140
(3-20)
Chapter 3
Graphs and Functions in the Cartesian Coordinate System
Selling price (in thousands of dollars)
a) Use the graph on the next page to estimate the average
retail price of a 2-year-old car in 1998. $12,000
b) Find the slope of the line shown in the figure. 1250
c) Use the slope to predict the price of a 2-year-old car.
$12,345
15
(1, 13,595)
(3, 11,095)
10
5
0
0.5
1.0
1.5
2.0 2.5 3.0
Age (in years)
3.5
4.0
4.5
FIGURE FOR EXERCISE 52
53. The points (3, ) and ( ,7) are on the line that passes
through (2, 1) and has slope 4. Find the missing coordinates
of the points. (3, 5), (0, 7)
54. If a line passes through (5, 2) and has slope 2, then what is
3
the value of y on this line when x 8, x 11, and x 12?
4, 6, 6 23
55. Find k so that the line through (2, k) and (3, 5) has
slope 1. 5
2
56. Find k so that the line through (k, 3) and (2, 0) has slope 3.
3 or 1
57. What is the slope of a line that is perpendicular to a line
with slope 0.247?
4.049
58. What is the slope of a line that is perpendicular to the line
through (3.27, 1.46) and (5.48, 3.61)?
1.726
GET TING MORE INVOLVED
59. Writing. What is the difference between zero slope and
undefined slope?
A horizontal line has a zero slope and a vertical line has undefined slope.
3.3
In this
section
●
Point-Slope Form
●
Slope-Intercept Form
●
Standard Form
●
Using Slope-Intercept Form
for Graphing
●
Linear Functions
61. Exploration. A rhombus is a quadrilateral with four equal
sides. Draw a rhombus with vertices (3, 1), (0, 3),
(2, 1), and (5, 3). Find the slopes of the diagonals of the
rhombus. What can you conclude about the diagonals of
this rhombus?
2, 1, perpendicular
2
MISCELL ANEOUS
2
60. Writing. Is it possible for a line to be in only one quadrant?
Two quadrants? Write a rule for determining whether a line
has positive, negative, zero, or undefined slope from knowing in which quadrants the line is found.
Every line goes through at least two quadrants. A nonhorizontal, nonvertical line that misses quadrant II or IV or
both has a positive slope. A nonhorizontal, nonvertical
line that misses quadrant I or III or both has a negative
slope.
62. Exploration. Draw a square with vertices (5, 3),
(3, 3), (1, 5), and (3, 1). Find the slopes of the diagonals
of the square. What can you conclude about the diagonals of
this square?
2, 1, perpendicular
2
GR APHING C ALCUL ATOR
EXERCISES
63. Graph y 1x, y 2x, y 3x, and y 4x together in
the standard viewing window. These equations are all of
the form y mx. What effect does increasing m have on
the graph of the equation? What are the slopes of these four
lines?
Increasing m makes the graph increase faster. The slopes of
these lines are 1, 2, 3, and 4.
64. Graph y 1x, y 2x, y 3x, and y 4x
together in the standard viewing window. These equations
are all of the form y mx. What effect does decreasing m have on the graph of the equation? What are the
slopes of these four lines?
Decreasing m makes the graph decrease faster. The slopes
of these lines are 1, 2, 3, and 4.
THREE FORMS FOR THE
EQUATION OF A LINE
In Section 3.1 you learned how to graph a straight line corresponding to a linear
equation. The line contains all of the points that satisfy the equation. In this section
we start with a line or a description of a line and write an equation corresponding to
the line.
Point-Slope Form
Figure 3.18 shows the line that has slope 2 and contains the point (3, 5). In Sec3
tion 3.2 you learned that the slope is the same no matter which two points of the line
3.3
Because the slope of this line is 23, we can
write
No two students learn in the
same way or at the same
speed. No one can tell you exactly how to study and learn.
Learning is personal.You must
discover what it takes for you
to learn mathematics and
then to do whatever it takes.
y–5
(3, 5)
x–3
2
1
–1
–1
Multiplying each side by x 3, we get
tip
(x, y)
8
7
6
5
4
y5
m .
x3
y5 2
.
x3 3
141
y
are used to calculate it. So if we find the
slope m for this line using an arbitrary
point of the line, say (x, y), and the specific point (3, 5), we get
study
(3-21)
Three Forms for the Equation of a Line
1
2
3
4
5
6
7
8
9
x
FIGURE 3.18
2
y 5 (x 3).
