed on x. Consider

the function which we’ll now write as since . y � f 1x243f 1x2 �x

2� 1y �

x

2� 1,

y � f 1x2

x � 1�54 , q 2x 7 �5

44x � 5 7 0

14x � 5 � 04x � 5 � 0y �2x

14x � 5

x � 1�3, q 2x 7 �3x � 3 7 0

x � 3 � 07

x � 3� 0y �

A

7

x � 3

y �2x

14x � 5y �A

7

x � 3

x � 1�q, q 2.

Y1 �X � 1

X2 � 90

�8 � 0

x � 1x � 1�q, �32 ´ 1�3, 32 ´ 13, q 2x2 � 9 � 0

�3x � 3�3

2 , q 2.x � �32 .

2x � 3 � 0,x � 1�q, �22 ´ 1�2, q 2.

�2

y � x2 � 5x � 7y �x � 1

x2 � 9

y � 12x � 3y �3

x � 2

Figure 1.51

Y1 � 1X � 12/ 1X2 � 92


we can determine the domain

and range of a function

Figure 1.52

Output

Sequence ofoperations

on x as defined by f

y � f (x)

f

Inputx


cob19537_ch01_033-049.qxd 1/31/11 5:53 PM

In words the function says, “divide inputs by 2, then add 1.” To evaluate the function at(Figure 1.53) we have:

input 4 input 4

Function notation enables us to summarize the three most important aspects of afunction using a single expression, as shown in Figure 1.54.

� 3 � 2 � 1

f 142 �4

2� 1

f 1x2 �x

2� 1

x � 4

Precalculus—

Output

Input4

Divideinputs by 2then add 1:

+ 124

3

f

S

Figure 1.53

f(x)

Sequence of operationsto perform on the input

Inputvalue

Output valuey = f(x)

Figure 1.54

Instead of saying, “. . . when the value of the function is 3,” we simply say“f of 4 is 3,” or write Note that the ordered pair (4, 3) is equivalent to (4, f(4)).

CAUTION � Although f(x) is the favored notation for a “function of x,” other letters can also be used.For example, g(x) and h(x) also denote functions of x, where g and h represent differentsequences of operations on the x-inputs. It is also important to remember that theserepresent function values and not the product of two variables: .

EXAMPLE 9 � Evaluating a FunctionGiven find

a. b. c. d.

Solution � a.

� �8 � 1�82 � �16 f 1�22 � �21�222 � 41�22

f 1x2 � �2x2 � 4x

f 1a � 12f 12a2f a7

2bf 1�22

f 1x2 � �2x2 � 4x,

f1x2 � f # 1x2

f 142 � 3.x � 4,

b.

��49

2� 14 �

�21

2 or �10.5

f a7

2b � �2 a

7

2b

2

� 4 a7

2b

f 1x2 � �2x2 � 4x

c.


� �8a2 � 8a � �214a22 � 8a

f 12a2 � �212a22 � 412a2 f 1x2 � �2x2 � 4x d.

� �2a2 � 2 � �2a2 � 4a � 2 � 4a � 4 � �21a2 � 2a � 12 � 4a � 4

f 1a � 12 � �21a � 122 � 41a � 12 f 1x2 � �2x2 � 4x


S

cob19537_ch01_033-049.qxd 1/31/11 5:54 PM

Precalculus—

D. Reading and Interpreting Information Given Graphically

Graphs are an important part of studying functions, and learning to read and interpretthem correctly is a high priority. A graph highlights and emphasizes the all-importantinput/output relationship that defines a function. In this study, we hope to firmly estab-lish that the following statements are synonymous:

1.2.3. is on the graph of f, and4. when

EXAMPLE 10A � Reading a GraphFor the functions f and g whose graphs are shown in Figures 1.58 and 1.59

a. State the domain of the function.b. Evaluate the function at .c. Determine the value(s) of x for which .d. State the range of the function.

y � 3x � 2

x � �2, f 1x2 � 51�2, 521�2, f 1�22 2 � 1�2, 52f 1�22 � 5

f

x

y

54321�5 �4 �3 �2 �1�1

�2

�3

2

3

4

5

1

Figure 1.58

g

x

y

54321�5 �4 �3 �2 �1�1

�2

�3

2

3

4

5

1

Figure 1.59


A graphing calculator can evaluate the function using the TABLEfeature, the feature, or function notation (on the home screen). The first two havebeen illustrated previously. To use function notation, we access the function namesusing the key and right arrow to select Y-VARS (Figure 1.55). The 1:Functionoption is the default, so pressing will enable us to make our choice (Figure 1.56).In this case, we selected 1:Y1, which the calculator then places on the home screen,enabling us to enclose the desired input value in parentheses (function notation).Pressing ENTER completes the evaluation (Figure 1.57), which verifies the resultfrom Example 9(b).

ENTER

VARS

TRACE

Y1 � �2X2 � 4XFigure 1.55

Solution � For f,a. The graph is a continuous line segment with endpoints at ( ) and (5, 3),

so we state the domain in interval notation. Using a vertical boundary line wenote the smallest input is and the largest is 5. The domain is x � 3�4, 5 4 .�4

�4, �3


C. You’ve just seen howwe can use function notationand evaluate functions

cob19537_ch01_033-049.qxd 1/31/11 5:54 PM

Precalculus—

b. The graph shows an input of corresponds to f (2) since (2, 1)is a point on the graph.

c. For f (x) � 3 (or ) the input value must be since (5, 3) is the pointon the graph.

d. Using a horizontal boundary line, the smallest output value is and thelargest is 3. The range is

For g,a. Since g is given as a set of plotted points, we state the domain as the set of first

coordinates: .b. An input of corresponds to g(2) since (2, 2) is on the graph.c. For g(x) (or ) the input value must be since (4, 3) is a point

on the graph.d. The range is the set of all second coordinates:

EXAMPLE 10B � Reading a GraphUse the graph of given to answer the following questions:

a. What is the value of ?b. What value(s) of x satisfy ?

Solution � a. The notation says to find the value of the function f when . Expressedgraphically, we go to and locate thecorresponding point on the graph (bluearrows). Here we find that .

b. For , we’re looking for x-inputs thatresult in an output of .From the graph, we note there are two pointswith a y-coordinate of 1, namely, ( , 1) and (0, 1). This shows

, and the required x-values are and .


In many applications involving functions, the domain and range can be determinedby the context or situation given.

EXAMPLE 11 � Determining the Domain and Range from the ContextPaul’s 2009 Voyager has a 20-gal tank and gets 18 mpg. The number of miles he can drive (hisrange) depends on how much gas is in the tank. As a function we have where M(g)represents the total distance in miles and grepresents the gallons of gas in the tank (seegraph). Find the domain and range.

Solution � Begin evaluating at since the tank cannothold less than zero gallons. With an empty tank, the(minimum) range is or 0 miles. On afull tank, the maximum range is or 360 miles. As shown in thegraph, the domain is g [0, 20] and the corresponding range is M(g) [0, 360].


��M1202 � 181202

M102 � 18102

x � 0,

M1g2 � 18g,

x � 0x � �3f 1�32 � 1, f 102 � 1�3

3since y � f 1x2 4y � 1f 1x2 � 1

f 1�22 � 4

x � �2x � �2

f 1�22

f 1x2 � 1f 1�22

f 1x2

R: 5�1, 0, 1, 2, 36.

x � 4,y � 3� 3� 2y � 2:x � 2

D: 5�4, �2, 0, 2, 46

y � 3�3, 3 4 .�3

x � 5y � 3

� 1y � 1:x � 2

(�2, 4)

(�3, 1) (0, 1)

y

5�5

�5

5

x

f

D. You’ve just seen howwe can read and interpretinformation given graphically

20100

240

360

120

480

(0, 0)

(20, 360)

M

g

600


cob19537_ch01_033-049.qxd 1/31/11 5:54 PM

Precalculus—

1.3 EXERCISES

2. A relation is a function if each element of theis paired with

element of the range.



1. If a relation is given in ordered pair form, we statethe domain by listing all of the coordinates in a set.


Determine whether the mappings shown representfunctions or nonfunctions. If a nonfunction, explain howthe definition of a function is violated.

7.

8.

9.

Air Jordan

The Mailman

The Doctor

The Iceman

The Shaq

Basketball star Reported height

7'1"

6'6"

6'7"

6'9"

7'2"

Hawaii

Roots

Shogun

20,000 LeaguesUnder the Sea

Where the RedFern Grows

Book Author

Rawls

Verne

Haley

Clavell

Michener

Indira Gandhi

Clara Barton

Margaret Thatcher

Maria Montessori

Susan B. Anthony

Woman Country

Britain

U.S.

Italy

India

10.

Determine whether the relations indicated representfunctions or nonfunctions. If the relation is a nonfunction,explain how the definition of a function is violated.

11. ( , 0), (1, 4), (2, ), (4, 2), ( , 6), (3, 6), (0, ), (4, ), and (6, 1)

12. ( ), ( , 3), (4, 0), ( ), (1, ), (0, 9), (2, ), (3, ), and ( , 7)

13. (9, ), ( , 6), (6, ), (4, ), (2, ), (1, 8), (0, ), ( ), and ( , 4)

14. (1, ), ( , 64), ( , 49), (5, ), ( , 25),(13, ), ( , 9), (34, ), and ( , 1)

15. 16.

x

y

5�5

�5

5(�3, 5)

(5, �3)(0, �2)

(3, 4)(1, 3)

(�4, �2)

x

y

5�5

�5

5

(�3, 4)

(�5, 0)(�1, 1)

(1, �4)

(2, 4)

(4, 2)

�55�4�21�16�8�36�3�2�81

�6�2, �7�2�2�1�10�7�10

�5�2�8�6�3, �5�5�7, �5

�5�1�5�5�3

Country Language

Canada

Japan

Brazil

Tahiti

Ecuador

Japanese

Spanish

French

Portuguese

English

3. The set of output values for a function is called theof the function.

4. Write using function notation: The function fevaluated at 3 is negative 5:

5. Discuss/Explain why the relation is afunction, while the relation is not. Justifyyour response using graphs, ordered pairs, and so on.

x � y2y � x2 6. Discuss/Explain the process of finding the domain

and range of a function given its graph, usingvertical and horizontal boundary lines. Include afew illustrative examples.


cob19537_ch01_033-049.qxd 1/31/11 5:54 PM

17. 18.

Determine whether or not the relations given representa function. If not, explain how the definition of afunction is violated.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

x

y

5�5

�5

5

x

y

�5

�5

5

5

x

y

5�5

�5

5

x5�5

�5

y5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5

5

�5

�5

x

y

5�5

�5

5

x

y

5�5

�5

5(�3, 4)

(3, �2)

(3, 3)

(1, 1)

(�5, �2)

(�1, �4)

x

y

5�5

�5

5

(�2, 3)

(�5, 1)

(4, �5)

(3, 4)

(1, 2)

(�2, �4)

Precalculus—

29. 30.

Graph each relation using a table, then use the verticalline test to determine if the relation is a function.

31. 32.

33. 34.

Use an inequality to write a mathematical model for eachstatement, then write the relation in interval notation.

35. To qualify for a secretarial position, a person musttype at least 45 words per minute.

36. The balance in a checking account must remainabove $1000 or a fee is charged.

37. To bake properly, a turkey must be kept betweenthe temperatures of and

38. To fly effectively, the airliner must cruise at orbetween altitudes of 30,000 and 35,000 ft.

Graph each inequality on a number line, then write therelation in interval notation.

39. 40.

41. 42.

43. 44.

45. 46.

Write the domain illustrated on each graph in setnotation and interval notation.

47.

48.

49.

50.

Determine whether or not the relations indicatedrepresent functions, then determine the domain andrange of each.

51. 52.

x

y

5�5

�5

5

x

y

5�5

�5

5

[ )31�1�2�3 2 40

[[1�1�2�3 20 3

)1�1�2�3 0 2 3

[1�1�2�3 2 30

�3 6 p � 45 7 x 7 2

x � �3x � 1

n � �4m � 5

x 7 �2p 6 3

450°.250°

x � �y � 2�y � 1x � 222y � 13 xy � x

x

y

5�5

�5

5

x

y

5�5

�5

5


cob19537_ch01_033-049.qxd 1/31/11 5:54 PM

Precalculus—

53. 54.

55. 56.

57. 58.

59. 60.

61. 62.

Determine the domain of the following functions, andwrite your response in interval notation.

63. 64.

65. 66.

67. 68.

69. 70.

71. 72. y �11

19x � 89y �

17

25x � 123

p �q � 7

q2 � 12u �

v � 5

v2 � 18

w1x2 �x � 4

x2 � 49v1x2 �

x � 2

x2 � 25

p1a2 � 15a � 2h1a2 � 13a � 5

g1x2 ��2

3 � xf 1x2 �

3

x � 5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

�5

�5

5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

5

�5

x

y

5�5

�5

5

x

y

5�5

�5

573. 74.

75. 76.

77. 78.

79. 80.

81. 82.

83. 84.

85. 86.

For Exercises 87 through 102, determine the value ofand , then simplify. Verify

results using a graphing calculator where possible.

87. 88.

89. 90.

Determine the value of h(3), , h(3a), and h(a � 2),then simplify.

91. 92.

93. 94.

Determine the value of g(4), g(2c), and g(c � 3),then simplify.

95. 96.

97. 98.

Determine the value of p(5), p(3a), and p(a � 1),then simplify.

99. 100.

101. 102.

Use the graph of each function given to (a) state thedomain, (b) state the range, (c) evaluate f (2), and (d) find the value(s) x for which (k a constant).Assume all results are integer-valued.

103. 104.

x

y

5�5

�5

5

x

y

5�5

�5

5

k � 3k � 4

f 1x2 � k

p1x2 �2x2 � 3

x2p1x2 �3x2 � 5

x2

p1x2 � 14x � 1p1x2 � 12x � 3

p132 2,

g1r2 � �r2hg1r2 � �r2

g1r2 � 2�rhg1r2 � 2�r

g 132 2,

h1x2 �4�x�x

h1x2 �5�x�x

h1x2 �2

x2h1x2 �3x

h1�23 2

f 1x2 � 2x2 � 3xf 1x2 � 3x2 � 4x

f 1x2 �2

3x � 5f 1x2 �

1

2x � 3

f 1c � 12f 1�62, f 132 2, f 12c2,

s1x2 �x2 � 4

111 � 2xr 1x2 �

2x � 1

13x � 7

p1x2 ��7

15 � xg1x2 �

A

�4

3 � x

f 1x2 �A

5

x � 2h1x2 �

�2

1x � 4

y �1x � 1

3x � 2y �

1x � 2

2x � 5

y2 �x � 4

x2 � 2x � 15y1 �

x

x2 � 3x � 10

y � �x � 2� � 3y � 2�x� � 1

s � t2 � 3t � 10m � n2 � 3n � 10


cob19537_ch01_033-049.qxd 1/31/11 5:55 PM

105. 106.

5�5

�5

x

y5

5�5

�5

5y

x

k � �3k � 1

Precalculus—

107. 108.

5�5

�5

5

x

y

x

y

5�5

�5

5

k � �1k � 2


109. Ideal weight for males: The ideal weight for an adult male can be modeledby the function shown, where W is his weight inpounds and H is his height in inches. (a) Find theideal weight for a male who is 75 in. tall. (b) If Iam 72 in. tall and weigh 210 lb, how much weightshould I lose?

110. Celsius to Fahrenheit conversions: The relationship between Fahrenheit degrees anddegrees Celsius is modeled by the function shown.(a) What is the Celsius temperature if ? (b) Use the formula to solve for F in terms of C,then substitute the result from part (a). What doyou notice?

°F � 41

C �59

1F � 322

W1H2 �92

H � 151 111. Pick’s theorem:

Pick’s theorem is an interesting yet little knownformula for computing the area of a polygon drawnin the Cartesian coordinate system. The formulacan be applied as long as the vertices of thepolygon are lattice points (both x and y areintegers). If B represents the number of latticepoints lying directly on the boundary of thepolygon (including the vertices), and I representsthe number of points in the interior, the area of thepolygon is given by the formula shown. Use somegraph paper to carefully draw a triangle withvertices at , (3, 9), and (7, 6), then use Pick’s theorem to compute the triangle’s area.

1�3, 12

A �12

B � I � 1

� APPLICATIONS

112. Gas mileage: John’s old ’87 LeBaron has a 15-galgas tank and gets 23 mpg. The number of miles hecan drive is a function of how much gas is in thetank. (a) Write this relationship in equation formand (b) determine the domain and range of thefunction in this context.

113. Gas mileage: Jackie has a gas-powered model boatwith a 5-oz gas tank. The boat will run for 2.5 minon each ounce. The number of minutes she canoperate the boat is a function of how much gas is inthe tank. (a) Write this relationship in equationform and (b) determine the domain and range ofthe function in this context.

114. Volume of a cube: The volume of a cube dependson the length of the sides. In other words, volumeis a function of the sides: . (a) In practicalterms, what is the domain of this function? (b) Evaluate V(6.25) and (c) evaluate the functionfor

115. Volume of a cylinder: For a fixed radius of 10 cm,the volume of a cylinder depends on its height. Inother words, volume is a function of height:

s � 2x2.

V1s2 � s3

. (a) In practical terms, what is thedomain of this function? (b) Evaluate V(7.5) and

(c) evaluate the function for .

116. Rental charges: Temporary Transportation Inc.rents cars (local rentals only) for a flat fee of$19.50 and an hourly charge of $12.50. This meansthat cost is a function of the hours the car is rentedplus the flat fee. (a) Write this relationship inequation form; (b) find the cost if the car is rentedfor 3.5 hr; (c) determine how long the car was rentedif the bill came to $119.75; and (d) determine thedomain and range of the function in this context, ifyour budget limits you to paying a maximum of$150 for the rental.

117. Cost of a service call: Paul’s Plumbing charges aflat fee of $50 per service call plus an hourly rateof $42.50. This means that cost is a function of thehours the job takes to complete plus the flat fee. (a) Write this relationship in equation form; (b) find the cost of a service call that takes ;(c) find the number of hours the job took if thecharge came to $262.50; and (d) determine the

212 hr

h �8�

V1h2 � 100�h


cob19537_ch01_033-049.qxd 1/31/11 5:55 PM

Precalculus—

domain and range of the function in this context, ifyour insurance company has agreed to pay for allcharges over $500 for the service call.

118. Predicting tides: The graphshown approximates theheight of the tides at FairHaven, New Brunswick, for a12-hr period. (a) Is this thegraph of a function? Why? (b) Approximately what timedid high tide occur? (c) Howhigh is the tide at 6 P.M.? (d) What time(s) will the tide be 2.5 m?

119. Predicting tides: The graphshown approximates theheight of the tides at Apia,Western Samoa, for a 12-hrperiod. (a) Is this the graph of a function? Why? (b) Approximately what timedid low tide occur? (c) Howhigh is the tide at 2 A.M.? (d) What time(s) will the tide be 0.7 m?

Met

ers

Time1 A.M. 311953 P.M. 7

2

3

4

1

5

Met

ers

Time2 A.M. 4121064 P.M. 8

1.0

0.5


120. A father challenges his son to a 400-m race,depicted in the graph shown here.

a. Who won and what was the approximatewinning time?

b. Approximately how many meters behind wasthe second place finisher?

c. Estimate the number of seconds the father wasin the lead in this race.

d. How many times during the race were thefather and son tied?

Time in seconds

Father: Son:

Dis

tanc

e in

met

ers

400

300

200

100

080706010 20 30 40 50

121. Sketch the graph of then discuss how youcould use this graph to obtain the graph of

without computing additional points.

What would the graph of look like?

122. Sketch the graph of then discuss howyou could use this graph to obtain the graph of

without computing additional points.

Determine what the graph of wouldlook like.

