college algebra lial et al. eleventh edition college

College Algebra

Lial Hornsby Schneider DanielsEleventh Edition

College A

lgebra Lial et al. Eleventh Edition

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A catalogue record for this book is available from the Brit ish Library

Printed in the United States of Am erica

ISBN 10: 1-292-02380-5ISBN 13: 978-1-292-02380-9

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Graphs and Functions

(Modeling) Cost, Revenue, and Profit Analysis A firm will break even (no profit and no

loss) as long as revenue just equals cost. The value of x (the number of items produced

and sold) where C1x2 = R1x2 is called the break-even point. Assume that each of the

following can be expressed as a linear function. Find

(a) the cost function, (b) the revenue function, and (c) the profit function.

(d) Find the break-even point and decide whether the product should be produced,

given the restrictions on sales.

See Example 9.

Fixed Cost Variable Cost Price of Item

87. $ 500 $ 10 $ 35 No more than 18 units can be sold.

88. $2700 $150 $280 No more than 25 units can be sold.

89. $1650 $400 $305 All units produced can be sold.

90. $ 180 $ 11 $ 20 No more than 30 units can be sold.

(Modeling) Break-Even Point The manager of a small company that produces roof tile

has determined that the total cost in dollars, C1x2, of producing x units of tile is given by

C1x2 = 200x + 1000,

while the revenue in dollars, R1x2, from the sale of x units of tile is given by

R1x2 = 240x.

91. Find the break-even point and the cost and revenue at the break-even point.

92. Suppose the variable cost is actually $220 per unit, instead of $200. How does this

affect the break-even point? Is the manager better off or not?

1. For A1-4, 22 and B1-8, -32 , find d1A, B2 , the distance between A and B .

2. Two-Year College Enrollment Enrollments

in two-year colleges for selected years are

shown in the table. Use the midpoint formula

to estimate the enrollments for 2002 and

2006.

3. Sketch the graph of y = -x2 + 4 by plotting points.

4. Sketch the graph of x2 + y2 = 16 .

5. Determine the radius and the coordinates of the center of the circle with equation

x2 + y2 - 4x + 8y + 3 = 0 .

Quiz (Sections 1–4)

Source: U.S. Center for Education

Statistics.

Year Enrollment (in millions)

2000 5.95

2004 6.55

2008 6.97

230

Point-Slope Form The graph of a linear function is a straight line. We now

develop various forms for the equation of a line.

Figure 43 shows the line passing through the fixed point 1x1, y12 having

slope m . (Assuming that the line has a slope guarantees that it is not vertical.) Let

1x, y2 be any other point on the line. Since the line is not vertical, x - x1 � 0.

Now use the definition of slope.

m =y - y1

x - x1

Slope formula (Section 4)

m1x - x12 = y - y1 Multiply each side by x - x1.

or y - y1 = m1x - x12 Interchange sides.

This result is the point-slope form of the equation of a line.

5 Equations of Lines and Linear Models

■ Point-Slope Form

■ Slope-Intercept Form

■ Vertical and Horizontal

Lines

■ Parallel and

Perpendicular Lines

■ Modeling Data

■ Solving Linear Equations

in One Variable by

Graphing

Figure 43

0x

y

(x, y)(x1, y1)

Fixed point

Slope = mAny otherpoint onthe line Point-Slope Form

The point-slope form of the equation of the line with slope m passing

through the point 1x1, y12 is

y � y1 � m 1x � x1 2 .

For Exercises 6–8, refer to the graph of ƒ1x2 = � x + 3 �.

6. Find ƒ1-12. 7. Give the domain and the range of ƒ.

8. Give the largest interval over which the function ƒ is

(a) decreasing, (b) increasing, (c) constant.

9. Find the slope of the line through the given points.

(a) 11, 52 and 15, 112 (b) 1-7, 42 and 1-1, 42 (c) 16, 122 and 16, -42 10. Motor Vehicle Sales The graph shows a straight line segment that approximates new

motor vehicle sales in the United States from 2005 to 2009. Determine the average rate

of change from 2005 to 2009, and interpret the results.

Num

ber

of

Veh

icle

s

(in t

housa

nds)

Source: U.S. Bureau of Economic Analysis.

Year

New Vehicle Sales

2005 2007 2009

20,000

15,000

10,000

0

5,000

17,445

10,601

x

y

(–3, 0)

(–1, 2)

(–5, 2)

0

f(x) = | x+ 3|

x

y


231


Slope-Intercept Form As a special case of the point-slope form of the equa-

tion of a line, suppose that a line passes through the point 10, b2, so the line has

y -intercept b . If the line has slope m , then using the point-slope form with x1 = 0

and y1 = b gives the following.

y - y1 = m1x - x12 Point-slope form

y - b = m1x - 02 x1 = 0, y1 = b

y = mx + b Solve for y.

Slope y- intercept

Since this result shows the slope of the line and the y -intercept, it is called the

slope-intercept form of the equation of the line.

EXAMPLE 1 Using the Point-Slope Form (Given a Point and the Slope)

Write an equation of the line through 1-4, 12 having slope -3.

SOLUTION Here x1 = -4, y1 = 1, and m = -3.

y - y1 = m1x - x12 Point-slope form

y - 1 = -33x - 1-424 x1 = -4, y1 = 1, m = -3

y - 1 = -31x + 42

y - 1 = -3x - 12 Distributive property

y = -3x - 11 Add 1.