3
Because (x, y) was an arbitrary point on the line, this equation is satisfied by every
point on the line.
If we use (x1, y1) as the specific point and (x, y) as an arbitrary point on a line
with slope m, we can write
yy
1 m.
x x1
Multiplying each side of this equation by x x1 gives us the point-slope form of
the equation of the line.
Point-Slope Form
The equation of the line through (x1, y1) with slope m in point-slope form is
y y1 m(x x1).
E X A M P L E
1
calculator
Solution
Use x1 2, y1 5, and m 3 in the point-slope form:
close-up
Graph y 3x 1 and
check that the line goes
through (2, 5) by using the
TRACE feature.
10
10
Writing an equation for a line given a point and the slope
Find an equation for the line through (2, 5) with slope 3 and solve it for y.
y 5 3[x (2)]
Now solve the equation for y:
y 5 3[x 2]
y 5 3x 6
y 3x 1
■
10
10
If you know two points on a line, then you can graph the line (two points determine a line). In the next example we will see that two points of a line also determine
an equation for the line.
142
(3-22)
Chapter 3
E X A M P L E
2
calculator
close-up
Graph y 34 x 1 and
4
check that the line goes
through (3, 2) and (1, 1) by
using the TRACE feature.
10
10
10
10
Graphs and Functions in the Cartesian Coordinate System
Writing an equation for a line given two points on the line
Find an equation for the line through (3, 2) and (1, 1) and solve it for y.
Solution
We are not given the slope, but we can find it because the points (3, 2) and (1, 1)
are on the line:
1 (2)
m
1 3
3
3
4
4
Now use this slope and one of the points, say (3, 2), to write the equation in pointslope form:
3
y (2) (x 3) Point-slope form
4
3
9
y 2 x Distributive property
4
4
3
1
9
9 8 1
y x Solve for y: 2 .
4
4 4 4
4
4
Note that we would get the same equation if we had used slope 3 and the other
4
■
point (1, 1). Try it.
For the next example, recall that if a line has slope m, then the slope of any line
perpendicular to it is 1, provided that m 0.
m
E X A M P L E
study
3
tip
When taking a test, put a
check mark beside every
question that you have answered and checked. When
you have finished the test,
then you can go back and
spend the remaining time on
the problems that are not yet
checked. You will not waste
time reworking problems that
you know are correct.
An equation of a line perpendicular to another line
Line l goes through (2, 0) and is perpendicular to the line through (5, 1) and
(1, 3). Find the equation of line l and then solve it for y.
Solution
First find the slope of the line through (5, 1) and (1, 3):
3 (1)
2
4
m 1 5
6
3
Because line l is perpendicular to this line, line l has slope 3. Now use (2, 0) and the
2
slope 32 in the point-slope formula to get the equation of line l:
3
y 0 (x 2)
2
3
y x 3 Distributive property
2
■
calculator close-up
With slope-intercept form and a graphing calculator, it is
easy to see how the slope affects the steepness of a line.
The graphs of y1 1x, y2 x, y3 2x, and y4 3x are
2
all shown on the accompanying screen.
10
10
10
10
3.3
Three Forms for the Equation of a Line
(3-23)
143
Slope-Intercept Form
The line y 3x 1 in Example 1 has slope 3. To find the y-intercept of
this line, let x 0 in y 3x 1: y 3(0) 1 1. The y-intercept is
(0, 1). Its y-coordinate appears in the equation:
y 3x 1
↑
Slope
↑
y-intercept (0, 1)
Because the slope and y-intercept can be read from the equation when it is solved
for y, this form of the equation of the line is called slope-intercept form.
Slope-Intercept Form
The equation of a line in slope-intercept form is
y mx b,
where m is the slope and (0, b) is the y-intercept.