123. If the equation of a function is given, the domain isimplicitly defined by input values that generatereal-valued outputs. But unless the graph is givenor can be easily sketched, we must attempt to findthe range analytically by solving for x in terms of y.We should note that sometimes this is an easy task,while at other times it is virtually impossible andwe must rely on other methods. For the followingfunctions, determine the implicit domain and findthe range by solving for x in terms of y.

a. b. y � x2 � 3y � x � 3x � 2

g1x2 ��x2 � 4�

x2 � 4

F1x2 � �x2 � 4�

f 1x2 � x2 � 4,

g1x2 ��x�x

F1x2 � �x�

f 1x2 � x,


124. (1.1) Find the equation of a circle whose center iswith a radius of 5. Then graph the circle.

125. (Appendix A.6) Compute the sum and productindicated:

a.b. 12 � 132 12 � 132

124 � 6154 � 16

14, �12126. (Appendix A.4) Solve the equation by factoring,

then check the result(s) using substitution:

127. (Appendix A.4) Factor the following polynomialscompletely:

a.b.c. 8x3 � 125

2x2 � 13x � 24

x3 � 3x2 � 25x � 75

3x2 � 4x � 7.


cob19537_ch01_033-049.qxd 1/31/11 5:55 PM

Precalculus—

1. Sketch the graph of the line. Plot and label

at least three points.

2. Find the slope of the linepassing through the givenpoints: and

.

3. In 2009, Data.com lost $2 million. In 2010, they lost $0.5 million. Will the slope ofthe line through these points bepositive or negative? Why?Calculate the slope. Were youcorrect? Write the slope as aunit rate and explain what itmeans in this context.

4. To earn some spending money,Sahara takes a job in a ski shopworking primarily with herspecialty—snowboards. She ispaid a monthly salary of $950plus a commission of $7.50 foreach snowboard she sells. (a) Write a function that modelsher monthly earnings E. (b) Usea graphing calculator todetermine her income if she sells20, 30, or 40 snowboards in one month. (c) Use theresults of parts a and b to set an appropriate viewingwindow and graph the line. (d) Use the feature todetermine the number of snowboards that must be soldfor Sahara’s monthly income to top $1300.

5. Write the equation for line L1 shown. Is this thegraph of a function? Discuss why or why not.

TRACE

14, �1021�3, 82

4x � 3y � 126. Write the equation for line L2 shown. Is this the

graph of a function? Discuss why or why not.

7. For the graph of function h(x) shown, (a) determinethe value of h(2); (b) state the domain; (c) determinethe value(s) of x for which and (d) statethe range.

8. Judging from the appearance of the graph alone,compare the rate of change (slope) from to

to the rate of change from to Which rate of change is larger? How is thatdemonstrated graphically?

9. Compute the slope of the lineshown, and explain what itmeans as a rate of change inthis context. Then use the slopeto predict the fox populationwhen the pheasant populationis 13,000.

10. State the domain and range for each function below.a. b.

c.

5�5 x

y5

�5

5�5 x

y5

�5

5�5

5

x

y

�5

x � 5.x � 4x � 2x � 1

h1x2 � �3;5�5

�5

L1

L2

x

y5

x

y

�5

�5

5

5

h(x)

Exercises 5 and 6

Exercises 7 and 8

105 6 7 8 910 2 3 4

10

Fox

popu

latio

n (i

n 10

0s)

Pheasant population (1000s)

2

3

1

4

5

6

7

8

9

P

F

F(p)

Exercise 9

REINFORCING BASIC CONCEPTS

Finding the Domain and Range of a Relation fr om Its GraphThe concepts of domain and range are an important and fundamental part of working with relations and functions. Inthis chapter, we learned to determine the domain of any relation from its graph using a “vertical boundary line,” and therange by using a “horizontal boundary line.” These approaches to finding the domain and range can be combined into asingle step by envisioning a rectangle drawn around or about the graph. If the entire graph can be “bounded” within therectangle, the domain and range can be based on the rectangle’s related length and width. If it’s impossible to bound thegraph in a particular direction, the related x- or y-values continue infinitely. Consider the graph in Figure 1.60. This isthe graph of an ellipse (Section 8.2), and a rectangle that bounds the graph in all directions is shown in Figure 1.61.

MID-CHAPTER CHECK


cob19537_ch01_033-049.qxd 1/31/11 5:56 PM

The domain of this relation is and the range is .

Use this approach to find the domain and range of the following relations and functions.

y � 3�5, q 2x � 3�7, q 2

Precalculus—

108642�10�8 �6 �4 �2

�10

�8

�6

�4

�2

8

4

10

x

y

2

6

10�10

�10

10

x

y

8642

�8�6�4�2 8642�8�6�4�2 10�10

�10

10

x

y

8642

�8�6�4�2 8642�8�6�4�2 10�10

�10

10

x

y

8642

�8�6�4�2 8642�8�6�4�2 10�10

�10

10

x

y

8642

�8�6�4�2 8642�8�6�4�2

108642�10�8 �6 �4 �2

�10

�8

�6

�4

�2

8

4

10

x

y

2

6

Figure 1.60


Figure 1.61

108642�10�8 �6 �4 �2

�10

�8

�6

�4

�2

8

4

10

x

y

2

6

108642�10�8 �6 �4 �2

�10

�8

�6

�4

�2

8

4

10

x

y

2

6

108642�10�8 �6 �4 �2

�10

�8

�6

�4

�2

8

4

10

x

y

2

6

108642�10�8 �6 �4 �2

�10

�8

�6

�4

�2

8

4

10

x

y

2

6


The rectangle extends from to in the horizontal direction, and from to in the verticaldirection. The domain of this relation is and the range is .

The graph in Figure 1.62 is a parabola, and no matter how large we draw the rectangle, an infinite extension of thegraph will extend beyond its boundaries in the left and right directions, and in the upward direction (Figure 1.63).

y � 31, 7 4x � 3�3, 9 4y � 7y � 1x � 9x � �3

The domain of this relation is and the range is .

Finally, the graph in Figure 1.64 is the graph of a square root function, and a rectangle can be drawn that bounds thegraph below and to the left, but not above or to the right (Figure 1.65).

y � 3�6, q 2x � 1�q, q 2

1–49 Reinforcing Basic Concepts 49

Exercise 1: Exercise 2: Exercise 3: Exercise 4:

cob19537_ch01_033-049.qxd 1/31/11 5:56 PM

EXAMPLE 1 � Solving for y in a Linear Equation Solve for y, then evaluate at , and

Solution � given equation

add 6x

divide by 2

Since the coefficients are integers, evaluate the function mentally. Inputs aremultiplied by 3, then increased by 2, yielding the ordered pairs (4, 14), (0, 2), and .


This form of the equation (where y has been written in terms of x) enables us toquickly identify what operations are performed on x in order to obtain y. Once again, for

multiply inputs by 3, then add 2.

EXAMPLE 2 � Solving for y in a Linear EquationSolve the linear equation for y, then identify the new coefficient of xand the constant term.


add 2x

divide by 3

The coefficient of x is and the constant term is 2.


When the coefficient of x is rational, it’s helpful to select inputs that are multiplesof the denominator if the context or application requires us to evaluate the equation.This enables us to perform most operations mentally. For possible inputsmight be and so on. See Exercises 19 through 24.

In Section 1.2, linear equations were graphed using the intercept method. When theequation is written with y in terms of x, we notice a powerful connection between thegraph and its equation—one that highlights the primary characteristics of a linear graph.

x � �9, �6, 0, 3, 6,y � 2

3x � 2,

23

y �2

3x � 2

3y � 2x � 6 3y � 2x � 6

3y � 2x � 6

y � 3x � 2:

1�13, 12

y � 3x � 2 2y � 6x � 4

2y � 6x � 4

x � �13.x � 4, x � 02y � 6x � 4

Precalculus—

1.4 Linear Functions, Special Forms, and More on Rates of Change

The concept of slope is an important part of mathematics, because it gives us a wayto measure and compare change. The value of an automobile changes with time,the circumference of a circle increases as the radius increases, and the tension in a spring grows the more it is stretched. The real world is filled with examples ofhow one change affects another, and slope helps us understand how these changesare related.

A. Linear Equations, Slope-Intercept Form and Function Form

In Section 1.2, we learned that a linear equation is one that can be written in the form. Solving for y in a linear equation offers distinct advantages to understand-

ing linear graphs and their applications.ax � by � c


how we can:

A. Write a linear equation inslope-intercept form andfunction form

B. Use slope-intercept formto graph linear equations

C. Write a linear equation inpoint-slope form

D. Apply the slope-interceptform and point-slopeform in context

WORTHY OF NOTEIn Example 2, the final form can bewritten as shown (inputsare multiplied by two-thirds, thenincreased by 2), or written as

(inputs are multiplied bytwo, the result divided by 3 and thisamount increased by 2). The twoforms are equivalent.

y � 2x3 � 2

y � 23x � 2

50 1–50

cob19537_ch01_050-063.qxd 1/28/11 8:21 PM

EXAMPLE 3 � Noting Relationships between an Equation and Its GraphFind the intercepts of and use them to graph the line. Then,

a. Use the intercepts to calculate the slope of the line, then identify the y-intercept.

b. Write the equation with y in terms of x and compare the calculated slope and y-intercept to the equation in this form. Comment on what you notice.

Solution � Substituting 0 for x in we find they-intercept is Substituting 0 for y gives anx-intercept of . The graph is displayed here.

a. The y-intercept is and by calculation or

counting the slope is [from the

intercept we count down 4, giving, and right 5, giving , to

arrive at the intercept ].b. Solving for y:

given equation

subtract 4x

divide by 5

The slope value seems to be the coefficient of x, while the y-intercept is theconstant term.


After solving a linear equation for y, an input of causes the “x-term” to becomezero, so the y-intercept automatically involves the constant term. As Example 3 illustrates,we can also identify the slope of the line—it is the coefficient of x. In general, a linearequation of the form is said to be in slope-intercept form, since the slopeof the line is m and the y-intercept is (0, b).

Slope-Intercept Form

For a nonvertical line whose equation is , the slope of the line is m and the y-intercept is (0, b).

Solving a linear equation for y in terms of x is sometimes called writing the equa-tion in function form, as this form clearly highlights what operations are performed onthe input value in order to obtain the output (see Example 1). In other words, this formplainly shows that “y depends on x,” or “y is a function of x,” and that the equations

and are equivalent.

Linear Functions

A linear function is one of the form

,

where m and b are real numbers.

Note that if , the result is a constant function . If and theresult is , called the identity function.f 1x2 � x

b � 0,m � 1f 1x2 � bm � 0

f 1x2 � mx � b

f 1x2 � mx � by � mx � b

y � mx � b

y � mx � b

x � 0

y ��4

5x � 4

5y � �4x � 20 4x � 5y � �20

10, �42¢x � 5¢y � �4

1�5, 02

m � �45

¢y

¢x,

10, �421�5, 0210, �42.

4x � 5y � �20,

4x � 5y � �20

1–51 Section 1.4 Linear Functions, Special Forms, and More on Rates of Change 51

Precalculus—

x

y

54321�5 �4 �3 �2 �1

�5

�4

�3

�2

�1

5

4

3

2

1(�5, 0)

�4

5

(0, �4)

cob19537_ch01_050-063.qxd 1/28/11 8:22 PM

EXAMPLE 4 � Finding the Function Form of a Linear EquationWrite each equation in both slope-intercept form and function form. Then identifythe slope and y-intercept of the line.

a. b. c.

Solution � a. b. c.

y-intercept y-intercept (0, 5) y-intercept (0, 0)


Note that we can analytically develop the slope-intercept form of a line using the slopeformula. Figure 1.66 shows the graph of a general line through the point (x, y) with ay-intercept of (0, b). Using these points in the slope formula, we have

slope formula

substitute: (0, b) for (x1, y1), (x, y) for (x2, y2)

simplify

multiply by x

add b to both sides

This approach confirms the relationship between the graphical characteristics of a lineand its slope-intercept form. Specifically, for any linear equation written in the form

, the slope must be m and the y-intercept is (0, b).

B. Slope-Intercept Form and the Graph of a Line

If the slope and y-intercept of a linear equation are known or can be found, we can con-struct its equation by substituting these values directly into the slope-intercept form

.

EXAMPLE 5 � Finding the Equation of a Line from Its GraphFind the slope-intercept equation of the line shown.

Solution � Using and in the slope formula,

or by simply counting , the slope is .

By inspection we see the y-intercept is (0, 4).Substituting for m and 4 for b in the slope-intercept form we obtain the equation


y � 2x � 4.

21

m � 42 or 21

¢y

¢x

1�1, 221�3, �22

y � mx � b

y � mx � b

y � mx � b

y � b � mx

y � b

x� m

y � b

x � 0� m

y2 � y1

x2 � x1� m

a0, �9

2b

m �1

2, b � 0m � �1, b � 5m �

3

2, b � �

9

2

f 1x2 �1

2 xf 1x2 � �1x � 5f 1x2 �

3

2x �

9

2

y �1

2xy � �1x � 5y �

3

2x �

9

2

y �x

2y � �x � 5�2y � �3x � 9

2y � xy � x � 53x � 2y � 9

2y � xy � x � 53x � 2y � 9


Precalculus—


we can write a linear equation

in slope-intercept form and

function form

y

5�5

�5

5

x

(�3, �2)

(�1, 2)

Figure 1.66

x

y

5�5

�5

5

(x, y)

(0, b)

cob19537_ch01_050-063.qxd 1/28/11 8:22 PM

Actually, if the slope is known and we have any point (x, y) on the line, we can stillconstruct the equation since the given point must satisfy the equation of the line. In thiscase, we’re treating as a simple formula, solving for b after substitutingknown values for m, x, and y.

EXAMPLE 6 � Using y � mx � b as a FormulaFind the slope-intercept equation of a linethat has slope and contains Verify results on a graphing calculator.

Solution � Use as a “formula,” withand

slope-intercept form

substitute for m, for x, and 2 for y

simplify

solve for b

The equation of the line is After entering the equation on the screen of a graphing calculator, we can evaluate on the home screen, oruse the feature. See the figures provided.


Writing a linear equation in slope-intercept form enables us to draw its graph witha minimum of effort, since we can easily locate the y-intercept and a second point using

the rate of change For instance, indicates that counting down 2 and

right 3 from a known point will locate another point on this line.

EXAMPLE 7 � Graphing a Line Using Slope-Intercept F orm and the Rate of ChangeWrite in slope-intercept form, then graph the line using the y-interceptand the rate of change (slope).


isolate y term

divide by 3

The slope is and the y-intercept is (0, 3).

Plot the y-intercept, then use (up 5 and

right 3—shown in blue) to find another point on theline (shown in red). Finish by drawing a linethrough these points.


For a discussion of what graphing method might be most efficient for a givenlinear equation, see Exercises 103 and 114.

Parallel and Perpendicular Lines

From Section 1.2 we know parallel lines have equal slopes: and perpendicular

lines have slopes with a product of or In some applications,

we need to find the equation of a second line parallel or perpendicular to a given line,through a given point. Using the slope-intercept form makes this a simple four-step process.

m1 � �1

m2.m1

# m2 � �1�1:

m1 � m2,

¢y

¢x�

5

3

m � 53

y � 53x � 3

3y � 5x � 9 3y � 5x � 9

3y � 5x � 9

¢y

¢x�

�2

3

¢y

¢x.

TRACE

x � �5Y=y � 4

5x � 6.

6 � b2 � �4 � b

�5452 � 4

5 1�52 � by � mx � b

y � 2.x � �5,m � 45,

y � mx � b

1�5, 22.m � 45

y � mx � b


Precalculus—

(0, 3)

(3, 8)Run 3

Rise 5

y � fx � 3

�y � f

�x

x

y

5�5

�2

WORTHY OF NOTENoting the fraction is equal to ,we could also begin at (0, 3) and

count (down 5 and left 3)

to find an additional point on the line: . Also, for any

negative slope note

�ab

��ab

�a

�b.

¢y

¢x� �

ab

,

1�3, �22

¢y

¢x�

�5�3

�5�3

53

�10

10

�10 10

(�5, 2) is on the line

cob19537_ch01_050-063.qxd 1/28/11 8:22 PM

Finding the Equation of a Line P arallel or Perpendicular to a Given Line

1. Identify the slope m1 of the given line.2. Find the slope m2 of the new line using the parallel or perpendicular

relationship.3. Use m2 with the point (x, y) in the “formula” and solve for b.4. The desired equation will be .y � m2 x � b

y � mx � b

EXAMPLE 8 � Finding the Equation of a P arallel LineFind the slope-intercept equation of a line that goes through and isparallel to

Solution � Begin by writing the equation in slope-intercept form to identify the slope.

given line

isolate y-term

result

The original line has slope and this will also be the slope of any lineparallel to it. Using with we have


simplify

solve for b

The equation of the new line is


Graphing the lines from Example 8 as Y1

and Y2 on a graphing calculator, we note thelines do appear to be parallel (they actually mustbe since they have identical slopes). Using the

8:ZInteger feature of the calculator, wecan quickly verify that Y2 indeed contains thepoint ( ).

For any nonlinear graph, a straight linedrawn through two points on the graph is calleda secant line. The slope of a secant line, and lines parallel and perpendicular to thisline, play fundamental roles in the further development of the rate-of-change concept.

EXAMPLE 9 � Finding Equations for Parallel and Perpendicular LinesA secant line is drawn using the points ( , 0) and (2, ) on the graph of thefunction shown. Find the equation of a line that is

a. parallel to the secant line through ( ).b. perpendicular to the secant line through ( ).

Solution � Either by using the slope formula or counting , we find the secant line has slope

.m ��2

6�

�1

3

¢y

¢x

�1, �4�1, �4

�2�4

�6, �1

ZOOM

y � �23 x � 5.

�5 � b �1 � 4 � b

�1 ��2

3 1�62 � b

y � mx � b

1x, y2 S 1�6, �12m2 � �23

m1 � �23

y � �23 x � 2

3y � �2x � 6 2x � 3y � 6

2x � 3y � 6.1�6, �12


Precalculus—

�47

31

�31

47

substitute for m,for x, and for y�1�6

�23

cob19537_ch01_050-063.qxd 2/1/11 10:35 AM

a. For the parallel line through ( ), .


substitute for m,for x, and for y

simplify

result

The equation of the parallel line (in blue) is .

b. For the perpendicular line through ( , ),.


substitute 3 for m, for x, and for y

simplify

result

The equation of the perpendicular line (in yellow)is .


C. Linear Equations in Point-Slope Form

As an alternative to using we can find the equation of the line using the

slope formula and the fact that the slope of a line is constant. For a given

slope m, we can let (x1, y1) represent a given point on the line and (x, y) represent any

other point on the line, and the formula becomes Isolating the “y” terms

on one side gives a new form for the equation of a line, called the point-slope form:

slope formula

multiply both sides by (x � x1)

simplify point-slope form

The Point-Slope Form of a Linear Equation

For a nonvertical line whose equation is , the slope of the line is m and (x1, y1) is a point on the line.

While using (as in Example 6) may appear to be easier, both theslope-intercept form and point-slope form have their own advantages and it willhelp to be familiar with both.

y � mx � b

y � y1 � m1x � x12

S y � y1 � m1x � x12

1x � x12

1 a

y � y1

x � x1b � m1x � x12

y � y1

x � x1� m

y � y1

x � x1� m.

y2 � y1

x2 � x1� m,

y � mx � b,

y � 3x � 1

�1 � b�4 � �3 � b

�4�1�4 � 31�12 � by � mx � b

m2 � 3�4�1

y ��1

3x �

13

3

�13

3� b

1�4 � �123 2�

12

3�

1

3� b

�4�1

�13�4 �

�1

31�12 � b

y � mx � b

m2 ��1

3�1, �4


Precalculus—

WORTHY OF NOTEThe word “secant” comes from theLatin word secare, meaning “tocut.” Hence a secant line is onethat cuts through a graph, asopposed to a tangent line, whichtouches the graph at only onepoint.

x

y

5�5

�5

5

(�1, �4)

x

y

5�5

�5

5

(�1, �4)


we can use the slope-intercept

form to graph linear equations

cob19537_ch01_050-063.qxd 1/28/11 8:22 PM


Precalculus—

EXAMPLE 10 � Using y � y1 � m(x � x1) as a FormulaFind the equation of the line in point-slope form, if and ( ) is on theline. Then graph the line.