Be careful with signs.

■✔ Now Try Exercise 23.

EXAMPLE 2 Using the Point-Slope Form (Given Two Points)

Write an equation of the line through 1-3, 22 and 12, -42. Write the result in

standard form Ax + By = C.

SOLUTION Find the slope first.

m =-4 - 2

2 - 1-32 = - 6

5 Definition of slope

The slope m is - 65 . Either 1-3, 22 or 12, -42 can be used for 1x1, y12. We

choose 1-3, 22. y - y1 = m1x - x12 Point-slope form

y - 2 = - 6

5 3x - 1-324 x1 = -3, y1 = 2, m = - 65

51y - 22 = -61x + 32 Multiply by 5.

5y - 10 = -6x - 18 Distributive property

6x + 5y = -8 Standard form (Section 4)

Verify that we obtain the same equation if we use 12, -42 instead of 1-3, 22 in

the point-slope form.


LOOKING AHEAD TO CALCULUS

A standard problem in calculus is to

find the equation of the line tangent to

a curve at a given point. The deriva-

tive (see Looking Ahead to Calculus in

Section 4 ) is used to find the slope of

the desired line, and then the slope and

the given point are used in the point-

slope form to solve the problem.

NOTE The lines in Examples 1 and 2 both have negative slopes. Keep in

mind that a slope of the form - AB may be interpreted as either -AB or A

-B .

232

Slope-Intercept Form

The slope-intercept form of the equation of the line with slope m and

y -intercept b is

y � mx � b.

NOTE Generalizing from Example 3, the slope m of the graph of

Ax + By = C

is - AB , and the y -intercept b is CB .

EXAMPLE 4 Using the Slope-Intercept Form (Given Two Points)

Write an equation of the line through 11, 12 and 12, 42 . Then graph the line

using the slope-intercept form.

SOLUTION In Example 2, we used the point-slope form in a similar problem.

Here we show an alternative method using the slope-intercept form . First, find

the slope.

m =4 - 1

2 - 1=

3

1= 3 Definition of slope

Now substitute 3 for m in y = mx + b and choose one of the given points, say

11, 12 , to find the value of b.

y = mx + b Slope-intercept form

1 = 3112 + b m = 3, x = 1, y = 1

y-intercept b = -2 Solve for b.

The slope-intercept form is

y = 3x - 2 .

The graph is shown in Figure 44 . We can plot 10, -22 and then use the defini-

tion of slope to arrive at 11, 12 . Verify that 12, 42 also lies on the line.

■✔ Now Try Exercise 13. Figure 44

(1, 1)

(2, 4)

(0, –2)

y = 3x – 2

y changes 3 units

x changes 1 unit

x0

y

EXAMPLE 3 Finding the Slope and y -Intercept from an Equation of a Line

Find the slope and y -intercept of the line with equation 4x + 5y = -10.

SOLUTION Write the equation in slope-intercept form.

4x + 5y = -10

5y = -4x - 10 Subtract 4x.

y = - 4

5 x - 2 Divide by 5.

m b

The slope is - 45 and the y -intercept is -2.



233


Vertical and Horizontal Lines We first saw graphs of vertical and horizon-

tal lines in Section 4. The vertical line through the point 1a, b2 passes through

all points of the form 1a, y2, for any value of y . Consequently, the equation of

a vertical line through 1a, b2 is x = a. For example, the vertical line through

1-3, 12 has equation x = -3. See Figure 46 (a). Since each point on the y -axis

has x -coordinate 0, the equation of the y-axis is x � 0.

The horizontal line through the point 1a, b2 passes through all points of the

form 1x, b2, for any value of x . Therefore, the equation of a horizontal line through

1a, b2 is y = b. For example, the horizontal line through 11, -32 has equation

y = -3. See Figure 46 (b). Since each point on the x -axis has y -coordinate 0,

the equation of the x-axis is y � 0.

EXAMPLE 5 Finding an Equation from a Graph

Use the graph of the linear function ƒ shown in

Figure 45 to complete the following.

(a) Find the slope, y -intercept, and x -intercept.

(b) Write the equation that defines ƒ .

SOLUTION

(a) The line falls 1 unit each time the x -value in-

creases by 3 units. Therefore, the slope is

-13 = - 13 . The graph intersects the y -axis at the

point 10, -12 and intersects the x -axis at the

point 1-3, 02. Therefore, the y -intercept is -1

and the x -intercept is -3.

(b) The slope is m = - 13 , and the y -intercept is b = -1.

y = ƒ1x2 = mx + b Slope-intercept form

ƒ1x2 = - 1

3 x - 1 m = - 13 , b = -1


Figure 45

y

y = f (x)

x0

3

1

–1–3

Equations of Vertical and Horizontal Lines

An equation of the vertical line through the point 1a, b2 is x � a.

An equation of the horizontal line through the point 1a, b2 is y � b.

(a)

x

y

(–3, 1)

0

x = –3

Verticalline

y

x0

1

y = –3

(1, –3)

Horizontalline

(b)

Figure 46

Parallel and Perpendicular Lines Since two parallel lines are equally “steep,”

they should have the same slope. Also, two distinct lines with the same “steepness”

are parallel. The following result summarizes this discussion. (The statement “ p if

and only if q ” means “if p then q and if q then p .”)

Parallel Lines

Two distinct nonvertical lines are parallel if and only if they have the same

slope.

234

college algebra lial et al. eleventh edition college

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