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Linear Functions and Slope

116. I graphed

and one piece of my graph is a single point.

117. I noticed that the difference quotient is always zero ifwhere is any constant.

118. Sketch the graph of using the following properties. (Morethan one correct graph is possible.) is a piecewise functionthat is decreasing on is increasing on

and the range of is

119. Define a piecewise function on the intervals (2, 5),and that does not “jump” at 2 or 5 such that onepiece is a constant function, another piece is an increasingfunction, and the third piece is a decreasing function.

120. Suppose that The function can be even,

odd, or neither. The same is true for the function

a. Under what conditions is definitely an even function?

b. Under what conditions is definitely an odd function?h

h

g.

fh1x2 =f1x2g1x2 .

35, q2 1- q , 24,30, q2.f12, q2, 1- q , 22, f122 = 0, ff

f

cf1x2 = c,

f1x2 = b2 if x Z 43 if x = 4

Group Exercise121. (For assistance with this exercise, refer to the discussion of

piecewise functions beginning on page 169, as well as to Exer-cises 79–80.) Group members who have cellular phoneplans should describe the total monthly cost of the plan asfollows:

$_____ per month buys _____ minutes. Additionaltime costs $_____ per minute.

(For simplicity, ignore other charges.) The group shouldselect any three plans, from “basic” to “premier.” For eachplan selected, write a piecewise function that describesthe plan and graph the function. Graph the threefunctions in the same rectangular coordinate system. Nowexamine the graphs. For any given number of callingminutes, the best plan is the one whose graph is lowest atthat point. Compare the three calling plans. Is one planalways a better deal than the other two? If not, determinethe interval of calling minutes for which each plan is thebest deal. (You can check out cellular phone plans byvisiting www.point.com.)

Preview ExercisesExercises 122–124 will help you prepare for the material coveredin the next section.

122. If and find

123. Find the ordered pairs (_____, 0) and (0, _____) satisfying

124. Solve for y: 3x + 2y - 4 = 0.

4 x - 3y - 6 = 0.

y2 - y1

x2 - x1.1x2 , y22 = 1-2, 4 2,1x1 , y12 = 1-3, 12

1.4 Linear Functions and Slope

Is there a relationshipbetween literacy and child

mortality? As the percentageof adult females who areliterate increases, does themortality of children underfive decrease? Figure 1.35 on

the next page indicatesthat this is, indeed, the

case. Each point inthe figure representsone country.

Objectives

! Calculate a line’s slope.

" Write the point-slope form of the equation of a line.

# Write and graph the slope-intercept form of the equationof a line.

$ Graph horizontal or vertical lines.

% Recognize and use the generalform of a line’s equation.

& Use intercepts to graph thegeneral form of a line’sequation.

' Model data with linear functionsand make predictions.

Sec t i on

“DYSLEXIA,” 1981

115. This work by artist Scott Kim (1955– ) has the same kind ofsymmetry as an even function.

178 Chapter 1 Functions and Graphs

P-BLTZMC01_135-276-hr 17-11-2008 11:46 78

The situation is illustrated in Figure 1.36. The slope of the line is 5. For everyvertical change, or rise, of 5 units, there is a corresponding horizontal change, orrun, of 1 unit. The slope is positive and the line rises from left to right.

1

12

45

2 3 4 5

1 2 3 4 5 1 2 4 5

y

x

( 3, 1)

Rise: 5 units

Run: 1 unit

( 2, 4)

Figure 1.36 Visualizing a slope of 5

Study TipWhen computing slope, it makes no difference which point you call and which pointyou call If we let and the slope is still 5:

However, you should not subtract in one order in the numerator and then in adifferent order in the denominator

Incorrect! The slope is not -5.-1 - 4

-2 - 1-32 = -51

= -5.

1x1 - x22. 1y2 - y12m =

Change in yChange in x

=y2 - y1

x2 - x1= -1 - 4

-3 - 1-22 = -5-1

= 5.