Solution � point-slope form

substitute for m; ( ) for (x1, y1)

simplify, point-slope form

To graph the line, plot ( ) and use to find additional points on the line.


D. Applications of Linear Equations

As a mathematical tool, linear equations rank among the most common, powerful, andversatile. In all cases, it’s important to remember that slope represents a rate of change.

The notation literally means the quantity measured along the y-axis, is chang-

ing with respect to changes in the quantity measured along the x-axis.

EXAMPLE 11 � Relating Temperature to AltitudeIn meteorological studies, atmospheric temperature depends on the altitudeaccording to the formula where T(h) represents theapproximate Fahrenheit temperature at height h (in thousands of feet, ).

a. Interpret the meaning of the slope in this context.b. Determine the temperature at an altitude of 12,000 ft.c. If the temperature is what is the approximate altitude?

Algebraic Solution � a. Notice that h is the input variable and T is the output. This shows

meaning the temperature drops for every 1000-ft increase in altitude.b. Since height is in thousands, use .

original formula

substitute 12 for h

result

Technology Solution �

At a height of 12,000 ft, the temperature is about .16.5°

� 16.5 T1122 � �3.51122 � 58.5

T1h2 � �3.5h � 58.5

h � 123.5°F

¢T

¢h�

�3.5

1,

�8°F

0 � h � 36T1h2 � �3.5h � 58.5,

m �¢y

¢x

¢y

¢x �

2

3�3, �3

y � 3 �2

3 1x � 32

�3, �323 y � 1�32 �

2

33x � 1�32 4

y � y1 � m1x � x12

�3, �3m � 23

�y�2

�x�3

(�3, �3)

y � 3 � s (x � 3)

y

�5

5

x5�5


we can write a linear equation

in point-slope form

cob19537_ch01_050-063.qxd 1/28/11 8:22 PM

c. Replacing T(h) with �8 and solving gives

Algebraic Solution � original formula

substitute �8 for T(h)

subtract 58.5

divide by

The temperature is about at a height of .

Graphical Solution � Since we’re given , we can set and . Atground level , the formula gives a temperature of , while at ,we have . This shows appropriate settings for the range would be

and (see figure). After setting ,we press and move the cursor until we find an output value near , whichoccurs when X is near 19. To check, we input 19 for x and the calculator displaysan output of , which corresponds with the algebraic result (at 19,000 ft, thetemperature is ).�8°F

�8

�8TRACE

Y1 � �3.5X � 58.5Ymax � 50Ymin � �50T1362 � �67.5

h � 3658.5°1x � 02Xmax � 40Xmin � 00 � h � 36

19 � 1000 � 19,000 ft�8°F

�3.5 19 � h �66.5 � �3.5h

�8 � �3.5h � 58.5 T1h2 � �3.5h � 58.5


Precalculus—

40

50

�50

0


In many applications, outputs that are integer or rational values are rare, making itdifficult to use the feature alone to find an exact solution. In the Section 1.5, we’lldevelop additional ways that graphs and technology can be used to solve equations.

In some applications, the relationship is known to be linear but only a few points onthe line are given. In this case, we can use two of the known data points to calculate theslope, then the point-slope form to find an equation model. One such application is lin-ear depreciation, as when a government allows businesses to depreciate vehicles andequipment over time (the less a piece of equipment is worth, the less you pay in taxes).

EXAMPLE 12A � Using Point-Slope Form to Find a Function ModelFive years after purchase, the auditor of a newspaper company estimates the valueof their printing press is $60,000. Eight years after its purchase, the value of thepress had depreciated to $42,000. Find a linear equation that models thisdepreciation and discuss the slope and y-intercept in context.

Solution � Since the value of the press depends on time, the ordered pairs have the form (time,value) or (t, v) where time is the input, and value is the output. This means theordered pairs are (5, 60,000) and (8, 42,000).

slope formula

simplify and reduce ��18,000

3�

�6000

1

1t2, v22 � 18, 42,00021t1, v12 � 15, 60,0002; �42,000 � 60,000

8 � 5

m �v2 � v1

t2 � t1

TRACE

cob19537_ch01_050-063.qxd 1/28/11 8:23 PM


WORTHY OF NOTEActually, it doesn’t matter which ofthe two points are used in Example12A. Once the point (5, 60,000) isplotted, a constant slope of

will “drive” the linethrough (8, 42,000). If we first graph(8, 42,000), the same slope would“drive” the line through (5, 60,000).Convince yourself by reworking theproblem using the other point.

m � �6000

The slope of the line is , indicating the printing press loses

$6000 in value with each passing year.

point-slope form

substitute for m; (5, 60,000) for (t1, v1)

simplify

solve for v

The depreciation equation is . The v-intercept (0, 90,000)indicates the original value (cost) of the equipment was $90,000.

Once the depreciation equation is found, it represents the (time, value) relationshipfor all future (and intermediate) ages of the press. In other words, we can now predictthe value of the press for any given year. However, note that some equation models arevalid for only a set period of time, and each model should be used with care.

EXAMPLE 12B � Using a Function Model to Gather Inf ormationFrom Example 12A,

a. How much will the press be worth after 11 yr?b. How many years until the value of the equipment is $9000?c. Is this function model valid for (why or why not)?

Solution � a. Find the value v when :

equation model

substitute 11 for t

result (11, 24,000)

After 11 yr, the printing press will only be worth $24,000.b. “. . . value is $9000” means :

value at time t

substitute for v (t )

subtract 90,000

divide by

After 13.5 yr, the printing press will be worth $9000.c. Since substituting 18 for t gives a negative quantity, the function model is not

valid for . In the current context, the model is only valid while and solving shows the domain of the function in thiscontext is .


t � 30, 15 4�6000t � 90,000 � 0

v � 0t � 18

�6000 t � 13.5 �6000t � �81,000

�6000t � 90,000 �6000t � 90,000 � 9000 v1t2 � 9000

v1t2 � 9000

� 24,000 v1112 � �60001112 � 90,000

v1t2 � �6000t � 90,000

t � 11

t � 18 yr

v1t2 � �6000t � 90,000

v � �6000t � 90,000 v � 60,000 � �6000t � 30,000

�6000 v � 60,000 � �60001t � 52 v � v1 � m1t � t12

¢value

¢time�

�6000

1

Precalculus—

D. You’ve just seen how

we can apply the slope-

intercept form and point-slope

form in context

cob19537_ch01_050-063.qxd 1/28/11 8:23 PM


Precalculus—

2. The notation indicates the is

changing in response to changes in .

¢cost

¢time



1. For the equation the slope is

and the y-intercept is .

y ��7

4x � 3,

3. Line 1 has a slope of The slope of any lineperpendicular to line 1 is .

�0.4. 4. The equation is called theform of a line.

y � y1 � m1x � x12

5. Discuss/Explain how to graph a line using only theslope and a point on the line (no equations).

6. Given and is on the line. Compareand contrast finding the equation of the line using

versus y � y1 � m1x � x12.y � mx � b

1�5, 62m � �35


Solve each equation for y and evaluate the result using

7. 8.

9. 10.

11. 12.

For each equation, solve for y and identify the newcoefficient of x and new constant term.

13. 14.

15. 16.

17. 18.

Evaluate each equation by selecting three inputs thatwill result in integer values. Then graph each line.

19. 20.

21. 22.

23. 24.

Find the x- and y-intercepts for each line, then (a) use these two points to calculate the slope of the line, (b) write the equation with y in terms of x (solve for y) and (c) compare the calculated slope and y-intercept to the equation from part (b). Comment on what you notice.

25. 26.

27. 28.

29. 30. 5y � 6x � �254x � 5y � �15

2x � 3y � 92x � 5y � 10

3y � 2x � �63x � 4y � 12

y � �13x � 3y � �1

6x � 4

y � 25x � 3y � �3

2x � 2

y � 54x � 1y � �4

3x � 5

712y � 4

15x � 76

56x � 1

7y � �47

�0.7x � 0.6y � �2.4�0.5x � 0.3y � 2.1

9y � 4x � 186x � 3y � 9

17y � 1

3x � 213x � 1

5y � �1

�0.2x � 0.7y � �2.1�0.4x � 0.2y � 1.4

3y � 2x � 94x � 5y � 10

x � �5, x � �2, x � 0, x � 1, and x � 3.Write each equation in slope-intercept form (solve for y)and function form, then identify the slope and y-intercept.

31. 32.

33. 34.

35. 36.

37. 38.

For Exercises 39 to 50, use the slope-intercept form tostate the equation of each line. Verify your solutions toExercises 45 to 47 using a graphing calculator.

39. 40.

41.

42. y-intercept

43. y-intercept (0, 2)

44. y-intercept

45. is on the line1�3, 22m � �4;

10, �42m � �32 ;

m � 3;

10, �32m � �2;

(�1, 0)

(�2, �3)

(0, 3)

54321�5�4�3�2�1�1�2�3�4�5

2345

1

y

x

x

y

54321�5�4�3�2�1�1�2�3�4�5

2345

1

(�5, 5)

(5, 1)

(0, 3)

x

y

54321�5�4�3�2�1�1�2�3�4�5

2345

1

(�3, �1)

(0, 1)

(3, 3)

5y � 3x � 20 � 03x � 4y � 12 � 0

2x � �5yx � 3y

y � 2x � 45x � 4y � 20

4y � 3x � 122x � 3y � 6

1.4 EXERCISES

cob19537_ch01_050-063.qxd 1/28/11 8:23 PM

46. is on the line

47. is on the line

48. 49.

50.

Write each equation in slope-intercept form, then use therate of change (slope) and y-intercept to graph the line.

51. 52.

53. 54.

Graph each linear equation using the y-intercept andrate of change (slope) determined from each equation.

55. 56.

57. 58.

59. 60.

61. 62.

Find the equation of the line using the informationgiven. Write answers in slope-intercept form.

63. parallel to through the point


65. perpendicular to through the point

66. perpendicular to through the point



69. parallel to , through the point (2, 5)

70. perpendicular to through the point (2, 5)y � �3

y � �3

13, �4215y � 8x � 50,

1�2, �1212x � 5y � 65,

1�5, 32x � 4y � 7,

16, �325y � 3x � 9,

1�3, �526x � 9y � 27,

1�5, 222x � 5y � 10,

y � �32 x � 2y � 1

2x � 3

y � �3x � 4y � 2x � 5

y � �45 x � 2y � �1

3 x � 2

y � 52x � 1y � 2

3x � 3

�3x � 2y � 42x � 3y � 15

2y � x � 43x � 5y � 20

1614128 10

800

400

1600

2000

1200

x

y

34323026 28

600

300

1200

1500

900

x

y

20181612 14

4000

2000

8000

10,000

6000

x

y

1�4, 72m � �32 ;

15, �32m � 2;


Precalculus—

Write the equations in slope-intercept form and statewhether the lines are parallel, perpendicular, or neither.

71. 72.

73. 74.

75. 76.

A secant line is one that intersects a graph at two ormore points. For each graph given, find an equation ofthe line (a) parallel and (b) perpendicular to the secantline, through the point indicated.

77. 78.

79. 80.

81. 82.

Find the equation of the line in point-slope form, thengraph the line.

83.

84.

85.

86.

87.

88. m � 1.5; P1 � 1�0.75, �0.1252

m � 0.5; P1 � 11.8, �3.12

P1 � 1�1, 62, P2 � 15, 12

P1 � 13, �42, P2 � 111, �12

m � �1; P1 � 12, �32

m � 2; P1 � 12, �52

(1, 3)

x

y

5�5

�5

5

(0, �2)

x

y

5�5

�5

5

(1, �2.5)

x

y

5�5

�5

5

(�1, 3)

x

y

5�5

�5

5

(1, 3)

x

y

5�5

�5

5

(2, �4)

x

y

5�5

�5

5

11y � 5x � �776x � 8y � 25y � 11x � 1353x � 4y � 12

2x � 3y � 64x � 3y � 18�4x � 6y � 122x � 5y � 20

2x � 3y � 65y � 4x � �153y � 2x � 64y � 5x � 8

cob19537_ch01_050-063.qxd 1/28/11 8:24 PM

Find the equation of the line in point-slope form, andstate the meaning of the slope in context—whatinformation is the slope giving us?

89. 90.

91. 92.

93. 94.

Egg

s pe

r he

n pe

r w

eek

Temperature in °F8075706560

4

2

0

8

10

6

x

y

Cat

tle r

aise

d pe

r ac

re

Rainfall per month(in inches)

54321

40

20

0

80

100

60

x

y

Onl

ine

brok

erag

e ho

uses

Independent investors (1000s)10864 5 7 92 31

4

2

3

5

7

9

1

0

8

10

6

x

y

Stud

ent’s

fin

al g

rade

(%

)(i

nclu

des

extr

a cr

edit)

Hours of television per day5432 2.5 3.5 4.51 1.50.50

40

20

30

50

70

90

10

80

100

60

x

yTy

pew

rite

rs in

ser

vice

(in

ten

thou

sand

s)

Year (1990 → 0)864 5 7 92 310

4

2

3

5

7

9

1

8

10

6

x

y

Inco

me

(in

thou

sand

s)

Sales (in thousands)x

y

864 5 7 92 310

4

2

3

5

7

9

1

8

10

6


Precalculus—

Using the concept of slope, match each description withthe graph that best illustrates it. Assume time is scaledon the horizontal axes, and height, speed, or distancefrom the origin (as the case may be) is scaled on thevertical axis.

95. While driving today, I got stopped by a statetrooper. After she warned me to slow down, Icontinued on my way.

96. After hitting the ball, I began trotting around thebases shouting, “Ooh, ooh, ooh!” When I saw itwasn’t a home run, I began sprinting.

97. At first I ran at a steady pace, then I got tired andwalked the rest of the way.

98. While on my daily walk, I had to run for a whilewhen I was chased by a stray dog.

99. I climbed up a tree, then I jumped out.

100. I steadily swam laps at the pool yesterday.

101. I walked toward the candy machine, stared at it fora while then changed my mind and walked back.

102. For practice, the girls’ track team did a series of25-m sprints, with a brief rest in between.

x

y H

x

y G

x

y F

x

y E

x

y A

x

y D

x

y C

x

y B


103. General linear equation:

The general equation of a line is shown here, wherea, b, and c are real numbers, with a and b notsimultaneously zero. Solve the equation for y andnote the slope (coefficient of x) and y-intercept(constant term). Use these to find the slope and y-intercept of the following lines, without solvingfor y or computing points.

a. b.c. d. 3y � 5x � 95x � 6y � �12

2x � 5y � �153x � 4y � 8

ax � by � c 104. Intercept-Intercept form of a linear

equation:

The x- and y-intercepts of a line can also be foundby writing the equation in the form shown (withthe equation set equal to 1). The x-intercept will be(h, 0) and the y-intercept will be (0, k). Find the x- and y-intercepts of the following lines using thismethod. How is the slope of each line related to thevalues of h and k?

a. b.c. 5x � 4y � 8

3x � 4y � �122x � 5y � 10

xh

�y

k� 1

cob19537_ch01_050-063.qxd 1/28/11 8:24 PM


Precalculus—

� APPLICATIONS

105. Speed of sound: The speed of sound as it travelsthrough the air depends on the temperature of theair according to the function whereV represents the velocity of the sound waves inmeters per second (m/s), at a temperature of

Celsius. (a) Interpret the meaning of the slopeand y-intercept in this context. (b) Determine thespeed of sound at a temperature of . (c) If thespeed of sound is measured at 361 m/s, what is thetemperature of the air?

106. Acceleration: A driver going down a straighthighway is traveling 60 ft/sec (about 41 mph) oncruise control, when he begins accelerating at a rateof 5.2 ft/sec2. The final velocity of the car is givenby where V is the velocity at time t.(a) Interpret the meaning of the slope and y-interceptin this context. (b) Determine the velocity of the car after 9.4 seconds. (c) If the car is traveling at 100 ft/sec, for how long did it accelerate?

107. Investing in coins: The purchase of a “collector’sitem” is often made in hopes the item will increasein value. In 1998, Mark purchased a 1909-S VDBLincoln Cent (in fair condition) for $150. By theyear 2004, its value had grown to $190. (a) Use therelation (time since purchase, value) with corresponding to 1998 to find a linear equationmodeling the value of the coin. (b) Discuss whatthe slope and y-intercept indicate in this context.(c) How much was the penny worth in 2009? (d) How many years after purchase will the penny’svalue exceed $250? (e) If the penny is now worth$170, how many years has Mark owned the penny?

108. Depreciation: Once a piece of equipment is putinto service, its value begins to depreciate. Abusiness purchases some computer equipment for$18,500. At the end of a 2-yr period, the value of theequipment has decreased to $11,500. (a) Use therelation (time since purchase, value) to find a linearequation modeling the value of the equipment. (b) Discuss what the slope and y-intercept indicate inthis context. (c) What is the equipment’s value after4 yr? (d) How many years after purchase will thevalue decrease to $6000? (e) Generally, companieswill sell used equipment while it still has valueand use the funds to purchase new equipment.According to the function, how many years will it take this equipment to depreciate in value to$1000?

t � 0

V � 265 t � 60,

20°C

T°

V � 35T � 331,

109. Internet connections: The number of householdsthat are hooked up to the Internet (homes that areonline) has been increasing steadily in recent years.In 1995, approximately 9 million homes wereonline. By 2001 this figure had climbed to about 51 million. (a) Use the relation (year, homes online)with corresponding to 1995 to find anequation model for the number of homes online.(b) Discuss what the slope indicates in this context.(c) According to this model, in what year did thefirst homes begin to come online? (d) If the rate ofchange stays constant, how many households wereon the Internet in 2006? (e) How many years after 1995 will there be over 100 millionhouseholds connected? (f) If there are 115 millionhouseholds connected, what year is it?Source: 2004 Statistical Abstract of the United States, Table 965

110. Prescription drugs: Retail sales of prescriptiondrugs have been increasing steadily in recent years.In 1995, retail sales hit $72 billion. By the year2000, sales had grown to about $146 billion. (a) Use the relation (year, retail sales of prescriptiondrugs) with corresponding to 1995 to find alinear equation modeling the growth of retail sales.(b) Discuss what the slope indicates in this context.(c) According to this model, in what year will salesreach $250 billion? (d) According to the model,what was the value of retail prescription drug salesin 2005? (e) How many years after 1995 will retailsales exceed $279 billion? (f) If yearly sales totaled$294 billion, what year is it?Source: 2004 Statistical Abstract of the United States, Table 122

111. Prison population: In 1990, the number of personssentenced and serving time in state and federalinstitutions was approximately 740,000. By the year2000, this figure had grown to nearly 1,320,000. (a) Find a linear function with correspondingto 1990 that models this data, (b) discuss the sloperatio in context, and (c) use the equation to estimatethe prison population in 2010 if this trend continues.Source: Bureau of Justice Statistics at www.ojp.usdoj.gov/bjs

112. Eating out: In 1990, Americans bought an average of143 meals per year at restaurants. This phenomenoncontinued to grow in popularity and in the year 2000,the average reached 170 meals per year. (a) Find alinear function with corresponding to 1990 thatmodels this growth, (b) discuss the slope ratio incontext, and (c) use the equation to estimate theaverage number of times an American will eat at arestaurant in 2010 if the trend continues.Source: The NPD Group, Inc., National Eating Trends, 2002

t � 0

t � 0

t � 0

t � 0

cob19537_ch01_050-063.qxd 1/28/11 8:24 PM


Precalculus—


113. Locate and read the following article. Then turn ina one-page summary. “Linear Function SavesCarpenter’s Time,” Richard Crouse, MathematicsTeacher, Volume 83, Number 5, May 1990: pp. 400–401.