E X A M P L E
4
Writing an equation given its slope and y-intercept
Write the slope-intercept form of the equation of the line shown in Fig. 3.19.
y
4
2
1
– 3 –2 –1
–1
1
2
3
x
4
Solution
From Fig. 3.19 we see that the y-intercept is (0, 3). If we start at the y-intercept and
move down 2 and 3 to the right, we get to another point on the line. So the slope is
2
3. The equation of this line in slope-intercept form is
2
y x 3.
3
FIGURE 3.19
■
Standard Form
study
If x students paid $5 each and y adults paid $7 each to attend a play for which the
ticket sales totaled $1900, then we can write the equation 5x 7y 1900. This
form of a linear equation is common in applications. It is called standard form.
tip
Get in the habit of checking
your work and having confidence in your answers.The answers to the odd-numbered
exercises are in the back of this
book, but you should look in
the answer section only after
you have checked on your
own. You will not always have
an answer section available.
E X A M P L E
5
Standard Form
The equation of a line in standard form is
Ax By C,
where A, B, and C are real numbers with A and B not both zero.
The numbers A, B, and C in standard form can be any real numbers, but it is a
common practice to write standard form using only integers and a positive coefficient for x.
Changing to standard form
Write the equation y 12 x 34 in standard form using only integers and a positive coefficient for x.
144
(3-24)
Chapter 3
helpful
hint
Solve Ax By C for y, to
get
A
C
y x .
B
B
So the slope of Ax By C is
A
. This fact can be used in
B
checking standard form. The
slope of 2x 4y 3 in Exam2
1
ple 5 is or , which is the
4
Graphs and Functions in the Cartesian Coordinate System
Solution
Use the properties of equality to get the equation in the form Ax By C:
3
1
y x Original equation
2
4
1
3
1
x y Subtract x from each side.
2
2
4
1
3
4 x y 4 Multiply each side by 4 to
2
4
get integral coefficients.
2x 4y 3
Distributive property
2x 4y 3
Multiply by 1 to make the
coefficient of x positive.
2
■
slope of the original equation.
To find the slope and y-intercept of a line written in standard form, we convert
the equation to slope-intercept form.
E X A M P L E
helpful
6
hint
Note that every term in a linear equation in two variables
is either a constant or a multiple of a variable. That is why
equations in one variable of
the form ax b 0 were
called linear equations in
Chapter 2.
E X A M P L E
7
Changing to slope-intercept form
Find the slope and y-intercept of the line 3x 2y 5.
Solution
Solve for y to get slope-intercept form:
3x 2y 5
2y 3x 5
3
5
y x 2
2
Original equation
Subtract 3x from each side.
Divide each side by 2.
The slope is 23 , and the y-intercept is 0, 52 .
■
You learned in Section 3.1 that the graph of the equation x 4 is a vertical line.
Because slope is undefined for vertical lines, the equation of this line cannot be
written in slope-intercept form or point-slope form. Only nonvertical lines can be
written in those forms. However, a vertical line can be written in standard form. For
example,
x4
can be written as
1 x 0 y 4.
Every line has an equation in standard form.
Finding the equation of a line
Write an equation in standard form with integral coefficients for the line l through
(2, 5) that is perpendicular to the line 2x 3y 1.
Solution
First solve the equation 2x 3y 1 for y to find its slope:
2x 3y 1
3y 2x 1
2
1
y x The slope is 23.
3
3
3.3
3
y 5 (x 2)
2
3
y 5 x 3
2
3
y x 2
2
3
x y 2
2
3x 2y 4
close-up
Graph y1 = 23 x 13 and
y2 32 x 2 to check that y2
is perpendicular to y1 and that
y2 goes through (2, 5).The lines
will look perpendicular only if
the same unit length is used
on both axes.
10
Some calculators have a feature that adjusts the window
to get the same unit length on
both axes.
E X A M P L E
Distributive property
Multiply each side by 2.
In the slope-intercept form, a point on the line (the y-intercept) and the slope are
readily available. To graph a line, we can start at the y-intercept and count off the
rise and run to get a second point on the line.
8
Using slope and y-intercept to graph
Graph the line 2x 3y 3.
Solution
First write the equation in slope-intercept form:
y
y=
2
—
x
3
+1
2x 3y 3
3y 2x 3
2
y x 1
3
Run 3
Rise 2
y-intercept
–1
–1
Point-slope form
Using Slope-Intercept Form for Graphing
10
–3
145
So 3x 2y 4 is the standard form of the equation of the line through (2, 5) that
■
is perpendicular to 2x 3y 1.