1x2 , y22 = 1-3, -12,1x1 , y12 = 1-2, 421x2 , y22. 1x1 , y12

b. To find the slope of the line passing through and we can letand The slope of the line is computed as

follows:

The situation is illustrated in Figure 1.37. The slope of the line is For everyvertical change of units (6 units down), there is a corresponding horizontalchange of 5 units.The slope is negative and the line falls from left to right.

Check Point 1 Find the slope of the line passing through each pair of points:

a. and b. and

Example 1 illustrates that a line with a positive slope is increasing and a linewith a negative slope is decreasing. By contrast, a horizontal line is a constantfunction and has a slope of zero. A vertical line has no horizontal change, so

in the formula for slope. Because we cannot divide by zero, the slope ofa vertical line is undefined. This discussion is summarized in Table 1.2.x2 - x1 = 0

1-1, 52.14, -221-4, -221-3, 42

-6-

65 .

m =Change in yChange in x

=y2 - y1

x2 - x1= -2 - 4

2 - 1-32 = -65

= - 65

.

1x2 , y22 = 12, -22.1x1 , y12 = 1-3, 42 12, -22,1-3, 42 1

12345

3 4 5

1 2 3 4 5 1 2 4 5

y

x

Rise: 6 units

( 3, 4)

(2, 2)

Run: 5 units

Figure 1.37 Visualizing a slope of - 65

Table 1.2 Possibilities for a Line’s Slope

Positive Slope Negative Slope Zero Slope Undefined Slope

x

y

Line rises from left to right.

m � 0

x

y

Line falls from left to right.

m � 0

x

y

Line is horizontal.

m � 0

x

y

Line is vertical.

m isundefined.

The Point-Slope Form of the Equation of a LineWe can use the slope of a line to obtain various forms of the line’s equation. Forexample, consider a nonvertical line that has slope and that contains the point1x1 , y12. m

! Write the point-slope form of the equation of a line.

Study TipAlways be clear in the way you uselanguage, especially in mathematics.For example, it’s not a good idea tosay that a line has “no slope.” Thiscould mean that the slope is zero orthat the slope is undefined.


P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 180

Section 1.4 Linear Functions and Slope 181

The line in Figure 1.38 has slope and contains the point Let represent any other point on the line.

Regardless of where the point is located, the steepness of the line inFigure 1.38 remains the same. Thus, the ratio for the slope stays a constant Thismeans that for all points along the line

We can clear the fraction by multiplying both sides by the least commondenominator.

This is the slope of the line in Figure 1.38.

Multiply both sides by

Simplify:

Now, if we reverse the two sides, we obtain the point-slope form of the equation ofa line.

y - y1

x - x1# (x - x1 ) = y - y1 . m1x - x12 = y - y1

x - x1 . m1x - x12 =y - y1

x - x1

# 1x - x12 m =

y - y1

x - x1

x - x1 ,


=y - y1

x - x1.

1x, y2 m.1x, y2 (x, y)(x1,y1).m

Study TipWhen writing the point-slope form ofa line’s equation, you will never sub-stitute numbers for and You willsubstitute values for and m.x1 , y1 ,

y.x

y

x

This point is arbitrary.

Slope is m.

(x1, y1)

(x, y)

y y1

This point is fixed.

x x1

Figure 1.38 A line passingthrough with slope m1x1 , y12

Point-Slope Form of the Equation of a LineThe point-slope form of the equation of a nonvertical line with slope that passesthrough the point is

y - y1 = m1x - x12.1x1 , y12 m

For example, the point-slope form of the equation of the line passing through(1, 5) with slope 2 is

We will soon be expressing the equation of a nonvertical line in functionnotation. To do so, we need to solve the point-slope form of a line’s equation for Example 2 illustrates how to isolate on one side of the equal sign.

Writing an Equation in Point-Slope Form for a Line

Write an equation in point-slope form for the line with slope 4 that passes throughthe point Then solve the equation for

Solution We use the point-slope form of the equation of a line withand

This is the point-slope form of the equation.

Substitute the given values: and

We now have an equation in point-slope formfor the given line.