114. The general form of a linear equation is, where a and b are not simultaneously

zero. (a) Find the x- and y-intercepts using thegeneral form (substitute 0 for x, then 0 for y).Based on what you see, when does the interceptmethod work most efficiently? (b) Find the slopeand y-intercept using the general form (solve for y).Based on what you see, when does the slope-intercept method work most efficiently?

ax � by � c

115. Match the correct graph to the conditions stated form and b. There are more choices than graphs.

a. b.c. d.e. f.g. h.

x

y

x

y

x

y(4) (5) (6)

x

y

x

y

x

y(1) (2) (3)

m � 0, b 6 0m 7 0, b � 0

m 6 0, b � 0m � 0, b 7 0

m 7 0, b 7 0m 6 0, b 7 0

m 7 0, b 6 0m 6 0, b 6 0


116. (1.3) Determine the domain:

a.

b.

117. (Appendix A.6) Simply without the use of a calculator.

a. b.

118. (Appendix A.3) Three equations follow. One is anidentity, another is a contradiction, and a third hasa solution. State which is which.

21x � 52 � 13 � 1 � 9 � 7 � 2x

21x � 42 � 13 � 1 � 9 � 7 � 2x

21x � 52 � 13 � 1 � 9 � 7 � 2x

281x22723

y �5

2x � 5

y � 12x � 5

119. (Appendix A.2) Compute the area of the circularsidewalk shown here . Use yourcalculator’s value of and round the answer (only)to hundredths.

�1A � �r22

8 yd

10 yd

cob19537_ch01_050-063.qxd 1/28/11 8:24 PM

Note the two expressions are equal (the equation is true) only when the input is .Solving equations graphically is a simple extension of this observation. By treatingthe expression on the left as the independent function Y1, we have andthe related linear graph will contain all ordered pairs shown in Table 1.4 (see Fig-ure 1.67). Doing the same for the right-hand expression yields and its related graph will likewise contain all ordered pairs shown in the Table 1.5 (seeFigure 1.68).

Y1 Y2

The solution is then found where , or in other words, at the point where thesetwo lines intersect (if it exists). See Figure 1.69.

Most graphing calculators have an intersect feature that can quickly find thepoint(s) where two graphs intersect. On many calculators, we access this ability usingthe sequence (CALC) and selecting option 5:intersect (Figure 1.70).TRACE2nd

Y1 � Y2

2x � 9 � �31x � 12 � 2

Y2 � �31X � 12 � 2,

Y1 � 2X � 9

x � 2

Precalculus—

1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving

In this section, we’ll build on many of the ideas developed in Appendix A.3 (SolvingLinear Equations and Inequalities), as we learn to manipulate formulas and employcertain problem-solving strategies. We will also extend our understanding of graphicalsolutions to a point where they can be applied to virtually any family of functions.

A. Solving Equations Graphically Using the Intersect Method

For some background on why a graphical solution is effective, consider the equation. By definition, an equation is a statement that two expres-

sions are equal for some value of the variable (Appendix A.3). To highlight this fact,the expressions and are evaluated independently for selectedintegers in Tables 1.4 and 1.5.

�31x � 12 � 22x � 9

2x � 9 � �31x � 12 � 2


how we can:

A. Solve equationsgraphically using theintersection-of-graphsmethod

B. Solve equationsgraphically using the x-intercept/zeroesmethod

C. Solve linear inequalitiesgraphically

D. Solve for a specifiedvariable in a formula orliteral equation

E. Use a problem-solvingguide to solve variousproblem types

x 2x � 9

�3 �15

�2 �13

�1 �11

0 �9

1 �7

2 �5

3 �3

Table 1.4

x �3(x � 1) � 2

�3 10

�2 7

�1 4

0 1

1 �2

2 �5

3 �8

Table 1.5

∂f

10

10

�10

�10

Figure 1.67 Figure 1.69Figure 1.68

10

10

�10

�10 10

10

�10

�10

64 1–64

cob19537_ch01_064-078.qxd 1/28/11 8:25 PM

Because the calculator can work with up to 10expressions at once, it will ask you to identifyeach graph you want to work with—even whenthere are only two. A marker is displayed oneach graph in turn, and named in the upper leftcorner of the window (Figure 1.71). You canselect a graph by pressing , or bypass a graphby pressing one of the arrow keys. For situa-tions involving multiple graphs or multiplesolutions, the calculator offers a “GUESS?”option that enables you to specify the approxi-mate location of the solution you’re interested in (Figure 1.72). For now, we’ll simplypress two times in succession to identify each graph, and a third time to bypass the“GUESS?” option. The calculator then finds and displays the point of intersection(Figure 1.73). Be sure to check the settings on your viewing window before you begin,and if the point of intersection is not visible, try 3:Zoom Out or other window-resizing features to help locate it.

EXAMPLE 1A � Solving an Equation Graphically

Solve the equation usinga graphing calculator.

Solution � Begin by entering the left-hand expression as Y1

and the right-hand expression as Y2 (Figure 1.74).To find points of intersection, press (CALC) and select option 5:intersect, whichautomatically places you on the graphingwindow, and asks you to identify the “First curve?.” As discussed, pressing three times in succession will identify eachgraph, bypass the “Guess?” option, then find and display the point of intersection(Figure 1.75). Here the point of intersectionis ( ), showing the solution to thisequation is (for which bothexpressions equal �3). This can be verifiedby direct substitution or by using theTABLE feature.

This method of solving equations is called the Intersection-of-Graphs method,and can be applied to many different equation types.

x � �2�2, �3

ENTER

TRACE2nd

21x � 32 � 7 �1

2x � 2

ZOOM

ENTER

ENTER

1–65 Section 1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving 65

Precalculus—

10

10

�10

�10 10

10

�10

�10

10

10

�10

�10

Figure 1.71Figure 1.70

Figure 1.72

Figure 1.73

Figure 1.74

Figure 1.75

10

10

�10

�10

cob19537_ch01_064-078.qxd 1/28/11 8:25 PM

Intersection-of-Graphs Method for Solving Equations

For any equation of the form f(x) � g(x),1. Assign f (x) as Y1 and g (x) as Y2.2. Graph both function and identify any point(s) of intersection, if they exist.

The x-coordinate of all such points is a solution to the equation.

Recall that in the solution of linear equations, we sometimes encounter equations thatare identities (infinitely many solutions) or contradictions (no solutions). These possi-bilities also have graphical representations, and appear as coincident lines and parallellines respectively. These possibilities are illustrated in Figure 1.76.

EXAMPLE 1B � Solving an Equation GraphicallySolve using a graphing calculator.

Solution � With and as Y2, we use the (CALC) option and select 5:intersect. The graphs appear to be parallel lines(Figure 1.77), and after pressing three times we obtain the error message shown(Figure 1.78), confirming there are no solutions.


B. Solving Equations Graphically Using

the x-Intercept/Zeroes Method

The intersection-of-graphs method works extremely well when the graphs of f (x) andg(x) (Y1 and Y2) are simple and “well-behaved.” Later in this course, we encounter anumber of graphs that are more complex, and it will help to develop alternative meth-ods for solving graphically. Recall that two equations are equivalent if they have thesame solution set. For instance, the equations and are equivalent(since is a solution to both), as are and (since2x � 6 � 03x � 1 � x � 5x � 3

2x � 6 � 02x � 6

ENTER

TRACE2nd0.511 � 1.5X2 � 30.75X � 2 as Y1

0.75x � 2 � 0.511 � 1.5x2 � 3


Precalculus—

y

One point of intersection(unique solution)

x

Infinitely manypoints of intersection

(identity)

x

y

x

No point of intersection(contradiction)

y

A. You’ve just seen howwe can solve equations usingthe Intersection-of-Graphsmethod

Figure 1.76

10

10

�10

�10


cob19537_ch01_064-078.qxd 1/28/11 8:25 PM

is a solution to both). Applying the intersection-of-graphs method to the last twoequivalent equations, gives

and

Y1 Y2 Y1 Y2

The intersection method will work equally well in both cases, but the equation on theright has only one variable expression, and will produce a single (visible) graph (since

is simply the x-axis). Note that here we seek an input value that will result inan output of 0. In other words, all solutions will have the form (x, 0), which is the x-intercept of the graph. For this reason, the method is alternatively called the zeroesmethod or the x-intercept method. The method employs the approach shown above,in which the equation is rewritten as , with assigned as Y1.

Zeroes/x-Intercept Method for Solving Equations

For any equation of the form 1. Rewrite the equation as .2. Assign as Y1.3. Graph the resulting function and identify any x-intercepts, if they exist.

Any x-intercept(s) of the graph will be a solution to the equation.

To locate the zero (x-intercept) for on a graphing calculator, enterfor Y1 and use the 2:zero option found on the same menu as the 5:intercept

option (Figure 1.79). Since some equations have more than one zero, the 2:zero optionwill ask you to “narrow down” the interval it should search, even though there is onlyone zero here. It does this by asking for a “Left Bound?”, a “Right Bound?”, and a“GUESS?” (the Guess? option can once again be bypassed). You can enter thesebounds by tracing along the graph or by inputting a chosen value, then pressing(note how the calculator posts a marker at eachbound). Figure 1.80 shows we entered as theleft bound and as the right, and the calculatorwill search for the x-intercept in this interval (notethat in general, the cursor will be either above orbelow the x-axis for the left bound, but must be on theopposite side of the x-axis for the right bound). Press-ing once more bypasses the Guess? option andlocates the x-intercept at (3, 0). The solution is x � 3(Figure 1.81).

ENTER

x � 4x � 0

ENTER

2X � 62x � 6 � 0

f 1x2 � g1x2f 1x2 � g1x2 � 0f 1x2 � g1x2,

f 1x2 � g1x2f 1x2 � g1x2 � 0f 1x2 � g1x2

Y2 � 0

2x � 6 � 03x � 1 � x � 5

x � 3


Precalculus—

WORTHY OF NOTEThe intersection-of-graphs methodfocuses on a point of intersection(a, b), and names as thesolution. The zeroes methodfocuses on the input , forwhich the output is 0 .All such values r are called thezeroes of the function. Much morewill be said about functions andtheir zeroes in later chapters.

3Y11r2 � 0 4x � r

x � a

f f f

5

10

10

�10

�10 10

10

�10

�10

Figure 1.79


cob19537_ch01_064-078.qxd 1/28/11 8:25 PM

EXAMPLE 2 � Solving an Equation Using the Zeroes MethodSolve the equation using the zeroes method.

Solution � As given, we have and . Rewriting the equationas gives

, where theexpression for g(x) is parenthesized to ensurethe equations remain equivalent. Entering

as Y1 andpressing (CALC) 2:zero producesthe screen shown in Figure 1.82, with thecalculator requesting a left bound. We caninput any x-value that is obviously to the leftof the x-intercept, or move the cursor to any position left of the x-intercept andpress (we input see Figure 1.83). The calculator then asks for a rightbound and as before we can input any x-value obviously to the right, or simplymove the cursor to any location on the opposite side of the x-axis and press (wechose , see Figure 1.84). After bypassing the Guess? option (press onceagain), the calculator locates the x-intercept at (�3.5, 0), and the solution to theoriginal equation is (Figure 1.85).


C. Solving Linear Inequalities Graphically

The intersection-of-graphs method can also be applied to solve linear inequalities. Thepoint of intersection simply becomes one of the boundary points for the solution inter-val, and is included or excluded depending on the inequality given. For the inequality

written as , it becomes clear the inequality is true for all inputs xwhere the outputs for Y1 are greater than the outputs for Y2, meaning the graph of f(x)is above the graph of g(x). A similar statement can be made for written as

.

Intersection-of-Graphs Method for Solving Inequalities

For any inequality of the form ,1. Assign f(x) as Y1 and g(x) as Y2.2. Graph both functions and identify any point(s) of intersection, if they exist.

The solution set is all real numbers x for which the graph of Y1

is above the graph of Y2.

For strict inequalities, the boundary of the solution interval is not included. A sim-ilar process is used for the inequalities , , and . Notethat we can actually draw the graphs of Y1 and Y2 differently (one more bold than the

f 1x2 � g1x2f 1x2 6 g1x2f 1x2 � g1x2

f 1x2 7 g1x2

Y1 6 Y2

f 1x2 6 g1x2

Y1 7 Y2f 1x2 7 g1x2

x � �3.5

ENTERx � �2

ENTER

x � �4,ENTER

TRACE2nd

�41X � 32 � 6 � 12X � 32

�41x � 32 � 6 � 12x � 32 � 0f 1x2 � g1x2 � 0

g1x2 � 2x � 3f 1x2 � �41x � 32 � 6

�41x � 32 � 6 � 2x � 3


Precalculus—

B. You’ve just seen howwe can solve equations usingthe x-intercept/zeroes method

10

10

�10

�10

10

10

�10

�10 10

10

�10

�10 10

10

�10

�10

Figure 1.82


cob19537_ch01_064-078.qxd 1/28/11 8:25 PM

other) to clearly tell them apart in the viewing window. This is done on the screen, by moving the cursor to the far left of the current function and pressing until a bold line appears. From the default setting, pressing one time producesthis result.

EXAMPLE 3 � Solving an Inequality Using the Inter section-of-Graphs MethodSolve using theintersection-of-graphs method.

Solution � To assist with the clarity of the solution, we setthe calculator to graph Y2 using a bolder line thanY1 (Figure 1.86). With as Y1 and

as Y2, we use the (CALC)option and select 5:intersect. Pressing threetimes serves to identify both graphs, bypassthe “Guess?” option, and display the pointof intersection (Figure 1.87). Sincethe graph of Y1 is below the graph of Y2

for all values of x to the right of, is the left boundary, with

the interval extending to positive infinity.Due to the less than or equal to inequality,we include and the solution intervalis .


D. Solving for a Specified Variable in Literal Equations

A formula is an equation that models a known relationship between two or more quan-tities. A literal equation is simply one that has two or more variables. Formulas are atype of literal equation, but not every literal equation is a formula. For example, the for-mula models the growth of money in an account earning simple interest,where A represents the total amount accumulated, P is the initial deposit, R is the annualinterest rate, and T is the number of years the money is left on deposit. To describe

, we might say the formula has been “solved for A” or that “A is writtenin terms of P, R, and T.” In some cases, before using a formula it may be convenient tosolve for one of the other variables, say P. In this case, P is called the object variable.

EXAMPLE 4 � Solving for Specified ariableGiven , write P in terms of A, R, and T (solve for P).

Solution � Since the object variable occurs in more than one term, we first apply thedistributive property.

focus on P—the object variable

factor out P

solve for P [divide by (1 � RT )]

result


A

1 � RT� P

A

1 � RT�

P11 � RT2

11 � RT2

A � P11 � RT2 A � P � PRT

A � P � PRT

A � P � PRT

A � P � PRT

x � 3�3, q 2x � �3

x � �31�3, �221Y1 � Y22

1�3, �22

ENTER

TRACE2nd2X � 4

0.513 � X2 � 5

0.513 � x2 � 5 � 2x � 4

ENTER

ENTER

Y=


Precalculus—

C. You’ve just seen howwe can solve linear inequalitiesgraphically

10

10

�10

�10

Figure 1.86

Figure 1.87

cob19537_ch01_064-078.qxd 1/28/11 8:25 PM

We solve literal equations for a specified variable using the same methods we usedfor other equations and formulas. Remember that it’s good practice to focus on the objectvariable to help guide you through the solution process, as again shown in Example 5.

EXAMPLE 5 � Solving for a Specified ariableGiven write y in terms of x (solve for y).

Solution � focus on the object variable

subtract 2x (isolate y-term)

multiply by (solve for y )

simplify and distribute


Literal Equations and General Solutions

Solving literal equations for a specified variable can help us develop the general solu-tion for an entire family of equations. This is demonstrated here for the family of linearequations written in the form A side-by-side comparison with a specificlinear equation demonstrates that identical ideas are used.

Specific Equation Literal Equation

focus on object variable

subtract constant

divide by coefficient

Of course the solution on the left would be written as and checked in theoriginal equation. On the right we now have a general formula for all equations of theform

EXAMPLE 6 � Solving Equations of the Form ax � b � c Using a General FormulaSolve using the formula just developed, and check your solution inthe original equation.

Solution � For this equation, and giving

Check:

✓


Developing a general solution for the linear equation seems to havelittle practical use. But in Section 3.2 we’ll use this idea to develop a general solutionfor quadratic equations, a result with much greater significance.

ax � b � c

�25 � �25 � �4

�24 � 1 � �25 ��24

6

61�42 � 1 � �25 ��25 � 1�12

6

6x � 1 � �25 x �c � b

a

c � �25,a � 6, b � �1,

6x � 1 � �25

ax � b � c.

x � 6

x �c � b

a x �

15 � 3

2

ax � c � b 2x � 15 � 3

ax � b � c 2x � 3 � 15

ax � b � c.

y � �23 x � 5

13 13 13y2 � 1

3 1�2x � 152 3y � �2x � 15

2x � 3y � 15

2x � 3y � 15,


Precalculus—

WORTHY OF NOTEIn Example 5, notice that in thesecond step we wrote thesubtraction of 2x as instead of For reasonsthat will become clearer as wecontinue our study, we generallywrite variable terms beforeconstant terms.

15 � 2x.�2x � 15


we can solve for a specified

variable in a formula or literal

equation

→

cob19537_ch01_064-078.qxd 1/28/11 8:26 PM

E. Using a Problem-Solving Guide

Becoming a good problem solver is an evolutionary process. Over time and with contin-ued effort, your problem-solving skills grow, as will your ability to solve a wider rangeof applications. Most good problem solvers develop the following characteristics:

• A positive attitude • Good mental-visual skills• A mastery of basic facts • Good estimation skills• Strong mental arithmetic skills • A willingness to persevere

These characteristics form a solid basis for applying what we call the Problem-SolvingGuide, which simply organizes the basic elements of good problem solving. Using thisguide will help save you from two common stumbling blocks—indecision and notknowing where to start.

Problem-Solving Guide

• Gather and organize information.Read the problem several times, forming a mental picture as you read. Highlightkey phrases. List given information, including any related formulas. Clearlyidentify what you are asked to find.

• Make the problem visual.Draw and label a diagram or create a table of values, as appropriate. This willhelp you see how different parts of the problem fit together.

• Develop an equation model.Assign a variable to represent what you are asked to find and build any relatedexpressions referred to in the problem. Write an equation model based on therelationships given in the problem. Carefully reread the problem to double-checkyour equation model.

• Use the model and given information to solve the problem.Substitute given values, then simplify and solve. State the answer in sentenceform, and check that the answer is reasonable. Include any units of measureindicated.

General Modeling Exercises

Translating word phrases into symbols is an important part of building equations frominformation given in paragraph form. Sometimes the variable occurs more than oncein the equation, because two different items in the same exercise are related. If therelationship involves a comparison of size, we often use line segments or bar graphs tomodel the relative sizes.

EXAMPLE 7 � Solving an Application Using the Problem-Solving GuideThe largest state in the United States is Alaska (AK), which covers an area that is230 square miles (mi2) more than 500 times that of the smallest state, Rhode Island(RI). If they have a combined area of 616,460 mi2, how many square miles doeseach cover?

Solution � Combined area is 616,460 mi2, AK covers gather and organize information

230 more than 500 times the area of RI. highlight any key phrases

make the problem visual


Precalculus—

Rhode Island’s area RAlaska

�230

500times

…

cob19537_ch01_064-078.qxd 1/28/11 8:26 PM

Let R represent the area of Rhode Island. assign a variable

Then represents Alaska’s area. build related expressions

write the equation model

combine like terms, subtract 230

divide by 501

Rhode Island covers an area of 1230 mi2, while Alaska covers an area of


Consecutive Integer Exercises

Exercises involving consecutive integers offer excellent practice in assigning vari-ables to unknown quantities, building related expressions, and the problem-solvingprocess in general. We sometimes work with consecutive odd integers or consecutiveeven integers as well.

EXAMPLE 8 � Solving a Problem Involving Consecutive Odd IntegersThe sum of three consecutive odd integers is 69. What are the integers?

Solution � The sum of three consecutive odd integers . . . gather/organize information

highlight any key phrases


Let n represent the smallest consecutive odd integer, assign a variable

then represents the second odd integer and build related expressions

represents the third.