15
5
4
3
(3-25)
The slope of line l is the opposite of the reciprocal of 2. So line l has slope 3 and
3
2
goes through (2, 5). Now use the point-slope form to write the equation:
calculator
15
Three Forms for the Equation of a Line
1
2
3
4
–2
–3
–4
FIGURE 3.20
5
x
The slope is 23, and the y-intercept is (0, 1). Start at (0, 1) on the y-axis, then rise
2 and run 3 to locate a second point on the line. Because there is only one line containing any two given points, these two points determine the line. See Fig. 3.20. ■
The three methods that we used for graphing linear equations are summarized as
follows.
Methods for Graphing a Linear Equation
1. Arbitrarily select some points that satisfy the equation, and draw a line
through them.
2. Find the x- and y-intercepts (provided that they are not the origin), and draw
a line through them.
3. Start at the y-intercept and use the slope to locate a second point, then draw
a line through the two points.
146
(3-26)
Chapter 3
Graphs and Functions in the Cartesian Coordinate System
If the y-coordinate of the y-intercept is an integer and the slope is a rational
number, then it is usually the easiest to use the y-intercept and slope.
Linear Functions
The linear equation y mx b with m 0 is a formula that shows how to determine a value of y from a value of x. We say that y is a linear function of x. Functions
in general will be discussed in Section 3.5. In the next example we use the pointslope formula to write Fahrenheit temperature as a linear function of Celsius
temperature.
E X A M P L E
9
Writing a linear function given two points
Fahrenheit temperature F is a linear function of Celsius temperature C. Water
freezes at 0°C or 32°F and boils at 100°C or 212°F. Find the linear equation that
expresses F as a linear function of C.
Solution
We want the equation of the line that contains the points (0, 32) and (100, 212) as
shown in Fig. 3.21. Use C as the independent variable (x) and F as the dependent
variable (y). The slope of the line is
F2 F1
212 32 180 9
m
.
C2 C1
100 0
100 5
Degrees
Fahrenheit
F
200
150
100
50
–50
Water boils
(100, 212)
Water freezes
(0, 32)
25 50 75 100
C
Degrees Celsius
FIGURE 3.21
Using a slope of 95 and the point (100, 212) in the point-slope formula, we get
9
F 212 (C 100).
5
We can solve this equation for F to get the familiar formula relating Celsius and
Fahrenheit temperature:
9
F C 32
5
Because we knew the intercept (0, 32), we could have used it and the slope 9 in
5
■
slope-intercept form to write F 95C 32.
WARM-UPS
True or false? Explain your answer.
1. There is exactly one line through a given point with a given slope. True
2. The line y a m(x b) goes through (a, b) and has slope m. False
3. The equation of the line through (a, b) with slope m is y mx b. False
3.3
WARM-UPS
(3-27)
147
(continued)
4.
5.
6.
7.
8.
9.
10.
3. 3
Three Forms for the Equation of a Line
The x-coordinate of the y-intercept of a nonvertical line is 0. True
The y-coordinate of the x-intercept of a nonhorizontal line is 0. True
Every line in the xy-plane has an equation in slope-intercept form. False
3
The line 2y 3x 7 has slope 2. True
The line y 3x 1 is perpendicular to the line y 1 x 1. False
3
The line 2y 3x 5 has a y-intercept of (0, 5). False
Every line in the xy-plane has an equation in standard form. True
EXERCISES
Reading and Writing After reading this section, write out the
answers to these questions. Use complete sentences.
1. What is point-slope form?
Point-slope form is y y1 m(x x1), where m is the
slope and (x1, y1) is a point on the line.
2. What is slope-intercept form?
Slope-intercept form is y mx b, where m is the slope
and (0, b) is the y-intercept.
3. What two bits of information must you have to write the
equation of a line from a description of the line?
To write an equation of a line, we need the slope and a point
on the line.
4. What is standard form?
Standard form is Ax Bx C, where A, B, and C are
real numbers with A and B not both zero.