Now we solve this equation for and write an equivalent equation that can beexpressed in function notation.


Use the distributive property.

Add 3 to both sides.

Check Point 2 Write an equation in point-slope form for the line with slope 6that passes through the point Then solve the equation for y.12, -52.

y = 4x + 7 y - 3 = 4x + 4

y-3=4(x+1)We need to isolate y.

y

y - 3 = 41x + 12 (x1, y1) = (- 1, 3) .m = 4 y - 3 = 43x - 1-124 y - y1 = m1x - x12y1 = 3.m = 4, x1 = -1,

y.1-1, 32.EXAMPLE 2

yy.

y - 5 = 21x - 12.1m = 22

P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 181







Simplify:


y - y1

x - x1# (x - x1 ) = y - y1 . m1x - x12 = y - y1

x - x1 . m1x - x12 =y - y1

x - x1

# 1x - x12 m =

y - y1

x - x1

x - x1 ,


=y - y1

x - x1.

1x, y2 m.1x, y2 (x, y)(x1,y1).m


y.x

y

x


Slope is m.

(x1, y1)

(x, y)

y y1


x x1



y - y1 = m1x - x12.1x1 , y12 m














y = 4x + 7 y - 3 = 4x + 4


y


y.1-1, 32.EXAMPLE 2

yy.

y - 5 = 21x - 12.1m = 22

P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 181







Simplify:


y - y1

x - x1# (x - x1 ) = y - y1 . m1x - x12 = y - y1

x - x1 . m1x - x12 =y - y1

x - x1

# 1x - x12 m =

y - y1

x - x1

x - x1 ,


=y - y1

x - x1.

1x, y2 m.1x, y2 (x, y)(x1,y1).m


y.x

y

x


Slope is m.

(x1, y1)

(x, y)

y y1


x x1



y - y1 = m1x - x12.1x1 , y12 m














y = 4x + 7 y - 3 = 4x + 4


y


y.1-1, 32.EXAMPLE 2

yy.

y - 5 = 21x - 12.1m = 22

P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 181

1.4


Data presented in a visual form as a set of points is called a scatter plot.Also shown in Figure 1.35 is a line that passes through or near the points. A linethat best fits the data points in a scatter plot is called a regression line. By writingthe equation of this line, we can obtain a model for the data and make predictionsabout child mortality based on the percentage of literate adult females in acountry.

Data often fall on or near a line. In this section, we will use functions to modelsuch data and make predictions. We begin with a discussion of a line’s steepness.

The Slope of a LineMathematicians have developed a useful measure of the steepness of a line, calledthe slope of the line. Slope compares the vertical change (the rise) to the horizontalchange (the run) when moving from one fixed point to another along the line. Tocalculate the slope of a line, we use a ratio that compares the change in (the rise)to the corresponding change in (the run).x

y

y

x

Percentage of Adult Females Who Are Literate

Und

er-F

ive

Mor

talit

y(p

er th

ousa

nd)

Literacy and Child Mortality

1009080706050403020100

350

300

250

200

150

100

50

Figure 1.35Source: United Nations

Definition of SlopeThe slope of the line through the distinct points

and is1x2 , y221x1 , y12

where x2 - x1 Z 0.

Runx 2 x 1

y

x 1

y 1

y 2

x 2

x

(x 1, y 1)

(x 2, y 2)Risey 2 y 1


RiseRun

=

= ,y2 -y1

x2 -x1

Vertical change

Horizontal change

It is common notation to let the letter represent the slope of a line. Theletter is used because it is the first letter of the French verb monter, meaning“to rise” or “to ascend.”

Using the Definition of Slope

Find the slope of the line passing through each pair of points:

a. and b. and

Solution

a. Let and We obtain the slope as follows:


=y2 - y1

x2 - x1=

4 - 1-12-2 - 1-32 = 5

1= 5.