In words: write the equation model

equation model

combine like terms

subtract 6

divide by 3

The odd integers are and ✓


Uniform Motion (Distance, Rate, Time) Exercises

Uniform motion problems have many variations, and it’s important to draw a gooddiagram when you get started. Recall that if speed is constant, the distance traveled isequal to the rate of speed multiplied by the time in motion: D � RT.

21 � 23 � 25 � 69n � 4 � 25.n � 21, n � 2 � 23,

n � 21 3n � 63

3n � 6 � 69 n � 1n � 22 � 1n � 42 � 69

first � second � third odd integer � 69

1n � 22 � 2 � n � 4n � 2

3 4 n n�1 n�2 n�3 n�42�1�2�3�4 10odd odd odd odd odd odd odd

�2 �2 �2 �2 �2

500112302 � 230 � 615,230 mi2.

R � 1230 501R � 616,230

R � 1500R � 2302 � 616,460 Rhode Island’s area � Alaska’s area � Total

500R � 230


Precalculus—

WORTHY OF NOTEThe number line illustration inExample 8 shows that consecutiveodd integers are two units apartand the related expressions werebuilt accordingly: n, and so on. In particular, we cannotuse n, . . . because nand are not two units apart. Ifwe know the exercise involves evenintegers instead, the same model isused, since even integers are alsotwo units apart. For consecutiveintegers, the labels are n,

and so on.n � 2,n � 1,

n � 1n � 1, n � 3,

n � 2, n � 4,

cob19537_ch01_064-078.qxd 1/28/11 8:26 PM

EXAMPLE 9 � Solving a Problem Involving Uniform MotionI live 260 mi from a popular mountain retreat. On my way there to do somemountain biking, my car had engine trouble—forcing me to bike the rest of theway. If I drove 2 hr longer than I biked and averaged 60 miles per hour driving and10 miles per hour biking, how many hours did I spend pedaling to the resort?

Solution � The sum of the two distances must be 260 mi. gather/organize information

The rates are given, and the driving time is highlight any key phrases

2 hr more than biking time. make the problem visual

Let t represent the biking time, assign a variable

then represents time spent driving. build related expressions

write the equation model

substitute for T, 60 for R, 10 for r

distribute and combine like terms

subtract 120

divide by 70

I rode my bike for , after driving .


Exercises Involving Mixtures

Mixture problems offer another opportunity to refine our problem-solving skills whileusing many elements from the problem-solving guide. They also lend themselves to avery useful mental-visual image and have many practical applications.

EXAMPLE 10 � Solving an Application Involving MixturesAs a nasal decongestant, doctors sometimes prescribe saline solutions with aconcentration between 6% and 20%. In “the old days,” pharmacists had to createdifferent mixtures, but only needed to stock these concentrations, since anypercentage in between could be obtained using a mixture. An order comes in for a15% solution. How many milliliters (mL) of the 20% solution must be mixed with10 mL of the 6% solution to obtain the desired 15% solution? Provide both analgebraic solution and a graphical solution.

t � 2 � 4 hrt � 2 hr

t � 2 70t � 140

70t � 120 � 260t � 2 601t � 22 � 10t � 260

RT � D1, rt � D2 RT � rt � 260 D1 � D2 � 260

T � t � 2

D1 � RT

D1 � D2 � Total distance

260 miles

Home

Driving Biking

ResortD2 � rt


Precalculus—

cob19537_ch01_064-078.qxd 1/28/11 8:26 PM

Algebraic Solution � Only 6% and 20% concentrations are available; gather/organize information

mix some 20% solution with 10 mL of the 6% solution. highlight any key phrases

(See Figure 1.88.)


Let x represent the amount of 20% solution, then assign a variable

represents the total amount of 15% solution. build related expressions

1st quantity times 2nd quantity times 1st�2nd quantity times

its concentration its concentration desired concentration

write equation model

distribute/simplify

subtract 0.6

subtract 0.15x

divide by 0.05

To obtain a 15% solution, 18 mL of the 20% solution must be mixed with 10 mL of the 6% solution.

Graphical Solution � Although both methods work equally well, here we elect to use the intersection-of-graphsmethod and enter as Y1 and

as Y2. Virtually all graphicalsolutions require a careful study of the context toset the viewing window prior to graphing. If 10 mLof liquid were used from each concentration, wewould have 20 mL of a 13% solution (see Worthyof Note), so more of the stronger solution isneeded. This shows that an appropriate Xmaxmight be close to 30. If all 30 mL were used,the output would be 30 , so anappropriate Ymax might be around 6 (see Figure 1.89). Using (CALC) 5:Intersect and pressing three times gives(18, 4.2) as the point of intersection, showing

mL of the stronger solution must be used(Figure 1.90).


x � 18

ENTER

TRACE2nd

10.152 � 4.5

110 � X2 10.1521010.062 � X10.22

x � 18 0.05x � 0.9

0.2x � 0.9 � 0.15x1.5 � 0.15x�0.2x�0.6110 � x2 10.152�x10.22�1010.062

10 � x

6%solution

? mL 10 mL

(10 � ?) mL15% solution

20%solution


Precalculus—

WORTHY OF NOTEFor mixture exercises, an estimateassuming equal amounts of eachliquid can be helpful. For example,assume we use 10 mL of the 6%solution and 10 mL of the 20%solution. The final concentrationwould be halfway in between,

13%. This is too low aconcentration (we need a 15%solution), so we know that morethan 10 mL of the stronger (20%)solution must be used.

6 � 202 �

30

6

0

0

Figure 1.88

Figure 1.90

Figure 1.89

E. You’ve just seen how

we can use the problem-

solving guide to solve various

problem types

cob19537_ch01_064-078.qxd 1/28/11 8:26 PM


Precalculus—

4. For the equation , we can saythat S is written in terms of and .

5. Discuss/Explain the similarities and differencesbetween the intersection and zeroes methods forsolving equations. How can the zeroes method beapplied to solving linear inequalities? Giveexamples in your discussion.

6. Discuss/Explain each of the four basic parts of theproblem-solving guide. Include a solved examplein your discussion.

S � 2�r2 � 2�rh


Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. When using the method, one sideof an equation is entered as and the other side as

on a graphing calculator. The of thepoint of is the solution of the equation.

2. To solve a linear inequality using the intersection-of-graphs method, first find the point of .The of this point is a boundary value ofthe solution interval and if the inequality is notstrict, this value is in the solution.

3. A(n) equation is an equation having ormore unknowns.

21.

22.

23.

24.

25.

26.

Solve the following inequalities using a graphingcalculator and the intersection-of-graphs method.Compare your results for Exercises 27 and 28 to those of Exercises 7 and 8.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36. 41x � 12 � 2x � 7 6 21x � 1.52

211.5x � 1.12 � 0.1x � 4x � 0.313x � 42

1.112 � x2 � 0.2 7 510.1 � 0.2x2 � 0.1x

�31x � 12 � 1 � x � 41x � 12

0.2514 � x2 � 1 6 1 � 0.5x

�0.31x � 22 � 1.1 6 0.2x � 3

413x � 52 � 2 � 312 � 4x2 � 24

2x � 13 � x2 � �215 � 2x2 � 7

�2x � 1 7 21x � 32 � 1

3x � 7 7 �21x � 12 � 10

3x � 210.2x � 1.42 � 410.8x � 0.72 � 0.2x

3x � 10.7x � 1.22 � 211.1x � 0.62 � 0.1x

312x � 12 � 1 � 2x � 11 � 4x2

21x � 22 � 1 � x � 11 � x2

0.813x � 12 � 0.2 � �2x � 3.8

�1.51x � 42 � 2.5 � 3x � 3.5


Solve the following equations using a graphing calculatorand the intersection-of-graphs method. For Exercises 7and 8, carefully sketch the graphs you designate as Y1

and Y2 by hand before using your calculator.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

Solve the following equations using a graphingcalculator and the x-intercept/zeroes method. Compareyour results for Exercises 17 and 18 to those of Exercises7 and 8.

17.

18.

19.

20. �31�3 � x2 � 4 � 2x � 5

213 � 2x2 � 5 � �3x � 3

�2x � 1 � 21x � 32 � 1

3x � 7 � �21x � 12 � 10

�13 1x � 62 � 5 � �x � 16 � 2

3x2

12 1x � 42 � 10 � x � 12 � 1

2x2

�45 x � 8 � 6

5x

13x � �2

3 x � 9

3x � 14 � x2 � 0.216 � 10x2 � 5.2

x � 13x � 12 � �0.514x � 62 � 2

0.5x � 2.5 � 0.7513 � 0.2x2 � 0.4

0.8x � 0.4 � 0.2512 � 0.4x2 � 2.8

�2x � 1 � 21x � 32 � 1

3x � 7 � �21x � 12 � 10

1.5 EXERCISES

cob19537_ch01_064-078.qxd 1/28/11 8:27 PM

Solve for the specified variable in each formula or literalequation.

37. for C (retail)

38. for P (retail)

39. for r (geometry)

40. for W (geometry)

41. for (science)

42. for (communication)

43. for h (geometry)


45. for n (sequences)


47. for P (geometry)S � B � 12PS

A �h1b1 � b22

2

Sn � naa1 � an

2b

V � 13�r2h

V � 43�r2h

P2C

P2�

P1

d2

T2P1V1

T1�

P2V2

T2

V � LWH

C � 2�r

S � P � PD

P � C � CM


Precalculus—

48. for g (physics)

49. for y

50. for y

51. for y

52. for y

53. for y

54. for y

The following equations are given in form.Solve by identifying the value of a, b, and c, then using

the formula .

55.

56.

57.

58.

59.

60. 3x � 4 � �25

7x � 13 � �27

�4x � 9 � 43

�6x � 1 � 33

7x � 5 � 47

3x � 2 � �19

x �c � b

a

ax � b � c

y � 4 � �215 1x � 102

y � 3 � �45 1x � 102

23x � 7

9y � 12

56x � 3

8y � 2

2x � 3y � 6

Ax � By � C

s � 12gt2 � vt


61. Surface area of a cylinder:

The surface area of a cylinder is given by theformula shown, where h is the height of the cylinderand r is the radius of the base. Find the height of acylinder that has a radius of 8 cm and a surface areaof 1256 cm2. Use .� � 3.14

SA � 2�r2 � 2�rh 62. Using the equation-solving process for Exercise 61 asa model, solve the formula for h.SA � 2�r2 � 2�rh

� APPLICATIONS

Solve by building an equation model and using the problem-solving guidelines as needed. Check all answers using agraphing calculator.

General Modeling Exercises

63. Two spelunkers (cave explorers) were exploringdifferent branches of an underground cavern. Thefirst was able to descend 198 ft farther than twicethe second. If the first spelunker descended 1218 ft,how far was the second spelunker able to descend?

64. The area near the joining of the Tigris andEuphrates Rivers (in modern Iraq) has often beencalled the Cradle of Civilization, since the area hasevidence of many ancient cultures. The length ofthe Euphrates River exceeds that of the Tigris by620 mi. If they have a combined length of 2880 mi,how long is each river?

65. U.S. postal regulationsrequire that a packagecan have a maximumcombined length andgirth (distance around)of 108 in. A shippingcarton is constructed sothat it has a width of 14 in., a height of 12 in.,and can be cut or folded to various lengths. What isthe maximum length that can be used?Source: www.USPS.com

W

LH

Girth

cob19537_ch01_064-078.qxd 1/28/11 8:27 PM

66. Hi-Tech Home Improvements buys a fleet ofidentical trucks that cost $32,750 each. Thecompany is allowed to depreciate the value of theirtrucks for tax purposes by $5250 per year. Ifcompany policies dictate that older trucks must be sold once their value declines to $6500,approximately how many years will they keepthese trucks?

67. The longest suspension bridge in the world is theAkashi Kaikyo (Japan) with a length of 6532 feet.Japan is also home to the Shimotsui Straightbridge. The Akashi Kaikyo bridge is 364 ft morethan twice the length of the Shimotsui bridge. Howlong is the Shimotsui bridge?Source: www.guinnessworldrecords.com

68. The Mars rover Spirit landed on January 3, 2004.Just over 1 yr later, on January 14, 2005, theHuygens probe landed on Titan (one of Saturn’smoons). At their closest approach, the distancefrom the Earth to Saturn is 29 million mi more than21 times the distance from the Earth to Mars. If thedistance to Saturn is 743 million mi, what is thedistance to Mars?

Consecutive Integer Exercises

69. Find two consecutive even integers such that thesum of twice the smaller integer plus the largerinteger is one hundred forty-six.

70. When the smaller of two consecutive integers isadded to three times the larger, the result is fifty-one. Find the smaller integer.

71. Seven times the first of two consecutive oddintegers is equal to five times the second. Find eachinteger.

72. Find three consecutive even integers where the sumof triple the first and twice the second is eight morethan four times the third.


Precalculus—

Uniform Motion Exercises

73. At 9:00 A.M., Linda leaves work on a business trip,gets on the interstate, and sets her cruise control at60 mph. At 9:30 A.M., Bruce notices she’s left herbriefcase and cell phone, and immediately startsafter her driving 75 mph. At what time will Brucecatch up with Linda?

74. A plane flying at 300 mph has a 3-hr head start ona “chase plane,” which has a speed of 800 mph.How far from the airport will the chase planeovertake the first plane?

75. Jeff had a job interview in a nearby city 72 miaway. On the first leg of the trip he drove anaverage of 30 mph through a long constructionzone, but was able to drive 60 mph after passingthrough this zone. If driving time for the trip was

hr, how long was he driving in the constructionzone?

76. At a high-school cross-country meet, Jared jogged8 mph for the first part of the race, then increasedhis speed to 12 mph for the second part. If the racewas 21 mi long and Jared finished in 2 hr, how fardid he jog at the faster pace?

Mixture Exercises

Give the total amount of the mix that results and thepercent concentration or worth of the mix.

77. Two quarts of 100% orange juice are mixed with 2 quarts of water (0% juice).

78. Ten pints of a 40% acid are combined with 10 pintsof an 80% acid.

79. Eight pounds of premium coffee beans worth $2.50per pound are mixed with 8 lb of standard beansworth $1.10 per pound.

80. A rancher mixes 50 lb of a custom feed blendcosting $1.80 per pound, with 50 lb of cheapcottonseed worth $0.60 per pound.

Solve each application of the mixture concept.

81. To help sell more of a lower grade meat, a butchermixes some premium ground beef worth $3.10/lb,with 8 lb of lower grade ground beef worth$2.05/lb. If the result was an intermediate grade ofground beef worth $2.68/lb, how much premiumground beef was used?

112

cob19537_ch01_064-078.qxd 1/28/11 8:27 PM


Precalculus—


85. Look up and read the following article. Then turnin a one page summary. “Don’t Give Up!,” WilliamH. Kraus, Mathematics Teacher, Volume 86,Number 2, February 1993: pages 110–112.

86. A chemist has four solutions of a very rare andexpensive chemical that are 15% acid (cost $120per ounce), 20% acid (cost $180 per ounce), 35%acid (cost $280 per ounce) and 45% acid (cost$359 per ounce). She requires 200 oz of a 29%acid solution. Find the combination of any two ofthese concentrations that will minimize the totalcost of the mix.

87. P, Q, R, S, T, and U represent numbers. The arrowsin the figure show the sum of the two or threenumbers added in the indicated direction

(Example: ). Find

88. Given a sphere circumscribed by acylinder, verify the volume of thesphere is that of the cylinder.2

3

S T

P

Q

U

R26

30

40 19 23 34

T � U.P � Q � R � S �Q � T � 23


89. (Appendix A.5) Solve for

90. (1.4) Solve for y, then state the slope and y-intercept of the line:

91. (Appendix A.4) Factor each expression:

a.b. x3 � 27

4x2 � 9

�6x � 7y � 42

x: 1

x � 2�

2x

�3

x2 � 2x

92. (1.3) Given , evaluate

g113 2, g1�22, and g152

g1x2 � x2 � 3x � 10

82. Knowing that the camping/hiking season hasarrived, a nutrition outlet is mixing GORP (GoodOld Raisins and Peanuts) for the anticipatedcustomers. How many pounds of peanuts worth$1.29/lb, should be mixed with 20 lb of deluxeraisins worth $1.89/lb, to obtain a mix that will sellfor $1.49/lb?

83. How many pounds of walnuts at 84¢/lb should bemixed with 20 lb of pecans at $1.20/lb to give amixture worth $1.04/lb?

84. How many pounds of cheese worth 81¢/lb must bemixed with 10 lb cheese worth $1.29/lb to make amixture worth $1.11/lb?

cob19537_ch01_064-078.qxd 1/28/11 8:27 PM

Precalculus—

Collecting and analyzing data is a tremendously important mathematical endeavor,having applications throughout business, industry, science, and government. The linkbetween classroom mathematics and real-world mathematics is called a regression, inwhich we attempt to find an equation that will act as a model for the raw data. In thissection, we focus on situations where the data is best modeled by a linear function.

A. Scatterplots and Positive/Negative Associations

In this section, we continue our study of ordered pairs and functions, but this time us-ing data collected from various sources or from observed real-world relationships. Youcan hardly pick up a newspaper or magazine without noticing it contains a large vol-ume of data presented in graphs, charts, and tables. In addition, there are many simpleexperiments or activities that enable you to collect your own data. We begin analyzingthe collected data using a scatterplot, which is simply a graph of all of the orderedpairs in a data set. Often, real data (sometimes called raw data) is not very “wellbehaved” and the points may be somewhat scattered—the reason for the name.

Positive and Negative Associations

Earlier we noted that lines with positive slope rise from left to right, while lines withnegative slope fall from left to right. We can extend this idea to the data from a scatter-plot. The data points in Example 1A seem to rise as you move from left to right, withlarger input values generally resulting in larger outputs. In this case, we say there is apositive association between the variables. If the data seems to decrease or fall as youmove left to right, we say there is a negative association.

EXAMPLE 1A � Drawing a Scatterplot and Observing AssociationsThe ratio of the federal debt to the total population is known as the per capita debt.The per capita debt of the United States is shown in the table for the odd-numberedyears from 1997 to 2007. Draw a scatterplot of the data and state whether theassociation is positive, negative, or cannot be determined.

Source: Data from the Bureau of Public Debt at www.publicdebt.treas.gov

1.6 Linear Function Models and Real Data

Deb

t ($1

000s

)

Year

1997 1999 2001 2003 2005

20

22

26

24

30

2007

28

Per Capita Year Debt ($1000s)

1997 20.0

1999 20.7

2001 20.5

2003 23.3

2005 27.6

2007 30.4

Solution � Since the amount of debt depends on the year, year is the input x and per capitadebt is the output y. Scale the x-axis from 1997 to 2007 and the y-axis from 20 to30 to comfortably fit the data (the “squiggly lines,” near the 20 and 1997 in thegraph are used to show that some initial values have been skipped). The graphindicates a positive association between the variables, meaning the debt isgenerally increasing as time goes on.


how we can:

A. Draw a scatterplot andidentify positive andnegative associations

B. Use a scatterplot toidentify linear andnonlinear associations

C. Use a scatterplot toidentify strong and weakcorrelations

D. Find a linear function thatmodels the relationshipsobserved in a set of data

E. Use linear regression tofind the line of best fit

1–79 79

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

EXAMPLE 1B � Drawing a Scatterplot and Observing AssociationsA cup of coffee is placed on a table and allowed to cool. The temperature of thecoffee is measured every 10 min and the data are shown in the table. Draw thescatterplot and state whether the association is positive, negative, or cannot bedetermined.

Solution � Since temperature depends on cooling time, time is the input x and temperatureis the output y. Scale the x-axis from 0 to 40 and the y-axis from 70 to 110 tocomfortably fit the data. As you see in the figure, there is a negative associationbetween the variables, meaning the temperature decreases over time.


B. Scatterplots and Linear/Nonlinear Associations

The data in Example 1A had a positive association, while the association in Example 1Bwas negative. But the data from these examples differ in another important way. InExample 1A, the data seem to cluster about an imaginary line. This indicates a linearequation model might be a good approximation for the data, and we say there is alinear association between the variables. The data in Example 1B could not accuratelybe modeled using a straight line, and we say the variables time and cooling tempera-ture exhibit a nonlinear association.