5. How do you find the slope of a line when its equation is
given in standard form?
To find the slope from standard form, solve the equation
for y to get the form y mx b, where m is the slope.
6. How do you graph a line when its equation is given in
slope-intercept form?
To graph a line knowing the slope and y-intercept, start at
the y-intercept and count off the rise and run to locate a second point. Then draw a line through the y-intercept and
your second point.
11. Line l goes through (2, 3) and (5, 6).
27
3
y x 7
7
12. Line l goes through (2, 1) and (3, 4).
y x 1
13. Line l goes through (3, 4) and is perpendicular to the line
through (3, 1) and (5, 1).
y 4x 8
14. Line l goes through (0, 0) and is perpendicular to the line
through (0, 6) and (5, 0).
5
y x
6
15. Line l goes through (0, 0) and is parallel to the line through
(9, 3) and (3, 6).
3
y x
4
16. Line l goes through (2, 4) and is parallel to the line
through (6, 2) and (2, 6).
1
y x 3
2
In Exercises 17–26, write an equation in slope-intercept form
(if possible) for each of the lines shown. See Example 4.
y
17.
Find the equation of line l in each case and solve it for y. See
Examples 1–3.
5
4
3
7. Line l goes through (2, 3) and has slope 2.
y 2x 7
8. Line l goes through (2, 5) and has slope 6.
y 6x 17
9. Line l goes through (2, 3) and has slope 12.
1
y x 2
2
10. Line l goes through (3, 5) and has slope 2.
3
2
y x 3
3
1
–3 –2 –1
–1
–2
–3
1
y x 2
2
1 2
3 4
x
148
(3-28)
Chapter 3
Graphs and Functions in the Cartesian Coordinate System
y
18.
y
19.
3
2
1
–4 –3
–1
–1
1 2
3
4
–3 –2 –1
–1
–2
–3
–4
–3
–4
–5
2
y x 2
3
1
2
3
4
x
23.
4
3
1
1
–2
–3
–4
1 2
3
x
–4 –3 –2 –1
–1
–2
–3
–4
1
2
3
4
x
1
3
4
5
x
1 2
3
4
y
–3 –2 –1
–1
–2
–3
26.
y
5
4
3
1
2 3
4
x
1
–3 –2 –1
–1
–2
–3
x
1
y x 2
2
y x 2
Write each equation in standard form using only integers and a
positive coefficient for x. See Example 5.
1
27. y x 2 x 3y 6
3
1
28. y x 7 x 2y 14
2
1
29. y 5 (x 3) x 2y 13
2
1
30. y 1 (x 6) x 4y 2
4
x
3
y x 3
2
4
3
–1
4
4
3
2
1
y
25.
3
y 2
yx
y
1 2
–4 – 3 – 2 – 1
–1
x
–3
–4
–5
–4 –3 –2 –1
–1
–2
–3
–4
y x
y 3x 3
5
4
3
2
1
–4 –3 –2 –1
–1
–2
–3
–4
–5 –4 –3 –2
4
y
22.
4
3
2
1
24.
2 3
x1
y
21.
3
2
1
4
3
2
1
x
y
20.
1 1
31. y (x 4) 2x 6y 11
2 3
1 1
32. y (x 3) 3x 12y 13
3 4
33. 0.05x 0.06y 8.9 0 34. 0.03x 0.07y 2
5x 6y 890
3x 7y 200
Write each equation in slope-intercept form, and identify the
slope and y-intercept. See Example 6.
35. 2x 5y 1
2
1 2
1
y x , , 0, 5
5 5
5
3.3
37. 3x y 2 0
(3-29)
149
61. Line l goes through (2, 5) and is parallel to the x-axis.
y5
2
2
y x , 1, 0, 3
3
36. 3x 3y 2
Three Forms for the Equation of a Line
62. Line l goes through (1, 6) and is parallel to the y-axis.
x 1
y 3x 2, 3, (0, 2)
39. y 3 5
1
5 1
5
y x , , 0, 2
2 2
2
y 2, 0, (0, 2)
Graph each line. Use the slope and y-intercept when possible.
See Example 8.