1x2 , y22 = 1-2, 42.1x1 , y12 = 1-3, -1212, -22.1-3, 421-2, 421-3, -12

EXAMPLE 1

mm


P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 179


Data presented in a visual form as a set of points is called a scatter plot.Also shown in Figure 1.35 is a line that passes through or near the points. A linethat best fits the data points in a scatter plot is called a regression line. By writingthe equation of this line, we can obtain a model for the data and make predictionsabout child mortality based on the percentage of literate adult females in acountry.

Data often fall on or near a line. In this section, we will use functions to modelsuch data and make predictions. We begin with a discussion of a line’s steepness.

The Slope of a LineMathematicians have developed a useful measure of the steepness of a line, calledthe slope of the line. Slope compares the vertical change (the rise) to the horizontalchange (the run) when moving from one fixed point to another along the line. Tocalculate the slope of a line, we use a ratio that compares the change in (the rise)to the corresponding change in (the run).x

y

y

x

Percentage of Adult Females Who Are Literate

Und

er-F

ive

Mor

talit

y(p

er th

ousa

nd)

Literacy and Child Mortality

1009080706050403020100

350

300

250

200

150

100

50

Figure 1.35Source: United Nations

Definition of SlopeThe slope of the line through the distinct points

and is1x2 , y221x1 , y12

where x2 - x1 Z 0.

Runx 2 x 1

y

x 1

y 1

y 2

x 2

x

(x 1, y 1)

(x 2, y 2)Risey 2 y 1


RiseRun

=

= ,y2 -y1

x2 -x1

Vertical change

Horizontal change

It is common notation to let the letter represent the slope of a line. Theletter is used because it is the first letter of the French verb monter, meaning“to rise” or “to ascend.”

Using the Definition of Slope

Find the slope of the line passing through each pair of points:

a. and b. and

Solution

a. Let and We obtain the slope as follows:


=y2 - y1

x2 - x1=

4 - 1-12-2 - 1-32 = 5

1= 5.

1x2 , y22 = 1-2, 42.1x1 , y12 = 1-3, -1212, -22.1-3, 421-2, 421-3, -12

EXAMPLE 1

mm


P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 179

Graph x = 3 Graph y = -2


Write an equation in point-slope form for the line passing through the pointsand (See Figure 1.39.) Then solve the equation for

Solution To use the point-slope form, we need to find the slope. The slope is thechange in the divided by the corresponding change in the

This is the definition of slope usingand

We can take either point on the line to be Let’s use Now, we are ready to write the point-slope form of the equation.


Substitute: and

Simplify.

We now have an equation in point-slope form for the line shown in Figure 1.39.Now, we solve this equation for



Subtract 3 from both sides.

Check Point 3 Write an equation in point-slope form for the line passingthrough the points and Then solve the equation for

The Slope-Intercept Form of the Equation of a LineLet’s write the point-slope form of the equation of a nonvertical line with slope and The line is shown in Figure 1.40. Because the is theline passes through We use the point-slope form with and

We obtain

Simplifying on the right side gives us

Finally, we solve for by adding to both sides.

Thus, if a line’s equation is written with isolated on one side, the is the line’s slope and the constant term is the This form of a line’sequation is called the slope-intercept form of the line.

y-intercept.coefficient of xy

y = mx + b

by

y - b = mx.

y - b = m1x - 02.y-y1=m(x-x1)

Let y1 � b. Let x1 � 0.

y1 = b.x1 = 010, b2. b,y-intercepty-intercept b.m

y.1-1, -62.1-2, -12 y = -

32 x + 3

y + 3 = - 32 x + 6

(x-4)y+3=–We need to isolate y.

32

y.

y + 3 = - 32 1x - 42 m = -

32 .1x1 , y12 = 14, -32 y - 1-32 = -

32 1x - 42 y - y1 = m1x - x12

1x1 , y12 = 14, -32.1x1 , y12.1-2, 62.14, -32m =6 - 1-32

-2 - 4= 9

-6= -

32

x-coordinates.y-coordinates

y.1-2, 62.14, -32EXAMPLE 3

1

1234

65

2 3 4

1 2 3 4 5 1 2 3 4 5

y

x

( 2, 6)

(4, 3)

Figure 1.39 Write an equationin point-slope form for this line.