EXAMPLE 2 � Drawing a Scatterplot and Observing AssociationsA college professor tracked her annual salary for 2002 to 2009 and the data areshown in the table. Draw the scatterplot and determine if there is a linear ornonlinear association between the variables. Also state whether the association ispositive, negative, or cannot be determined.


Precalculus—

Elapsed Time Temperature(minutes) (ºF)

0 110

10 89

20 76

30 72

40 71

Tem

p (°

F)

Time (minutes)

50 10 15 20 3530 4025

70

80

90

100

110

120


we can draw a scatterplot and

identify positive and negative

associations

SalaryYear ($1000s)

2002 30.5

2003 31

2004 32

2005 33.2

2006 35.5

2007 39.5

2008 45.5

2009 52

Sala

ry (

$100

0s)

Appearsnonlinear

2002 2004 2008 20102006

30

35

40

45

50

55

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

Solution � Since salary earned depends on a given year, year is the input x and salary is theoutput y. Scale the x-axis from 2002 to 2010, and the y-axis from 30 to 55 tocomfortably fit the data. A line doesn’t seem to model the data very well, and theassociation appears to be nonlinear. The data rises from left to right, indicating apositive association between the variables. This makes good sense, since we expectour salaries to increase over time.


C. Identifying Strong and Weak Correlations

Using Figures 1.91 and 1.92 shown, we can make one additional observation regard-ing the data in a scatterplot. While both associations shown appear linear, the data inFigure 1.91 seems to cluster more tightly about an imaginary straight line than the datain Figure 1.92.

1–81 Section 1.6 Linear Function Models and Real Data 81

Precalculus—

B. You’ve just seen

how we can use a scatterplot

to identify linear and nonlinear

associations

1050

10

x

y

5

1050

10

x

y

5

0

Perfectnegative

correlation

Strongnegative

correlation

Moderatenegative

correlation Weaknegative

correlation

Nocorrelation

Weakpositive

correlation

Moderatepositive

correlation Strongpositive

correlation

Perfectpositive

correlation

�1.00 �1.00

Figure 1.91

Figure 1.93

Figure 1.92

We refer to this “clustering” as the “goodness of fit,” or in statistical terms, thestrength of the correlation. To quantify this fit we use a measure called the correla-tion coefficient r, which tells whether the association is positive or negative: or , and quantifies the strength of the association: . Actually, thecoefficient is given in decimal form, making . If the data points form aperfectly straight line, we say the strength of the correlation is either or 1,depending on the association. If the data points appear clustered about the line, butare scattered on either side of it, the strength of the correlation falls somewherebetween and 1, depending on how tightly/loosely they’re scattered. This is sum-marized in Figure 1.93.

�1

�10r 0 � 1

0r 0 � 100%r 6 0r 7 0

The following scatterplots help to further illustrate this idea. Figure 1.94 shows alinear and negative association between the value of a car and the age of a car, with astrong correlation. Figure 1.95 shows there is no apparent association between familyincome and the number of children, and Figure 1.96 appears to show a linear andpositive association between a man’s height and weight, with a weak correlation.

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

Until we develop a more accurate method of calculating a numerical value for thiscorrelation, the best we can do are these broad generalizations: weak correlation,strong correlation, or no correlation.

EXAMPLE 3A � High School and College GPAsMany colleges use a student’s high school GPA as a possible indication of theirfuture college GPA. Use the data from Table 1.6 (high school/college GPA) to drawa scatterplot. Then

a. Sketch a line that seems to approximate the data, meaning it has the samegeneral direction while passing through the observed “center” of the data.

b. State whether the association is positive, negative, or cannot be determined.c. Decide whether the correlation is weak or strong.


Precalculus—

Age of auto

Val

ue o

f au

to

Number of children

Fam

ily in

com

e

Male heights

Mal

e w

eigh

ts


High School CollegeGPA GPA

1.8 1.8

2.2 2.3

2.8 2.5

3.2 2.9

3.4 3.6

3.8 3.9

Col

lege

GPA

High School GPA

1.5 2.0 3.5 4.02.5 3.0

1.5

2.0

2.5

3.0

3.5

4.0

Solution � a. A line approximating the data set as a whole is shown in the figure.b. Since the line has positive slope, there is a positive association between a

student’s high school GPA and their GPA in college.c. The correlation appears strong.

EXAMPLE 3B � Natural Gas ConsumptionThe amount of natural gas consumed by homes and offices varies with the season,with the highest consumption occurring in the winter months. Use the data fromTable 1.7 (outdoor temperature/gas consumed) to draw a scatterplot. Then

a. Sketch an estimated line of best fit.b. State whether the association is positive, negative, or cannot be determined.c. Decide whether the correlation is weak or strong.

Table 1.6

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

Solution � a. We again use appropriate scales and sketch a line that seems to model the data(see figure).

b. There is a negative association between temperature and the amount of naturalgas consumed.

c. The correlation appears to be strong.


D. Linear Functions That Model Relationships

Observed in a Set of Data

Finding a linear function model for a set of data involves visually estimating andsketching a line that appears to best “fit” the data. This means answers will varyslightly, but a good, usable model can often be obtained. To find the function, we selecttwo points on this imaginary line and use either the slope-intercept form or the point-slope formula to construct the function. Points on this estimated line but not actually inthe data set can still be used to help determine the function.

EXAMPLE 4 � Finding a Linear Function to Model the Relationship Betw een GPAsUse the scatterplot from Example 3A to find a function model for the line a collegemight use to project an incoming student’s future GPA.

Solution � Any two points on or near the estimated best-fit line can be used to help determinethe linear function (see the figure in Example 3A). For the slope, it’s best to picktwo points that are some distance apart, as this tends to improve the accuracy of themodel. It appears (1.8, 1.8) and (3.8, 3.9) are both on the line, giving

slope formula

substitute (x2, y2) for S (3.8, 3.9), (x1, y1) for S (1.8, 1.8)

slope

point-slope form

substitute 1.05 for m, (1.8, 1.8) for (x1, y1)

distribute

add 1.8 (solve for y )

One possible function model for this data is . Slightly differentfunctions may be obtained, depending on the points chosen.


f 1x2 � 1.05x � 0.09

y � 1.05x � 0.09 y � 1.8 � 1.05x � 1.89 y � 1.8 � 1.051x � 1.82 y � y1 � m1x � x12

� 1.05

�3.9 � 1.8

3.8 � 1.8

m �y2 � y1

x2 � x1


Precalculus—

Outdoor Gas ConsumedTemperature (ºF) (cubic feet)

30 800

40 620

50 570

60 400

70 290

80 220

Gas

con

sum

ptio

n (f

t3 )

Outdoor temperature (�F)

300 40 70 8050 60

200

400

600

800


we can use a scatterplot to

identify strong and weak

correlations

Table 1.7

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

The function from Example 4 predicts that a student with a high school GPA of 3.2will have a college GPA of almost 3.3: , yet the datagives an actual value of only 2.9. When working with data and function models, weshould expect some variation when the two are compared, especially if the correlationis weak.

Applications of data analysis can be found in virtually all fields of study. In Exam-ple 5 we apply these ideas to an Olympic swimming event.

EXAMPLE 5 � Finding a Linear Function to Model the Relationship (Y ear, Gold Medal Times)The men’s 400-m freestyle times (gold medal times—to the nearest second) forthe 1976 through 2008 Olympics are given in Table 1.8 (1900S0). Let the year bethe input x, and winning race time be the output y. Based on the data, draw ascatterplot and answer the following questions.

a. Does the association appear linear or nonlinear?b. Is the association positive or negative?c. Classify the correlation as weak or strong.d. Find a function model that approximates the data, then use it to predict the

winning time for the 2012 Olympics.

f 13.22 � 1.0513.22 � 0.09 � 3.3


Precalculus—

WORTHY OF NOTESometimes it helps to draw a straightline on an overhead transparency,then lay it over the scatterplot. Byshifting the transparency up anddown, and rotating it left and right,the line can more accurately beplaced so that it’s centered amongand through the data.

Year (x) Time ( y)(1900S0) (sec)

76 232

80 231

84 231

88 227

92 225

96 228

100 221

104 223

108 223

Table 1.8

Tim

e (s

ec)

Year

76 84 10892 100

210

218

226

234

242

Solution � Begin by choosing appropriate scales for the axes. The x-axis (year) could bescaled from 76 to 112, and the y-axis (swim time) from 210 to 246. This will allowfor a “frame” around the data. After plotting the points, we obtain the scatterplotshown in the figure.

a. The association appears to be linear.b. The association is negative, showing that finishing times tend to decrease

over time.c. There is a moderate to strong correlation.d. The points (76, 232) and (104, 223) appear to be on a line approximating the

data, and we select these to develop our equation model.

slope formula

,

slope (rounded to tenths)

point-slope form

distribute

add 232 (solve for y ) y � �0.32x � 256.32 y � 232 � �0.32x � 24.32 y � 232 � �0.321x � 762

� �0.32

1x2, y22S 1104, 22321x1, y12S 176, 2322 �223 � 232

104 � 76

m �y2 � y1

x2 � x1

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

One model for this data is . Based on this model, thepredicted time for the 2012 Olympics would be

function model

substitute 112 for x (2012)

result

In 2012 the winning time is projected to be about 220.5 sec.


As a reminder, great care should be taken when using equation models obtainedfrom real data. It would be foolish to assume that in the year 2700, swim times for the400-m freestyle would be near 0 sec—even though that’s what the model predicts for

. Most function models are limited by numerous constraining factors, and datacollected over a much longer period of time might even be better approximated usinga nonlinear model.

E. Linear Regression and the Line of Best Fit

There is actually a sophisticated method for calculating the equation of a line that bestfits a data set, called the regression line. The method minimizes the vertical distancebetween all data points and the line itself, making it the unique line of best fit. Mostgraphing calculators have the ability to perform this calculation quickly. The processinvolves these steps: (1) clearing old data, (2) entering new data, (3) displaying thedata, (4) calculating the regression line, and (5) displaying and using the regressionline. We’ll illustrate by finding the regression line for the data shown in Table 1.8 inExample 5, which gives the men’s 400-m freestyle gold medal times (in seconds) forthe 1976 through the 2008 Olympics, with 1900S0.

Step 1: Clear Old Data

To prepare for the new data, we first clear out any old data. Press the key andselect option 4:ClrList. This places the ClrList command on the home screen. Wetell the calculator which lists to clear by pressing 1 to indicate List1 (L1), thenenter a comma using the key, and continue entering other lists we want to clear:

2 3 will clear List1 (L1), List2 (L2), and List3 (L3).

Step 2: Enter New Data

Press the key and select option 1:Edit. Move thecursor to the first position of List1, and simply enterthe data from the first column of Table 1.8 in order: 76 80 84 , and so on. Then use the right arrow to navigate to List2, and enter the data fromthe second column: 232 231 231 , and so on.When finished, you should obtain the screen shown inFigure 1.97.

Step 3: Display the Data

With the data held in these lists, we can now displaythe related ordered pairs on the coordinate grid. Firstpress the key and any existing equations.Then press to access the “STATPLOTS”screen. With the cursor on 1:Plot1, press and besure the options shown in Figure 1.98 are highlighted.If you need to make any changes, navigate the cursor to

ENTER

Y=2nd

CLEARY=

ENTERENTERENTER

ENTERENTERENTER

STAT

ENTER2nd,2nd

,

2nd

STAT

x � 800

� 220.48 f 11122 � �0.3211122 � 256.32

f 1x2 � �0.32x � 256.32

y � �0.32x � 256.32


Precalculus—


we can find a linear function

that models relationships

observed in a set of data

Figure 1.97

Figure 1.98WORTHY OF NOTEAs a rule of thumb, the tick marksfor Xscl can be set by mentally

estimating and

using a convenient number in theneighborhood of the result (thesame goes for Yscl). As analternative to manually setting thewindow, the 9:ZoomStatfeature can be used.

ZOOM

�Xmax� � �Xmin�

10

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM

the desired option and press . Note the data inL1 ranges from 76 to 108, while the data in L2ranges from 221 to 232. This means an appro-priate viewing window might be [70, 120] for thex-values, and [210, 240] for the y-values. Pressthe key and set up the window accordingly.After you’re finished, pressing the keyshould produce the graph shown in Figure 1.99.

Step 4: Calculate the Regression Equation

To have the calculator compute the regressionequation, press the and keys to movethe cursor over to the CALC options (see Fig-ure 1.100). Since it appears the data is best mod-eled by a linear equation, we choose option4:LinReg( ). Pressing the number 4places this option on the home screen, andpressing computes the values of a and b (thecalculator automatically uses the values in L1and L2 unless instructed otherwise). Rounded tohundredths, the linear regression model is

(Figure 1.101).

Step 5: Display and Use the Results

Although graphing calculators have the abilityto paste the regression equation directly intoY1 on the screen, for now we’ll enter

by hand. Afterward,pressing the key will plot the data points (ifPlot1 is still active) and graph the line. Yourdisplay screen should now look like the one inFigure 1.102. The regression line is the best esti-mator for the set of data as a whole, but there willstill be some difference between the values itgenerates and the values from the set of raw data(the output in Figure 1.102 shows the estimatedtime for the 2000 Olympics was about 224 sec,when actually it was the year Ian Thorpe ofAustralia set a world record of 221 sec).

EXAMPLE 6 � Using Regression to Model Employee PerformanceRiverside Electronics reviews employee performancesemiannually, and awards increases in their hourly rateof pay based on the review. The table shows Thomas’hourly wage for the last 4 yr (eight reviews). Find theregression equation for the data and use it to project hishourly wage for the year 2011, after his fourteenthreview.

Solution � Following the prescribed sequence produces theequation For we obtain

or a wage of $15.81. According to thismodel, Thomas will be earning $15.81 per hour in 2011.


y�0.48 1142�9.09x � 14y � 0.48x � 9.09.

GRAPH

Y1 � �0.33x � 257.06Y=

y � �0.33x � 257.06

ENTER

ax � b

STAT

GRAPH

WINDOW

ENTER


Precalculus—

210

240

70 120

210

240

70 120

Review (x) Wage ( y)

(2004) 1 $9.58

2 $9.75

(2005) 3 $10.54

4 $11.41

(2006) 5 $11.60

6 $11.91

(2007) 7 $12.11

8 $13.02

Figure 1.99

Figure 1.100

Figure 1.101

Figure 1.102

cob19537_ch01_079-093.qxd 2/1/11 10:37 AM

With each linear regression, the calculator can be set to compute a correlation co-efficient that is a measure of how well the equation fits the data (see Subsection C). Todisplay this “r-value” use (CATALOG) and activate DiagnosticOn. Fig-ure 1.103 shows a scatterplot with perfect negative correlation and notice alldata points are on the line. Figure 1.104 shows a strong positive correlation of the data from Example 6. See Exercise 35.

1r � 0.9821r � �12

02nd


Precalculus—

E. You’ve just seen how

we can use a linear regression

to find the line of best fit


2. If the data points seem to form a curved pattern orif no pattern is apparent, the data is said to have a

association.



1. When the ordered pairs from a set of data areplotted on a coordinate grid, the result is called a

.

1.6 EXERCISES

3. If the data points seems to cluster along animaginary line, the data is said to have a association.

4. If the pattern of data points seems to increase asthey are viewed left to right, the data is said to havea association.

5. Compare/Contrast: One scatterplot is linear, with aweak and positive association. Another is linear,with a strong and negative association. Give awritten description of each scatterplot.

6. Discuss/Explain how this is possible: Workingfrom the same scatterplot, Demetrius obtained theequation for his equation model,while Jessie got the equation .y � �0.59x � 42

y � �0.64x � 44

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM


Precalculus—


7. For mail with a high priority,“Express Mail” offers next daydelivery by 12:00 noon to mostdestinations, 365 days of theyear. The service was firstoffered by the U.S. PostalService in the early 1980s andhas been growing in use eversince. The cost of the service(in cents) for selected years isshown in the table. (a) Draw ascatterplot of the data, then (b) decide if the association ispositive, negative, or cannot be determined.Source: 2004 Statistical Abstract of the United States; USPS.com

8. After the Surgeon General’sfirst warning in 1964,cigarette consumption began asteady decline as advertisingwas banned from televisionand radio, and publicawareness of the dangers ofcigarette smoking grew. Thepercentage of the U.S. adultpopulation who consideredthemselves smokers is shownin the table for selected years.(a) Draw a scatterplot of thedata, then (b) decide if theassociation is positive, negative, or cannot bedetermined.Source: 1998 Wall Street Journal Almanac and 2009 Statistical Abstract of theUnited States, Table 1299

9. Since the 1970s women havemade tremendous gains in thepolitical arena, with more andmore female candidates runningfor, and winning seats in theU.S. Senate and U.S. Congress.The number of womencandidates for the U.S. Congressis shown in the table for selectedyears. (a) Draw a scatterplot of the data, (b) decide ifthe association is linear or nonlinear and (c) if theassociation is positive, negative, or cannot bedetermined.Source: Center for American Women and Politics atwww.cawp.rutgers.edu/Facts3.html

10. The number of shares traded on the New YorkStock Exchange experienced dramatic change inthe 1990s as more and more individual investorsgained access to the stock market via the Internet

and online brokerage houses.The volume is shown in thetable for 2002, and the oddnumbered years from 1991 to2001 (in billions of shares). (a) Draw a scatterplot of thedata, (b) decide if theassociation is linear ornonlinear, and (c) if theassociation is positive,negative, or cannot be determined.Source: 2000 and 2004 Statistical Abstract of the United States, Table 1202

The data sets in Exercises 11 and 12 are known to be linear.

11. The total value of the goodsand services produced by anation is called its grossdomestic product or GDP. TheGDP per capita is the ratio ofthe GDP for a given year to thepopulation that year, and is one of many indicators ofeconomic health. The GDP percapita (in $1000s) for theUnited States is shown in thetable for selected years. (a) Draw a scatterplot usingscales that appropriately fit the data, then sketch anestimated line of best fit, (b) decide if theassociation is positive or negative, then (c) decidewhether the correlation is weak or strong.Source: 2004 Statistical Abstract of the United States, Tables 2 and 641

12. Real estate brokers carefullytrack sales of new homeslooking for trends in location,price, size, and other factors.The table relates the averageselling price within a pricerange (homes in the $120,000to $140,000 range arerepresented by the $130,000figure), to the number of newhomes sold by Homestead Realty in 2004. (a) Draw a scatterplot using scales thatappropriately fit the data, then sketch an estimatedline of best fit, (b) decide if the association ispositive or negative, then (c) decide whether thecorrelation is weak or strong.

x y

1981 935

1985 1075

1988 1200

1991 1395

1995 1500

1999 1575

2002 1785

2010 1830

x y

1965 42.4

1974 37.1

1979 33.5

1985 29.9

1990 25.3

1995 24.6

2000 23.1

2002 22.4

2005 16.9

x y

1972 32

1978 46

1984 65

1992 106

1998 121

2004 141

x y

1991 46

1993 67

1995 88

1997 134

1999 206

2001 311

2002 369

x( ) y

0 5.1

5 7.6

10 12.3

15 17.7

20 23.3

25 27.7

30 35.0

33 37.8

1970 S 0

Price Sales

130’s 126

150’s 95

170’s 103

190’s 75

210’s 44

230’s 59

250’s 21

cob19537_ch01_079-093.qxd 1/28/11 8:29 PM


Precalculus—

For the scatterplots given: (a) Arrange them in orderfrom the weakest to the strongest correlation, (b) sketcha line that seems to approximate the data, (c) statewhether the association is positive, negative, or cannotbe determined, and (d) choose two points on (or near)the line and use them to approximate its slope (roundedto one decimal place).

13. A. B.

C. D.

14. A. B.

C. D.

For the scatterplots given, (a) determine whether a linearor nonlinear model would seem more appropriate. (b) Determine if the association is positive or negative. (c) Classify the correlation as weak or strong. (d) Iflinear, sketch a line that seems to approximate the dataand choose two points on the line and use them toapproximate its slope.