40. y 9 0
y 9, 0, (0, 9)
1
63. y x
2
2
64. y x
3
65. y 2x 3
66. y x 1
2
67. y x 2
3
68. y 3x 4
69. 3y x 0
70. 4y x 0
38. 5 x 2y 0
41. y 2 3(x 1)
y 3x 1, 3, (0, 1)
42. y 4 2(x 5)
y5 3
43. x4 2
y 2x 6, 2, (0, 6)
3
3
y x 11, , (0, 11)
2
2
y6
3
44. x2
5
3
36 3
36
y x , , 0, 5
5
5
5
7 1
7
1
y x , , 0, 12 3
12
3
1 1
1
45. y x 2 3
4
1
1
1
46. y x 3
2
4
1
11 1
11
y x , , 0, 2
24 2
24
47. y 6000 0.01(x 5700)
y 0.01x 6057, 0.01, (0, 6057)
48. y 5000 0.05(x 1990)
y 0.05x 4900.5, 0.05, (0, 4900.5)
Find the equation of line l in each case and then write it in
standard form with integral coefficients. See Example 7.
49. Line l has slope 1 and goes through (0, 5).
2
x 2y 10
50. Line l has slope 5 and goes through 0, 1 .
2
10x 2y 1
51. Line l has x-intercept (2, 0) and y-intercept (0, 4).
2x y 4
52. Line l has y-intercept (0, 5) and x-intercept (4, 0).
5x 4y 20
53. Line l goes through (2, 1) and is parallel to y 2x 6.
2x y 5
54. Line l goes through (1, 3) and is parallel to y 3x 5.
3x y 0
55. Line l is parallel to 2x 4y 1 and goes through (3, 5).
x 2y 7
56. Line l is parallel to 3x 5y 7 and goes through (2, 4).
3x 5y 14
57. Line l goes through (1, 1) and is perpendicular to
y 1 x 3. 2x y 3
2
58. Line l goes through (1, 2) and is perpendicular to
y 3x 7. x 3y 5
59. Line l goes through (2, 3) and is perpendicular to
x 3y 4. 3x y 9
60. Line l is perpendicular to 2y 5 3x 0 and goes
through (2, 7). 2x 3y 25
150
(3-30)
Chapter 3
Graphs and Functions in the Cartesian Coordinate System
71. y x 3 0
72. x y 4
Solve each problem. See Example 9.
85. Heating water. Suppose the temperature, t, of a cup of
water is a linear function of the number of seconds, s, that
it is in the microwave. If the temperature at s 0 second is
t 60°F and the temperature at s 120 seconds is 200°F,
find the linear equation that expresses t as a function of s.
What should the temperature be after 30 seconds? (Hint:
Write the equation of the line containing the points (0, 60)
and (120, 200) in the form t ms b.) Draw a graph of
this linear function. t 7 s 60, 95F
73. 3x 2y 6
74. 3x 5y 10
86. Making circuit boards. The accountant at Apollo Manufacturing has determined that the cost, C, per week in
dollars for making circuit boards is a linear function of the
number, n, of circuit boards produced in a week. If
C $1500 when n 1000, and C $2000 when
n 2000, find the linear equation that expresses C in
terms of n. What is the cost if Apollo produces only one circuit board in a week? Draw a graph of this linear function.
C 12 n 1000, $1000.50
6
76. y 3 0
77. x 3 0
78. x 5 0
Determine whether each pair of lines is parallel or
perpendicular.
79. y 3x 8, x 3y 7 Perpendicular
1
1
1
80. y x 4, x y 1 Perpendicular
2
2
4
1
2
81. 2x 4y 9, x y 8 Parallel
3
3
1
1
1 1
1
82. x y , y x 2 Parallel
4
6
3 3
2
83. x 6 9, y 4 12 Perpendicular
1
84. 9 x 3, x 8 Parallel
2
Flow (thousands of ft3/sec)
75. y 2 0
87. Carbon dioxide emission. Worldwide emission of carbon
dioxide (CO2) increased from 14 billion tons in 1970 to
24 billion tons in 1995 (World Resources Institute,
www.wri.org).