DiscoveryYou can use either point for when you write a point-slopeequation for a line. Rework Example3 using for Once yousolve for you should still obtain

y = - 32 x + 3.

y,1x1 , y12.1-2, 62

1x1 , y12

! Write and graph the slope-intercept form of the equation of a line.

Slope-Intercept Form of the Equation of a LineThe slope-intercept form of the equation of a nonvertical line with slope and

is

y = mx + b.

y-intercept bm

x

y-intercept is b. Fixed point,(x1, y1), is (0, b).

This pointis arbitrary.

Slope is m.(0, b)

(x, y)

y

Figure 1.40 A line with slope and y-intercept b

m


P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 182


The slope-intercept form of a line’s equation, can be expressedin function notation by replacing with

We have seen that functions in this form are called linear functions. Thus, in theequation of a linear function, the is the line’s slope and the constantterm is the Here are two examples:

If a linear function’s equation is in slope-intercept form, we can use theand the slope to obtain its graph.y-intercept

The slope is 2. The y -intercept is 4. The y -intercept is 2.

y=2x-4 f(x)= x+2.12

12The slope is .

y-intercept.coefficient of x

f1x2 = mx + b.

f1x2:yy = mx + b,

Graphing Using the Slope and

1. Plot the point containing the on the This is the point 2. Obtain a second point using the slope, Write as a fraction, and use rise

over run, starting at the point containing the to plot this point.3. Use a straightedge to draw a line through the two points. Draw arrowheads

at the ends of the line to show that the line continues indefinitely in bothdirections.

y-intercept,mm.

10, b2.y-axis.y-intercept

y-Intercepty ! mx " b

Study TipWriting the slope, as a fractionallows you to identify the rise (thefraction’s numerator) and the run(the fraction’s denominator).

m,


Graph the linear function:

Solution The equation of the line is in the form We can find theslope, by identifying the coefficient of We can find the byidentifying the constant term.

Now that we have identified the slope, and the 2, we use the three-step procedure to graph the equation.

Step 1 Plot the point containing the on the The is 2.We plot (0, 2), shown in Figure 1.41.

Step 2 Obtain a second point using the slope, Write as a fraction, and use riseover run, starting at the point containing the to plot this point. Theslope, is already written as a fraction.

We plot the second point on the line by starting at (0, 2), the first point. Based on theslope, we move 3 units down (the rise) and 2 units to the right (the run). This puts usat a second point on the line, shown in Figure 1.41.

Step 3 Use a straightedge to draw a line through the two points. The graph ofthe linear function is shown as a blue line in Figure 1.41.

Check Point 4 Graph the linear function: f1x2 = 35 x + 1.

f1x2 = - 32 x + 2

12, -12,m = -

32

= -32

= RiseRun

- 32 ,

y-intercept,mm.

y-intercepty-axis.y-intercept

y-intercept,- 32 ,

The slope is 32 .

f(x)=– x+232

The y -intercept is 2.

y-intercept, b,x.m,f1x2 = mx + b.

f1x2 = - 32

x + 2.

y-InterceptEXAMPLE 4

1

12345

2 3 4 5

1 3 4 5 1 2 3 4 5

y

x

Rise � 3

Run � 2

y -intercept: 2

Figure 1.41 The graph off1x2 = -

32 x + 2

P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 183

Equations of Horizontal and Vertical LinesIf a line is horizontal, its slope is zero: Thus, the equation becomes where is the All horizontal lines have equations ofthe form

Graphing a Horizontal Line

Graph in the rectangular coordinate system.y = -4

EXAMPLE 5

y = b.y-intercept.by = b,

y = mx + bm = 0.

1

1234

2 3

5 6

1 2 3 4 5 1 2 3 4 5

y

x

( 2, 4)(0, 4)

(3, 4)

y-intercept is 4.

! Graph horizontal or vertical lines.