15. 16.

17. 18.

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

19. 20.

21. In most areas of the country,law enforcement has becomea major concern. The numberof law enforcement officersemployed by the federalgovernment and having theauthority to carry firearmsand make arrests is shown inthe table for selected years.(a) Draw a scatterplot using scales thatappropriately fit the data and sketch an estimatedline of best fit and (b) decide if the association ispositive or negative. (c) Choose two points on ornear the estimated line of best fit, and use them tofind a function model and predict the number offederal law enforcement officers in 1995 and theprojected number for 2011. Answers may vary.Source: U.S. Bureau of Justice, Statistics at www.ojp.usdoj.gov/bjs/fedle.htm

22. Due to atmospheric pressure,the temperature at whichwater will boil variespredictably with the altitude.Using special equipmentdesigned to duplicateatmospheric pressure, a labexperiment is set up to studythis relationship for altitudesup to 8000 ft. The set of datacollected is shown in the table,with the boiling temperature yin degrees Fahrenheit,depending on the altitude x infeet. (a) Draw a scatterplot using scales thatappropriately fit the data and sketch an estimatedline of best fit, (b) decide if the association ispositive or negative. (c) Choose two points on ornear the estimated line of best fit, and use them tofind a function model and predict the boiling pointof water on the summit of Mt. Hood in WashingtonState (11,239 ft height), and along the shore of theDead Sea (approximately 1312 ft below sea level).Answers may vary.

23. For the data given in Exercise 11 (Gross DomesticProduct per Capita), choose two points on or nearthe line you sketched and use them to find a functionmodel for the data. Based on this model, what is theprojected GDP per capita for the year 2010?

1 x

y6055504540353025

2 3 4 5 6 7 8 901 x

y6055504540353025

2 3 4 5 6 7 8 90

x y(1990→0) (1000s)

3 68.8

6 74.5

8 83.1

10 88.5

14 93.4

x y

213.8

0 212.0

1000 210.2

2000 208.4

3000 206.5

4000 204.7

5000 202.9

6000 201.0

7000 199.2

8000 197.4

�1000

cob19537_ch01_079-093.qxd 1/28/11 8:30 PM


Precalculus—

24. For the data given in Exercise 12 (Sales by RealEstate Brokers), choose two points on or near theline you sketched and use them to find a function

model for the data. Based on this model, how manysales can be expected for homes costing $275,000?$300,000?


25. Circumference of a Circle: : Theformula for the circumference of a circle can bewritten as a function of C in terms of .(a) Set up a table of values for through 6 anddraw a scatterplot of the data. (b) Is the associationpositive or negative? Why? (c) What can you sayabout the strength of the correlation? (d) Sketch aline that “approximates” the data. What can you sayabout the slope of this line?

26. Volume of a Cylinder: : As part of aproject, students cut a long piece of PVC pipe witha diameter of 10 cm into sections that are 5, 10,15, 20, and 25 cm long. The bottom of each is thenmade watertight and each section is filled to the

V � �r2h

r � 1r: C1r2 � 2�r

C � 2�r brim with water. The volumeis then measured using a flaskmarked in cm3 and the resultscollected into the tableshown. (a) Draw a scatterplotof the data. (b) Is theassociation positive ornegative? Why? (c) What canyou say about the strength ofthe correlation? (d) Would thecorrelation here be stronger or weaker than thecorrelation in Exercise 25? Why? (e) Run a linearregression to verify your response.

� APPLICATIONS

Use the regression capabilities of a graphing calculator to complete Exercises 27 through 34.

28. Patent applications: Every year the U.S. Patentand Trademark Office (USPTO) receives thousandsof applications from scientists and inventors. Thetable given shows the number of applicationsreceived for the odd years from 1993 to 2003( ). Use the data to (a) draw the scatterplot;(b) determine whether the association is linear ornonlinear; (c) determine whether the association ispositive or negative; and (d) find the regressionequation and use it to predict the number ofapplications that will be received in 2011.Source: United States Patent and Trademark Office at www.uspto.gov/web

1990 S0

Height Volume(cm) (cm3)

5 380

10 800

15 1190

20 1550

25 1955

Height (x) Wingspan ( y)

61 60.5

61.5 62.5

54.5 54.5

73 71.5

67.5 66

51 50.75

57.5 54

52 51.5

27. Height versus wingspan: Leonardo da Vinci’sfamous diagram is an illustration of how the humanbody comes in predictable proportions. One suchcomparison is a person’s height to their wingspan(the maximum distance from the outstretched tip of one middle finger to the other). Carefulmeasurements were taken on eight students and theset of data is shown here. Using the data, (a) drawthe scatterplot; (b) determine whether theassociation is linear or nonlinear; (c) determinewhether the association is positive or negative; and(d) find the regression equation and use it to predictthe wingspan of a student with a height of 65 in.

Year Applications(1990S0) (1000s)

3 188.0

5 236.7

7 237.0

9 278.3

11 344.7

13 355.4

cob19537_ch01_079-093.qxd 1/28/11 8:30 PM

29. Patents issued: Anincrease in thenumber of patentapplications (seeExercise 28), typicallybrings an increase inthe number of patentsissued, though manyapplications aredenied due toimproper filing, lackof scientific support, and other reasons. The tablegiven shows the number of patents issued for theodd years from 1993 to 2003 ( ). Use thedata to (a) draw the scatterplot; (b) determinewhether the association is linear or nonlinear; (c) determine whether the association is positive ornegative; and (d) find the regression equation anduse it to predict the number of applications thatwill be approved in 2011. Which is increasingfaster, the number of patent applications or thenumber of patents issued? How can you tell for sure?Source: United States Patent and Trademark Office at www.uspto.gov/web

1999 S 0

30. High jump records: In thesport of track and field, thehigh jumper is an unusualathlete. They seem to defygravity as they launch theirbodies over the high bar.The winning height at thesummer Olympics (to thenearest unit) has steadilyincreased over time, asshown in the table forselected years. Using thedata, (a) draw thescatterplot, (b) determinewhether the association is linear or nonlinear, (c) determine whether theassociation is positive ornegative, and (d) find theregression equation using

corresponding to 1900 and predict thewinning height for the 2004 and 2008 Olympics.How close did the model come to the actualheights?Source: athens2004.com

t � 0

31. Females/males in the workforce: Over the last 4 decades, the percentage of the female populationin the workforce has been increasing at a fairlysteady rate. At the same time, the percentage of themale population in the workforce has been declining.The set of data is shown in the tables. Using the data,(a) draw scatterplots for both data sets, (b) determinewhether the associations are linear or nonlinear, (c) determine whether the associations are positive ornegative, and (d) determine if the percentage offemales in the workforce is increasing faster than thepercentage of males is decreasing. Discuss/Explainhow you can tell for sure.Source: 1998 Wall Street Journal Almanac, p. 316

32. Height versus male shoesize: While it seemsreasonable that tallerpeople should have largerfeet, there is actually awide variation in therelationship betweenheight and shoe size. Thedata in the table show theheight (in inches)compared to the shoe sizeworn for a random sampleof 12 male chemistrystudents. Using the data,(a) draw the scatterplot, (b) determine whether theassociation is linear ornonlinear, (c) determinewhether the association is positive or negative, and(d) find the regression equation and use it to predictthe shoe size of a man 80 in. tall and another that is60 in. tall. Note that the heights of these two menfall outside of the range of our data set (seecomment after Example 5 on page 85).


Precalculus—

Exercise 31 (women) Exercise 31 (men)

Year Patents(1990S0) (1000s)

3 107.3

5 114.2

7 122.9

9 159.2

11 187.8

13 189.6

Year Height (1900S0) in.

0 75

12 76

24 78

36 80

56 84

68 88

80 93

88 94

92 92

96 94

100 93

104

108

Year (x)(1950S0) Percent

5 36

10 38

15 39

20 43

25 46

30 52

35 55

40 58

45 59

50 60

Year (x)(1950S0) Percent

5 85

10 83

15 81

20 80

25 78

30 77

35 76

40 76

45 75

50 73

Height Shoe Size

66 8

69 10

72 9

75 14

74 12

73 10.5

71 10

69.5 11.5

66.5 8.5

73 11

75 14

65.5 9

cob19537_ch01_079-093.qxd 1/28/11 8:30 PM

33. Plastic money: The totalamount of businesstransacted using creditcards has been changingrapidly over the last 15 to20 years. The totalvolume (in billions ofdollars) is shown in thetable for selected years.(a) Use a graphingcalculator to draw ascatterplot of the dataand decide whether theassociation is linear or nonlinear. (b) Calculate aregression equation with corresponding to1990 and display the scatterplot and graph on thesame screen. (c) According to the equation model,how many billions of dollars were transacted in2003? How much will be transacted in the year 2011?Source: Statistical Abstract of the United States, various years

x � 0


Precalculus—

34. Sales of hybridcars: Since theirmass introductionnear the turn ofthe century, thesales of hybridcars in the UnitedStates grewsteadily until late 2007, whenthe price ofgasoline beganshowing signs ofweakening and eventually dipped below $3.00/gal.Estimates for the annual sales of hybrid cars aregiven in the table for the years 2002 through 2009

. (a) Use a graphing calculator to draw ascatterplot of the data and decide if the associationis linear or nonlinear. (b) If linear, calculate aregression model for the data and display thescatterplot and data on the same screen. (c) Assuming that sales of hybrid cars recover, how many hybrids does the model project will besold in the year 2012?Source: http://www.hybridcar.com

12000 S02


35. It can be very misleading torely on the correlationcoefficient alone whenselecting a regression model.To illustrate, (a) run a linearregression on the data setgiven (without doing ascatterplot), and note thestrength of the correlation (the correlation coefficient).(b) Now run a quadraticregression ( CALC 5:QuadReg) and note thestrength of the correlation. (c) What do younotice? What factors other than the correlationcoefficient must be taken into account whenchoosing a form of regression?

STAT

36. In his book Gulliver’s Travels, Jonathan Swiftdescribes how the Lilliputians were able tomeasure Gulliver for new clothes, even though hewas a giant compared to them. According to thetext, “Then they measured my right thumb, anddesired no more . . . for by mathematicalcomputation, once around the thumb is twicearound the wrist, and so on to the neck and waist.”Is it true that once around the neck is twice aroundthe waist? Find at least 10 willing subjects and takemeasurements of their necks and waists inmillimeters. Arrange the data in ordered pair form(circumference of neck, circumference of waist).Draw the scatterplot for this data. Does theassociation appear to be linear? Find the equationof the best fit line for this set of data. What is theslope of this line? Is the slope near ?m � 2

x(1990S0) y

1 481

2 539

4 731

7 1080

8 1157

9 1291

10 1458

12 1638

x y

50 67

100 125

150 145

200 275

250 370

300 550

350 600

Year Hybrid Sales (2000S0) (in thousands)

2 35

3 48

4 88

5 200

6 250

7 352

8 313

9 292

cob19537_ch01_079-093.qxd 1/28/11 8:30 PM

1–93 Making Connections 93

Precalculus—

MAKING CONNECTIONS

Making Connections: Graphically , Symbolically , Numerically, and V erballyEight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs.

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

1. ____

2. ____

3. ____

4. ____

5. ____

6. ____ ,

7. ____

8. ____ m �2

3

m � �2

b 6 0m 6 0

y � �2

x � �1

b 6 0m 7 0,

y � �x � 1

y �1

3x � 1

(a) (b) (c) (d)

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y(e) (f) (g) (h)

9. ____

10. ____ ,

11. ____ for

12. ____

13. ____ for

14. ____ m is zero

15. ____ function is increasing, y-intercept is negative

16. ____ function is decreasing, y-intercept is negative

x � 31, q 2f 1x2 � 0

x � 3

x � 3�3, q 2f 1x2 � 0

f 142 � 3f 1�42 � 3

f 112 � 0f 1�32 � 4,


37. (1.3) Is the graph shown here, the graph of afunction? Discuss why or why not.

38. (Appendix A.2/A.3)

Determine the area ofthe figure shown

39. (1.5) Solve for r:

40. (Appendix A.3) Solve for w (if possible):�216w2 � 52 � 1 � 7w � 413w2 � 12

A � P � Prt

1A � LW, A � �r22.24 cm

18 cm

cob19537_ch01_079-093.qxd 1/28/11 8:30 PM

Precalculus—

SECTION 1.1 Rectangular Coor dinates; Graphing Cir cles and Other Relations

KEY CONCEPTS• A relation is a collection of ordered pairs (x, y) and can be stated as a set or in equation form.

• As a set of ordered pairs, we say the relation is pointwise-defined. The domain of the relation is the set of all firstcoordinates, and the range is the set of all corresponding second coordinates.

• A relation can be expressed in mapping notation , indicating an element from the domain is mapped to(corresponds to or is associated with) an element from the range.

• The graph of a relation in equation form is the set of all ordered pairs (x, y) that satisfy the equation. We plot asufficient number of points and connect them with a straight line or smooth curve, depending on the patternformed.

• The x- and y-variables of linear equations and their graphs have implied exponents of 1.

• With a relation entered on the screen, a graphing calculator can provide a table of ordered pairs and therelated graph.

• The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is .

• The distance between the points (x1, y1) and (x2, y2) is .

• The equation of a circle centered at (h, k) with radius r is .

EXERCISES1. Represent the relation in mapping notation, then state the domain and range.

2. Graph the relation by completing the table, then state the domain and range of the relation.

3. Use a graphing calculator to graph the relation . Then use the TABLE feature to determine thevalue of y when , and the value(s) of x when , and write the results in ordered pair form.

Mr. Northeast and Mr. Southwest live in Coordinate County and are good friends. Mr. Northeast lives at 19 East and25 North or (19, 25), while Mr. Southwest lives at 14 West and 31 South or ( ). If the streets in CoordinateCounty are laid out in one mile squares,

4. Use the distance formula to find how far apart they live.

5. If they agree to meet halfway between their homes, what are the coordinates of their meeting place?

6. Sketch the graph of .

7. Sketch the graph of .

8. Find an equation of the circle whose diameter has the endpoints ( ) and (0, 4).�3, 0

x2 � y2 � 6x � 4y � 9 � 0

x2 � y2 � 16

�14, �31

y � 0x � 05x � 3y � �15

y � 225 � x2

5 1�7, 32, 1�4, �22, 15, 12, 1�7, 02, 13, �22, 10, 82 6

1x � h22 � 1y � k22 � r2

d � 21x2 � x122 � 1y2 � y12

2

ax1 � x2

2,

y1 � y2

2b

Y=

x S y

SUMMAR Y AND CONCEPT REVIEW


x y

0

245

�2

�4�5

cob19537_ch01_094-102.qxd 1/28/11 8:54 PM

1–95 Summary and Concept Review 95

Precalculus—

SECTION 1.2 Linear Equations and Rates of Change

KEY CONCEPTS• A linear equation can be written in the form , where a and b are not simultaneously equal to 0.

• The slope of the line through (x1, y1) and (x2, y2) is , where .

• Other designations for slope are .

• Lines with positive slope rise from left to right; lines with negative slope ( ) fall from left to right.

• The equation of a horizontal line is ; the slope is .

• The equation of a vertical line is ; the slope is undefined.

• Lines can be graphed using the intercept method. First determine (x, 0) (substitute 0 for y and solve for x), then (0, y) (substitute 0 for x and solve for y). Then draw a straight line through these points.

• Parallel lines have equal slopes ( ); perpendicular lines have slopes that are negative reciprocals

( or ).

EXERCISES9. Plot the points and determine the slope, then use the ratio to find an additional point on the line:

a. and and b. (3, 4) and

10. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicular, or neither:

a. L1: and (0, 6); L2: (1, 8) and (0, 5)b. L1: (1, 10) and : L2: and

11. Graph each equation by plotting points: (a) (b)

12. Find the intercepts for each line and sketch the graph: (a) (b)

13. Identify each line as either horizontal, vertical, or neither, and graph each line.

a. b. c.

14. Determine if the triangle with the vertices given is a right triangle: (7, 2), (0, 16).

15. Find the slope and y-intercept of the line shown and discuss the slope ratio in this context.

SECTION 1.3 Functions, Function Notation, and the Graph of a Function

KEY CONCEPTS• A function is a relation, rule, or equation that pairs each element from the domain with exactly one element of the

range.

• The vertical line test says that if every vertical line crosses the graph of a relation in at most one point, the relationis a function.

• The domain and range can be stated using set notation, graphed on a number line, or expressed using intervalnotation.

82 4 6

2

4

6

8

10

Haw

k po

pula

tion

(100

s)

Rodent population (1000s)

x

y

1�5, �42,

2y � x � 5y � �4x � 5

y � 43x � 2.2x � 3y � 6

y � �32x � 1.y � 3x � 2

11, �321�2, �121�1, 721�2, 02

1�6, 12.15, �221�4, 32

¢y

¢x�

riserun

m1# m2 � �1m1 � �

1m2

m1 � m2

x � hm � 0y � k

m 6 01m 7 02

m �riserun

�change in y

change in x�

¢y

¢x�

vertical change

horizontal change

x1 � x2m �y2 � y1

x2 � x1

ax � by � c

cob19537_ch01_094-102.qxd 1/28/11 8:54 PM

• On a graph, vertical boundary lines can be used to identify the domain, or the set of “allowable inputs” for afunction.

• On a graph, horizontal boundary lines can be used to identify the range, or the set of y-values (outputs) generatedby the function.

• When a function is stated as an equation, the implied domain is the set of x-values that yield real number outputs.

• x-values that cause a denominator of zero or that cause the radicand of a square root expression to be negativemust be excluded from the domain.

• The phrase “y is a function of x,” is written as . This notation enables us to summarize the three mostimportant aspects of a function with a single expression (input, sequence of operations, output).

EXERCISES16. State the implied domain of each function:

a. b.

17. Determine , and h(3a) for

18. Determine if the mapping given represents a function. If not, explain how the definition of a function is violated.

19. For the graph of each function shown, (a) state the domain and range, (b) find the value of f(2), and (c) determinethe value(s) of x for which .

I. II. III.

SECTION 1.4 Linear Functions, Special For ms, and Mor e on Rates of Change

KEY CONCEPTS• The equation of a nonvertical line in slope-intercept form is or . The slope of the line

is m and the y-intercept is (0, b).

• To graph a line given its equation in slope-intercept form, plot the y-intercept, then use the slope ratio tofind a second point, and draw a line through these points.

• If the slope m and a point (x1, y1) on the line are known, the equation of the line can be written in point-slopeform: .y � y1 � m1x � x12

m �¢y

¢x

f 1x2 � mx � by � mx � b

5�5

�5

5

x

y

5�5

�5

5

x

y

5�5

�5

5

x

y

f 1x2 � 1

Apollo

Jupiter

Ares

Neptune

Mercury

Venus

Ceres

Mars

messenger

war

craftsman

love and beauty

music and healing

oceans

all things

agriculture

Mythological deities

Primaryconcern

h1x2 � 2x2 � 3x.h1�22, h1�23 2

g1x2 �x � 4

x2 � x � 6f 1x2 � 24x � 5

y � f 1x2


Precalculus—

cob19537_ch01_094-102.qxd 1/28/11 8:54 PM

• A secant line is the straight line drawn through two points on a nonlinear graph.

• The notation literally means the quantity measured along the y-axis is changing with respect to changes

in the quantity measured along the x-axis.

EXERCISES20. Write each equation in slope-intercept form, then identify the slope and y-intercept.

a. b.21. Graph each equation using the slope and y-intercept.

a. b.22. Graph the line with the given slope through the given point.

a. b.23. What are the equations of the horizontal line and the vertical line passing through

Which line is the point (7, 5) on?

24. Find the equation of the line passing through (1, 2) and Write your finalanswer in slope-intercept form.

25. Find the equation for the line that is parallel to and passes throughthe point (3, 4). Write your final answer in slope-intercept form.

26. Determine the slope and y-intercept of the line shown. Then write the equation of

the line in slope-intercept form and interpret the slope ratio in the contextof this exercise.