a) Find the equation of the line through (1970, 14) and
(1995, 24). y 0.4x 774
b) Use the equation to predict the worldwide emission of
CO2 in 2005. 28 billion tons
88. World energy use. Worldwide energy use in all forms increased from the equivalent of 3.5 billion tons of oil in 1970
to the equivalent of 6 billion tons of oil in 1995 (World
Resources Institute, www.wri.org).
a) Find the equation of the line through (70, 3.5) and (95, 6).
y 0.1x 3.5
b) Use the equation to predict the worldwide energy use in
2005. 7 billion tons
89. Depth and flow. On May 1, 1998 the depth of the water in
the Tangipahoa River at Robert, Louisiana was 8.24 feet
and the flow was 1015.5 cubic feet per second (ft3/sec). On
May 8 the depth was 7.26 feet and the flow was 717.1 cubic
feet per second (U.S. Geological Survey, Water Resources
Data for Louisiana, 1998). The flow w is a linear function
of the depth d.
5
4
3
2
1
5
15
10
Depth (feet)
20
FIGURE FOR EXERCISE 89
3.4
a) Write the equation of the line through (8.24, 1015.5) and
(7.26, 717.1) and express w as a linear function of d.
b) What is the flow when the depth is 7.81 feet?
c) Is the flow increasing or decreasing as the depth
increases?
a) w 304.5d 1493.5 b) 884.6 ft3/sec
c) increasing
90. Buying stock. On July 2, 1998 a mutual fund manager
spent $5,031,250 on x shares of Ford Motor Stock at $58.25
per share and y shares of General Motors stock at $47.50
per share.
a) Write a linear equation that models this situation.
b) If 35,000 shares of Ford were purchased, then how
many shares of GM were purchased?
c) What are the intercepts of the graph of the linear equation? Interpret the intercepts.
d) As the number of shares of Ford increases, does the
number of shares of GM increase or decrease?
a) 58.25x 47.50y 5,031,250 b) 63,000
c) (0, 105,921.1), (86,373.4, 0), The intercepts give the
number of shares if all of the money was spent on
only one type of stock. d) decrease
GM shares (in thousands)
150
100
50
0
0
50
100
Ford shares (in thousands)
FIGURE FOR EXERCISE 90
3.4
In this
section
●
Definition
●
Graphing Linear Inequalities
●
The Test Point Method
●
Graphing Compound
Inequalities
●
Applications
Linear Inequalities and Their Graphs
(3-31)
151
GET TING MORE INVOLVED
91. Exploration. Plot the points (1, 1), (2, 3), (3, 4), (4, 6), and
(5, 7) on graph paper. Use a ruler to draw a straight line that
“best fits” the five points. The line drawn does not necessarily have to go through any of the five points.
a) Estimate the slope and y-intercept for the line drawn and
write an equation for the line in slope-intercept form.
b) For each x-coordinate from 1 through 5, find the
difference between the given y-coordinate and the ycoordinate on your line.
c) To determine how well you have done, square each
difference that you found in part (b) and then find the
sum of those squares. Compare your sum with your
classmates’ sums. The person with the smallest sum has
done the best job of fitting a line to the five given points.
G R A P H I N G C ALC U L ATO R
EXERCISES
92. Graph the equation y 0.5x 1 using the standard viewing window. Adjust the range of y-values so that the line
goes from the lower left corner of your viewing window to
the upper right corner.
93. Graph y x 3000, using a viewing window that shows
both the x-intercept and the y-intercept.
94. Graph y 2x 400 and y 0.5x 1 on the same
screen, using the viewing window 500 x 500 and
1000 y 1000. Should these lines be perpendicular?
Explain.
The lines are perpendicular and will appear so in a window
in which the length of one unit on the x-axis is equal to the
length of one unit on the y-axis.
95. The lines y 2x 3 and y 1.9x 2 are not parallel.
Find a viewing window in which the lines intersect.
Estimate the point of intersection.
The lines intersect at (50, 97).
LINEAR INEQUALITIES AND
THEIR GRAPHS
In the first three sections of this chapter you studied linear equations. We now turn
our attention to linear inequalities.
Definition
A linear inequality is a linear equation with the equal sign replaced by an inequality
symbol.
Linear Inequality
If A, B, and C are real numbers with A and B not both zero, then
Ax By C
is called a linear inequality. In place of , we can also use , , or .