Figure 1.42 The graph ofor f1x2 = -4y = -4

Solution All ordered pairs that aresolutions of have a value of that isalways Any value can be used for In thetable on the right, we have selected three ofthe possible values for and 3. Thetable shows that three ordered pairs that aresolutions of are and

Drawing a line that passes through thethree points gives the horizontal line shown inFigure 1.42.

Check Point 5 Graph in the rectangular coordinate system.y = 3

13, -42. 1-2, -42, 10, -42,y = -4

x: -2, 0,

x.-4.yy = -4

y ! "4 (x, y)

–2

0

3

–4

–4

–4

(–2, –4)

(0, –4)

(3, –4)

x

For allchoices of x,

y is aconstant 4.

1

12345

2 3 4 5

1 2 3 4 5 6 7 1 2 3

y

x

(2, 3)

(2, 2)

(2, 0)

x-interceptis 2.

Figure 1.43 The graph of x = 2

Equation of a Horizontal LineA horizontal line is given by an equation of the form

where is the of the line. The slope of ahorizontal line is zero.

y-interceptb

y = b,

x

y

y-intercept: b

(0, b)

Because any vertical line can intersect the graph of a horizontal line only once, a horizontal line is the graph of a function. Thus, we can expressthe equation as This linear function is often called a constantfunction.

Next, let’s see what we can discover about the graph of an equation of theform by looking at an example.

Graphing a Vertical Line

Graph the linear equation:

Solution All ordered pairs thatare solutions of have a valueof that is always 2. Any value canbe used for In the table on theright, we have selected three of thepossible values for and 3.The table shows that three orderedpairs that are solutions of are

(2, 0), and (2, 3). Drawing aline that passes through the threepoints gives the vertical line shownin Figure 1.43.

12, -22, x = 2

y: -2, 0,

y.x

x = 2

x = 2.

EXAMPLE 6

x = a

f1x2 = b.y = b

y = b

x ! 2 (x, y)

2

2

2

–2

0

3

(2, –2)

(2, 0)

(2, 3)

y

For all choices of y,

x is always 2.


P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 184

Equations of Horizontal and Vertical LinesIf a line is horizontal, its slope is zero: Thus, the equation becomes where is the All horizontal lines have equations ofthe form

Graphing a Horizontal Line

Graph in the rectangular coordinate system.y = -4

EXAMPLE 5

y = b.y-intercept.by = b,

y = mx + bm = 0.

1

1234

2 3

5 6

1 2 3 4 5 1 2 3 4 5

y

x

( 2, 4)(0, 4)

(3, 4)

y-intercept is 4.

! Graph horizontal or vertical lines.

Figure 1.42 The graph ofor f1x2 = -4y = -4

Solution All ordered pairs that aresolutions of have a value of that isalways Any value can be used for In thetable on the right, we have selected three ofthe possible values for and 3. Thetable shows that three ordered pairs that aresolutions of are and

Drawing a line that passes through thethree points gives the horizontal line shown inFigure 1.42.

Check Point 5 Graph in the rectangular coordinate system.y = 3

13, -42. 1-2, -42, 10, -42,y = -4

x: -2, 0,

x.-4.yy = -4

y ! "4 (x, y)

–2

0

3

–4

–4

–4

(–2, –4)

(0, –4)

(3, –4)

x

For allchoices of x,

y is aconstant 4.

1

12345

2 3 4 5

1 2 3 4 5 6 7 1 2 3

y

x

(2, 3)

(2, 2)

(2, 0)

x-interceptis 2.

Figure 1.43 The graph of x = 2

Equation of a Horizontal LineA horizontal line is given by an equation of the form

where is the of the line. The slope of ahorizontal line is zero.

y-interceptb

y = b,

x

y

y-intercept: b

(0, b)

Because any vertical line can intersect the graph of a horizontal line only once, a horizontal line is the graph of a function. Thus, we can expressthe equation as This linear function is often called a constantfunction.

Next, let’s see what we can discover about the graph of an equation of theform by looking at an example.

Graphing a Vertical Line

Graph the linear equation:

Solution All ordered pairs thatare solutions of have a valueof that is always 2. Any value canbe used for In the table on theright, we have selected three of thepossible values for and 3.The table shows that three orderedpairs that are solutions of are

(2, 0), and (2, 3). Drawing aline that passes through the threepoints gives the vertical line shownin Figure 1.43.

