College Algebra
Lial Hornsby Schneider DanielsEleventh Edition
College A
lgebra Lial et al. Eleventh Edition
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Printed in the United States of Am erica
ISBN 10: 1-292-02380-5ISBN 13: 978-1-292-02380-9
ISBN 10: 1-292-02380-5ISBN 13: 978-1-292-02380-9
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
Graphs and Functions
231
Graphs and Functions
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.
■✔ Now Try Exercise 13.
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.
■✔ Now Try Exercise 35.
Graphs and Functions
233
Graphs and Functions
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
■✔ Now Try Exercise 39.
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