27. For the graph given, (a) find the equation of the line in point-slope form, (b) use theequation to predict the x- and y-intercepts, (c) write the equation in slope-interceptform, and (d) find y when , and the value of x for which

SECTION 1.5 Solving Equations and Inequalities Graphically; Formulas and Pr oblem Solving

KEY CONCEPTS• To use the intersection-of-graphs method for solving equations, assign the left-hand expression as Y1 and the

right-hand as Y2. The solution(s) of the original equation are the x-coordinate(s) of the point(s) of intersection of the graphs of Y1 and Y2.

• When an equation is written in the form , the solutions can be found using the x-Intercept/Zeroes method.Assign h(x) as Y1, and find the x-intercepts of its graph.

• Linear inequalities can be solved by first applying the intersection-of-graphs method to identify the boundaryvalue of the solution interval. Next, the solution is determined by a careful observation of the relative positions ofthe graphs (is Y1 above or below Y2) and the given inequality.

• To solve formulas for a specified variable, focus on the object variable and apply properties of equality to writethis variable in terms of all others.

• The basic elements of good problem solving include:1. Gathering and organizing information2. Making the problem visual3. Developing an equation model4. Using the model to solve the application

For a complete review, see the problem-solving guide on page 71.

h1x2 � 0

y � 15.x � 20

m �¢W

¢R

4x � 3y � 12

1�3, 52.

1�2, 52?

m � �12; 1�2, 32m � 2

3; 11, 42

h1x2 � 52 x � 3f 1x2 � �2

3 x � 1

5x � 3y � 154x � 3y � 12 � 0

m �¢y

¢x

1–97 Summary and Concept Review 97

Precalculus—

10862

2

4

6

8

10

4

Wol

f po

pula

tion

(100

s)

Rabbit population (100s)R

W

Exercise 26

108620

20

40

60

80

100

4 x

y

Exercise 27

cob19537_ch01_094-102.qxd 1/28/11 8:55 PM

EXERCISES28. Solve the following equation using the intersection-of-graphs method.

29. Solve the following equation using the x-intercept/zeroes method.

30. Solve the following inequality using the intersection-of-graphs method.

Solve for the specified variable in each formula or literal equation.31. for h 32.

33. for x 34. for y

Use the problem-solving guidelines (page 71) to solve the following applications.35. At a large family reunion, two kegs of lemonade are available. One is 2% sugar (too sour) and the second is 7% sugar

(too sweet). How many gallons of the 2% keg, must be mixed with 12 gallons of the 7% keg to get a 5% mix?

36. A rectangular window with a width of 3 ft and a height of 4 ft is topped by a semi-circular window. Find the totalarea of the window.

37. Two cyclists start from the same location and ride in opposite directions, one riding at 15 mph and the other at 18 mph.If their radio phones have a range of 22 mi, how many minutes will they be able to communicate?

SECTION 1.6 Linear Function Models and Real Data

KEY CONCEPTS• A scatterplot is the graph of all the ordered pairs in a real data set.

• When drawing a scatterplot, be sure to scale the axes to comfortably fit the data.

• If larger inputs tend to produce larger output values, we say there is a positive association.

• If larger inputs tend to produce smaller output values, we say there is a negative association.

• If the data seem to cluster around an imaginary line, we say there is a linear association between the variables.

• If the data clearly cannot be approximated by a straight line, we say the variables exhibit a nonlinear association(or sometimes no association).

• The correlation coefficient r measures how tightly a set of data points cluster about an imaginary curve. Thestrength of the correlation is given as a value between �1 and 1. Measures close to �1 or 1 indicate a very strongcorrelation. Measures close to 0 indicate a very weak correlation.

• We can attempt to model linear data sets using an estimated line of best fit.

• A regression line minimizes the vertical distance between all data points and the graph itself, making it the uniqueline of best fit.

EXERCISES38. To determine the value of doing homework, a student in college algebra collects data

on the time spent by classmates on their homework in preparation for a quiz. Her datais entered in the table shown. (a) Use a graphing calculator to draw a scatterplot ofthe data. (b) Does the association appear linear or nonlinear? (c) Is the association positive or negative?

39. If the association in Exercise 38 is linear, (a) use a graphing calculator to find a linear function that models the relation (study time, grade), then (b) graph the data and the line on the same screen. (c) Does the correlation appear weak or strong?

40. According to the function model from Exercise 39, what grade can I expect if I study for 120 minutes?

2x � 3y � 6ax � b � c

P � 2L � 2W for LV � �r2h

31x � 22 � 2.2 6 �2 � 410.2 � 0.5x2

21x � 12 � 32 � 5125x � 1

5 2 � 32

31x � 22 � 10 � 16 � 213 � 2x2


Precalculus—

Exercise 38

x y(min study) (quiz score)

45 70

30 63

10 59

20 67

60 73

70 85

90 82

75 90

cob19537_ch01_094-102.qxd 1/28/11 8:55 PM

1–99 Practice Test 99

Precalculus—

PRACTICE TEST

1. Solve each equation.

a.

b.c. ; for Cd. ; for W

2. How much water that is must be mixed with 25 gal of water at , so that the resultingtemperature of the water will be .

3. To make the bowling team, Jacques needs a three-game average of 160. If he bowled 141 and 162 for thefirst two games, what score S must be obtained in thethird game so that his average is at least 160?

4. In the 2009 movie Star Trek (Chris Pine, ZacharyQuinto, Zoy Zaldana, Eric Bana), Sulu falls off ofthe drill platform without a parachute, and Kirkdives off the platform to save him. To slow his fall,Sulu uses a spread-eagle tactic, while Kirk keeps hisbody straight and arms at his side, to maximize hisfalling speed. If Sulu is falling at a rate of 180 ft/sec,while Kirk is falling at 250 ft/sec, how long would ittake Kirk to reach Sulu, if it took Kirk a full 2 sec toreact and dive after Sulu?

5. Two relations here are functions and two are not.Identify the nonfunctions (justify your response).a. b.c. d.

6. Determine whether the lines are parallel,perpendicular, or neither:

and .

7. Graph the line using the slope and y-intercept:

8. Find the center and radius of the circle defined by, then sketch its graph.

9. After 2 sec, a car is traveling 20 mph. After 5 sec, itsspeed is 40 mph. Assuming the relationship is linear,find the velocity equation and use it to determine thespeed of the car after 9 sec.

10. Find the equation of the line parallel to containing the point . Answer in slope-intercept form.

12, �226x � 5y � 3,

x2 � y2 � 4x � 6y � 3 � 0

x � 4y � 8

L2: y � 25 x � 7L1: 2x � 5y � �15

y � x2 � 2x�y� � 1 � xy � 25 � 2xx � y2 � 2y

97°F91°F

102°F

P � 2L � 2WP � C � k C�5.7 � 3.1x � 14.5 � 41x � 1.52

�2

3x � 5 � 7 � 1x � 32

11. My partner and I are at coordinates on amap. If our destination is at coordinates ,(a) what are the coordinates of the rest stationlocated halfway to our destination? (b) How faraway is our destination? Assume that each unit is 1 mi.

12. Write the equations for lines L1 and L2 shown.

13. State the domain and range for the relations shownon graphs 13(a) and 13(b).

Exercise 13(a) Exercise 13(b)

14. For the linear function shown,a. Determine the value of

W(24) from the graph.b. What input h will give an

output of c. Find a linear function for

the graph.d. What does the slope indicate in this context?e. State the domain and range of h.

15. Given :

a. b. f 1a � 32f 123 2

f 1x2 �2 � x2

x2 , evaluate and simplify

W1h2 � 375?

6�4

�5

5y

xx

y

6�4

�5

5

5�5

�5

5L1 L2

y

x

135, �1221�20, 152

4080 16 24 32

500

300

400

200

100

Hours worked

Wag

es e

arne

d

W(h)

h

cob19537_ch01_094-102.qxd 1/28/11 8:55 PM

16. In 2007, there were 3.3 million Apple iPhones sold worldwide.By 2009, this figure hadjumped to approximately 30.3 million [Source:http://brainstormtech.blogs.fortune.cnn.com/2009/03/12/].Assume that for a time, thisgrowth could be modeled by alinear function. (a) Determine

the rate of change , and

(b) interpret it in this context. Then use the rate of change to (c) approximate the number of sales in2008, and what the projected sales would be for2010 and 2011.

17. Solve the following equations using the x-intercept/zeroes method.

a.

b.

18. Solve the following inequalities using theintersection-of-graphs method.a.b. 210.75x � 12 6 0.7 � 0.513x � 12

3x � 15 � x2 � 215 � x2 � 3

210.7x � 1.32 � 2.6 � 2x � 310.2x � 22

2x � a4 �1

3 xb � �120 � x2

¢sales

¢time


Precalculus—

19. To study how annual rainfall affects the ability to attain certain levels of livestock production, a local university collects data on the average annual rainfall for a particular area and compares this to the average number of free-ranging cattle per acre for ranchers in that area. The data collected are shown in the table. (a) Use a graphing calculatorto draw a scatterplot of the data. (b) Does theassociation appear linear or nonlinear? (c) Is theassociation positive or negative?

20. If the association in Exercise 19 is linear, (a) use agraphing calculator to find a linear function thatmodels the relation (rainfall, cattle per acre), (b) usethe function to find the number of cattle per acre thatmight be possible for an area receiving 50 in. ofrainfall per year, and (c) state whether the correlationis weak or strong.

STRENGTHENING CORE SKILLS

The Various Forms of a Linear EquationLearning mathematics is very much like the construction of a skyscraper. The final height of the skyscraper ultimatelydepends on the strength of the foundation and quality of the frame supporting each new floor as it is built. Ourprevious work with linear functions and their graphs, while having a number of useful applications, is actually thefoundation on which much of our future work is built. For this reason, it’s important you gain a certain fluency withlinear functions and relationships—even to a point where things come to you effortlessly and automatically. As noted mathematician Henri Lebesque once said, “An idea reaches its maximum level of usefulness only when youunderstand it so well that it seems like you have always known it. You then become incapable of seeing the idea asanything but a trivial and immediate result.” These formulas and concepts, while simple, have an endless number ofsignificant and substantial applications.

Forms and Formulas

slope formula point-slope form slope-intercept form standard form

given any two points given slope m and given slope m and A, B, and C are integerson the line any point y-intercept (0, b) (used in linear systems)1x1, y12

Ax � By � Cy � mx � by � y1 � m1x � x12m �y2 � y1

x2 � x1

Exercise 19

Rainfall Cattle (in.) per Acre

12 2

16 3

19 7

23 9

28 11

32 22

37 23

40 26

Exercise 16

cob19537_ch01_094-102.qxd 1/28/11 8:55 PM

Characteristics of Lines

increasing decreasing

(0, y) (x, 0)

let , let , line slants upward line slants downwardsolve for y solve for x from left to right form left to right

Relationships between Lines

intersecting parallel perpendicular dependent

lines intersect lines do not lines intersect lines intersectat one point intersect at right angles at all points

Special Lines

horizontal vertical identity

horizontal line vertical line the input valuethrough k through h identifies the output

Use the formulas and concepts reviewed here to complete the following exercises.For the two points given: (a) compute the slope of the line through the points and state whether the line is increasing

or decreasing, (b) find the equation of the line in point-slope form, then write the equation in slope-intercept form, and (c) find the and intercepts and graph the line.

Exercise 1: P1(0, 5) P2(6, 7) Exercise 4: P2(3, 2)

Exercise 2: P1(3, 2) P2(0, 9) Exercise 5:

Exercise 3: P1(3, 2) P2(9, 5) Exercise 6: P2 1�8, �22P112, �72

P2 16, �12P11�2, 52

P11�5, �42

y-x-

y � xx � hy � k

b1 � b2 b1 � b2

m1 � m2,m1m2 � �1 m1 � m2,m1 � m2

y � 0x � 0

m 6 0m 7 0

x-intercepty-intercept

Evaluating Expr essions and Looking for Patter nsThese “explorations” are designed to explore the full potential of a graphing calculator, as well as to use this potentialto investigate patterns and discover connections that might otherwise be overlooked. In this exploration and discovery,we point out the various ways an expression can be evaluated on a graphing calculator. Some ways seem easier, faster,and/or better than others, but each has advantages and disadvantages depending on the task at hand, and it will help tobe aware of them all for future use.

One way to evaluate an expression is to use the TABLE feature of a graphing calculator, with the expressionentered as Y1 on the screen. If you want the calculator to generate inputs, use the (TBLSET) screento indicate a starting value (TblStart�) and an increment value ( ), and set the calculator in Indpnt:

mode (to input specific values, the calculator should be in Indpnt: AUTO mode). After pressing (TABLE), the calculator shows the correspondinginput and output values.

Expressions can also be evaluated on the home screen for a single value or a seriesof values. Enter the expression on the screen (see Figure 1.105) anduse (QUIT) to get back to the home screen. To evaluate this expression,access Y1 using (Y-VARS), and use the first option 1:Function . Thisbrings us to a submenu where any of the equations Y1 through Y0 (actually Y10) can be accessed. Since the default setting is the one we need (1:Y1), simply press andY1 appears on the home screen. To evaluate a single input, simply enclose it in

ENTER

ENTERVARS

MODE2nd

Y=�34x � 5

GRAPH2nd

ASKASKAUTO�Tbl �

WINDOW2ndY=

1–101 Calculator Exploration and Discovery 101

Precalculus—

CALCULATOR EXPLORA TION AND DISCOVER Y

Figure 1.105

cob19537_ch01_094-102.qxd 2/1/11 10:53 AM

parentheses. To evaluate more than one input, enter the numbers as a set of values withthe set enclosed in parentheses. In Figure 1.106, Y1 has been evaluated for ,then simultaneously for , and 2.

A third way to evaluate expressions is using a list, with the desired inputs enteredin List 1 (L1), then List 2 (L2) defined in terms of L1. For example, will return the same values for inputs of and 2 seen previously on the homescreen (remember to clear the lists first). Lists are accessed by pressing 1:Edit.Enter the numbers , and 2 in L1, then use the right arrow to move to L2. Itis important to note that you next press the up arrow key so that the cursor overliesL2. The bottom of the screen now reads “ ” and the calculator is waiting for us todefine L2. After entering (see Figure 1.107) and pressing weobtain the same outputs as before (see Figure 1.108). The advantage of using the “list”method is that we can further explore or experiment with the output values in a searchfor patterns.

Exercise 1: Evaluate the expression on the list screen, using consecutiveinteger inputs from to 6 inclusive. What do you notice about the outputs?

Exercise 2: Evaluate the expression on the list screen, usingconsecutive integer inputs from to 6 inclusive. We suspect there is a pattern to theoutput values, but this time the pattern is very difficult to see. On the home screen,compute the difference between a few successive outputs from L2 [for example,

)]. What do you notice?L2112 � L212

�612L1 � 19.1

�60.2L1 � 3

ENTERL2 � �34L1 � 5

L2 �

�4, �2, 0STAT

�4, �2, 0,L2 � �3

4L1 � 5

�2, 0,x � �4x � �4


Precalculus—

Figure 1.106

Figure 1.107

Figure 1.108

cob19537_ch01_094-102.qxd 2/1/11 10:46 AM

Precalculus—

1–103 103

CONNECTIONS TO CALCULUS

Understanding and internalizing concepts related to linear functions is one of the mainobjectives of Chapter 1 and its Strengthening Core Skills feature. The ability to quicklyand correctly write the equation of a line given sufficient information has a number ofsubstantial applications in the calculus sequence. ln calculus, we will extend ourunderstanding of secant lines to help understand lines drawn tangent to a curve and fur-ther to lines and planes drawn tangent to three-dimensional surfaces. These relation-ships will lead to significant and meaningful applications in topography, meteorology,engineering, and many other areas.

Tangent Lines

In Section 1.4 we learned how to write the equation of a line given its slope and any pointon the line. ln that section, Example 6 uses the slope-intercept form y � mx � b as a for-mula, while Example 10 applies the point-slope form y � y1 � m(x � x1). In calculus,we regularly use both forms to write the equations of secant lines and tongent lines.

Consider the graph of a parabola that opens upward, with y-intercept (0, �3) and vertex at (1, �4). Using the tools of calculus, we can show thatthe function gives the slope of any line drawn tangent to this graph at agiven x. For example, to find the slope of the line tangent to this curve when x � 3, weevaluate g(x) at x � 3, and find the slope will be g(3) � 4. Note how this informationis used in Example 1.

EXAMPLE 1 � Finding the Equation of a Tangent LineFind an equation of the line drawn tangent to the graph of at

and write the result in slope-intercept form.

Solution � To begin, we first determine the slope of the tangent line. For we have

given equation and slope of tangent line

substitute 0 for x

result

The slope of the line drawn tangent to this curve at x � 0, is m � �2. However, recallthat to find the equation of this line we also need a point on the line. Using x � 0 onceagain, we note (0, �3) is the y-intercept of the graph of f(x) and is also a point on the tan-gent line. Using the point (0, �3) and the slope m � �2,we have

point-slope form

substitute �2 for m, (o, �3) for (x1, y1)

simplify


In the figure shown, note the graph of is tangent to the graph ofat the point (0, �3).


Similar to our work here with tangent lines, many applications of advanced mathematicsrequire that we find the equation of a line drawn perpendicular to this tangent line. In thiscase, the line is called a normal to the curve at point (x, y). See Exercises 9 through 12.

y � x2 � 2x � 3y � �2x � 3

y � �2x � 3 y � 3 � �2x

y � 1�32 � �21x � 02 y � y1 � m1x � x12

� �2

g102 � 2102 � 2

g1x2 � 2x � 2

x � 0,

x � 0,f 1x2 � x2 � 2x � 3

g1x2 � 2x � 2

f 1x2 � x2 � 2x � 3,

x

y

54321�5 �4 �3 �2 �1

�5

�4

�3

�2

�1

5

4

3

2

1

(0, �3)

y � x2 � 2x � 3

y � � 2x � 3

cob19537_ch01_103-104.qxd 1/28/11 8:57 PM

For Exercises 1 through 4, the function g(x) � 2x � 4gives the slope of any line drawn tangent to the graph off(x) � x2 � 4x � 5 at a given x. Find an equation of theline drawn tangent to the graph of f (x) at the following x-values, and write the result in slope-intercept form.

1. 2.3. 4.

For Exercises 5 through 8, the function v(x) � gives the slope of any line drawntangent to the graph of atgiven x. Find an equation of the line drawn tangent tothe graph of s(x) at the following x-values, and writethe result in slope-intercept form.

5. 6.7. 8.

For Exercises 9 through 12, find the equation of thenormal to each tangent shown, or the equation of thetangent line for each normal given.

9.

10.

11.

x

y

54321�5 �4 �3 �2 �1

�5

�4

�3

�2

�1

5

4

3

2

1

y � �2x � 1

(�1, 1)

x

y

54321�5 �4 �3 �2 �1

�5

�4

�3

�2

�1

5

4

3

2

1

y � �0.5x � 3

(3, 1.5)

x

y

54321�5 �4 �3 �2 �1

�5

�4

�3

�2

�1

5

4

3

2

1

(�1, �2)

y � 2x

x � 4x � 1x � 3x � �2

s1x2 � 2x3 � 3x2 � 36x6x2 � 6x � 36

x � 1x � �2x � 0x � �4

104 Connections to Calculus 1–104

Precalculus—

12.

13. A theorem from elementry geometary states, “Aline drawn tangent to a circle is perpendicular tothe radius at the point of tangency.” Find theequation of the tangent line shown in the figure.

14. Find the equation of the line containing theperpendicular bisector of the line segment withendpoints and shown.

x

y

543�5 �4

�5

�4

5

4

3

21�3 �2 �1

�3

�2

�1

2

1

(4, 1)(�2, �3)

x

y

543�5 �4

�5

�4

5

4

3

21�3 �2 �1

�3

�2

2

1

(�0.5, �1)

x

y

54321�5 �4 �3 �2 �1

�5

�4

�3

�2

�1

5

4

3

2

1

y � �x � 4

(1, 3)

Connections to Calculus Exercises

cob19537_ch01_103-104.qxd 1/28/11 8:57 PM

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