12, -22, x = 2

y: -2, 0,

y.x

x = 2

x = 2.

EXAMPLE 6

x = a

f1x2 = b.y = b

y = b

x ! 2 (x, y)

2

2

2

–2

0

3

(2, –2)

(2, 0)

(2, 3)

y

For all choices of y,

x is always 2.


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Checkpoint:!!Graph!each!equation!in!standard!form.!

2.!!C3x!+!4y!=!12!! ! ! ! !

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3.!!y!=!C2x!C!6!

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2.!!C3x!+!4y!=!12!! ! ! ! !

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3.!!y!=!C2x!C!6!

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Does a vertical line represent the graph of a linear function? No. Look at thegraph of in Figure 1.43. A vertical line drawn through (2, 0) intersects thegraph infinitely many times. This shows that infinitely many outputs are associatedwith the input 2. No vertical line represents a linear function.

Check Point 6 Graph the linear equation:

The General Form of the Equation of a LineThe vertical line whose equation is cannot be written in slope-intercept form,

because its slope is undefined. However, every line has an equationthat can be expressed in the form For example, can beexpressed as or The equation iscalled the general form of the equation of a line.

Ax + By + C = 0x - 5 = 0.1x + 0y - 5 = 0,x = 5Ax + By + C = 0.

y = mx + b,x = 5

x = -3.

x = 2

General Form of the Equation of a LineEvery line has an equation that can be written in the general form

where and are real numbers, and and are not both zero.BACA, B,

Ax + By + C = 0,

If the equation of a line is given in general form, it is possible to find the slope,and the for the line. We solve the equation for transforming it

into the slope-intercept form In this form, the coefficient of is theslope of the line and the constant term is its

Finding the Slope and the

Find the slope and the of the line whose equation is

Solution The equation is given in general form. We begin by rewriting it in theform We need to solve for

2y=–3x+43x+2y-4=0

Our goal is to isolate y.

2y2

–3x+42

=

32

y=– x+2

slope y-intercept

y.y = mx + b.

3x + 2y - 4 = 0.y-intercept


y-intercept.xy = mx + b.

y,y-intercept, b,m,

This is the given equation.Isolate the term containing by adding to both sides.

Divide both sides by 2.

On the right, divide each term in the numerator by 2 to obtain slope-intercept form.

-3x + 4y

The coefficient of is the slope and the constant term, 2, is the This is the form of the equation that we graphed in Figure 1.41 on page 183.

y-intercept.x, - 32 ,

! Recognize and use the generalform of a line’s equation.

Equation of a Vertical LineA vertical line is given by an equation of the form

where is the of the line. The slope of avertical line is undefined.

x-intercepta

x = a,

x

y

x-intercept: a

(a, 0)

P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 185

Using Intercepts to Graph a Linear Equation

Graph using intercepts:

SolutionStep 1 Find the Let and solve for

Replace with 0 in

Simplify.


Divide both sides by 4.

The is so the line passes through or (1.5, 0), as shown inFigure 1.44.

Step 2 Find the Let and solve for

Replace with 0 in

Simplify.


Divide both sides by

The is so the line passes through as shown in Figure 1.44.Step 3 Graph the equation by drawing a line through the two points containing theintercepts. The graph of is shown in Figure 1.44.

Check Point 8 Graph using intercepts: 3x - 2y - 6 = 0.

4x - 3y - 6 = 0

10, -22,-2,y-intercept

-3. y = -2

-3y = 6

-3y - 6 = 0

4x - 3y - 6 = 0.x 4 # 0 - 3y - 6 = 0

y.x ! 0y-intercept.

A32 , 0 B32 ,x-intercept

x = 64

= 32

4x = 6

4x - 6 = 0

4x - 3y - 6 = 0.y 4x - 3 # 0 - 6 = 0

x.y ! 0x-intercept.

4x - 3y - 6 = 0.

EXAMPLE 8

! Use intercepts to graph thegeneral form of a line’s equation.

Using Intercepts to Graph 1. Find the Let and solve for Plot the point containing the

on the 2. Find the Let and solve for Plot the point containing the

on the 3. Use a straightedge to draw a line through the two points containing the

intercepts. Draw arrowheads at the ends of the line to show that the linecontinues indefinitely in both directions.

y-axis.y-intercepty.x = 0y-intercept.

x-axis.x-interceptx.y = 0x-intercept.

Ax " By " C ! 0

1

12345

2 3 4 5

2 3 4 5 1 2 3 4 5

y

x

y-intercept:� 2

( , 0)x-intercept:

32

(0, 2)

32

Figure 1.44 The graph of4x - 3y - 6 = 0

Check Point 7 Find the slope and the of the line whose equationis Then use the and the slope to graph theequation.

Using Intercepts to Graph Example 7 and Check Point 7 illustrate that one way to graph the general form of aline’s equation is to convert to slope-intercept form, Then use theslope and the to obtain the graph.

A second method for graphing uses intercepts. Thismethod does not require rewriting the general form in a different form.

Ax + By + C = 0y-intercept

y = mx + b.

Ax " By " C ! 0

y-intercept3x + 6y - 12 = 0.y-intercept


P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 186


The slope-intercept form of a line’s equation, can be expressedin function notation by replacing with

We have seen that functions in this form are called linear functions. Thus, in theequation of a linear function, the is the line’s slope and the constantterm is the Here are two examples:

If a linear function’s equation is in slope-intercept form, we can use theand the slope to obtain its graph.y-intercept

The slope is 2. The y -intercept is 4. The y -intercept is 2.

y=2x-4 f(x)= x+2.12

12The slope is .

y-intercept.coefficient of x

f1x2 = mx + b.

f1x2:yy = mx + b,


1. Plot the point containing the on the This is the point 2. Obtain a second point using the slope, Write as a fraction, and use rise

over run, starting at the point containing the to plot this point.3. Use a straightedge to draw a line through the two points. Draw arrowheads

at the ends of the line to show that the line continues indefinitely in bothdirections.

y-intercept,mm.

10, b2.y-axis.y-intercept

y-Intercepty ! mx " b

Study TipWriting the slope, as a fractionallows you to identify the rise (thefraction’s numerator) and the run(the fraction’s denominator).

m,


Graph the linear function:

Solution The equation of the line is in the form We can find theslope, by identifying the coefficient of We can find the byidentifying the constant term.

Now that we have identified the slope, and the 2, we use the three-step procedure to graph the equation.

Step 1 Plot the point containing the on the The is 2.We plot (0, 2), shown in Figure 1.41.

Step 2 Obtain a second point using the slope, Write as a fraction, and use riseover run, starting at the point containing the to plot this point. Theslope, is already written as a fraction.

We plot the second point on the line by starting at (0, 2), the first point. Based on theslope, we move 3 units down (the rise) and 2 units to the right (the run). This puts usat a second point on the line, shown in Figure 1.41.

Step 3 Use a straightedge to draw a line through the two points. The graph ofthe linear function is shown as a blue line in Figure 1.41.

Check Point 4 Graph the linear function: f1x2 = 35 x + 1.

f1x2 = - 32 x + 2

12, -12,m = -

32

= -32

= RiseRun

- 32 ,

y-intercept,mm.

y-intercepty-axis.y-intercept

y-intercept,- 32 ,

The slope is 32 .

f(x)=– x+232

The y -intercept is 2.

y-intercept, b,x.m,f1x2 = mx + b.

f1x2 = - 32

x + 2.


1

12345

2 3 4 5

1 3 4 5 1 2 3 4 5

y

x

Rise � 3

Run � 2

y -intercept: 2

Figure 1.41 The graph off1x2 = -

32 x + 2

P-BLTZMC01_135-276-hr 17-11-2008 11:46 Page 183

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2.!!C3x!+!4y!=!12!! ! ! ! !

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