Hungerford T.W., Shaw D.J. Contemporary Precalculus: A Graphing Approach
Подождите немного. Документ загружается.
There are four possibilities for a line L with slope m.
Properties of Slope
Points Slopes
(1, 1) and (0, 2) L
1
:
2
0
(
(
1
1
)
)
3
1
3
(0, 2) and (2, 4) L
2
:
4
2
2
0
2
2
1
(6, 2) and (3, 2) L
3
:
3
2
(
2
6)
0
9
0
(3, 5) and (3, 1) L
4
:
3
1
(
5
3)
6
6
1
(1, 0) and (2, 2) L
5
:
2
2
1
0
1
2
2 ■
As Examples 1 and 2 illustrate, the slope is a number m that measures how
steeply a line rises or falls, as summarized below.
56 CHAPTER 1 Basics
y
−1
−1
−2
−2−3−4−5−6
1
2
3
4
5
1
L
5
L
4
L
3
L
2
L
1
23
x
Figure 1–34
m ⬎ 0 m ⬍ 0 m ⫽ 0 m is not defined
The line rises from left The line falls from left The line is horizontal. The line is vertical.
to right. to right.
The larger m is, the more The larger m is, the more
steeply the line rises. steeply the line falls.
[See Example 1(a) and [See Example 1(b) and [See Example 1(c) and [See Example 1(d)]
lines L
1
and L
2
in lines L
4
and L
5
in line L
3
in Example 2.]
Example 2.] Example 2.]
x
y
x
y
x
y
x
y
SLOPE-INTERCEPT FORM
Let L be a nonvertical line with slope m and y-intercept b. Then (0, b) is a point on
L. Let (x, y) be any other point on L. Using the points (0, b) and (x, y) to compute
the slope of L (see Figure 1–35), we have
Slope of L
y
x
b
0
.
Since the slope of L is m, this equation becomes
m
y
x
b
Multiply both sides by x: mx y b
Rearrange terms: y mx b
y
(x, y)
y − b
x − 0
(0, b)
x
Figure 1–35
Thus the coordinates of any point on L satisfy the equation y mx b. So we
have the following fact.
EXAMPLE 3
List the slope and y-intercept of the line whose equation is given and describe its
graph.
(a) y 3x 5 (b) 2x y 7
SOLUTION
(a) The equation has the form in the preceding box, with m 3 and b 5,
so the line has slope 3 and y-intercept 5. This shows that the line rises from
left to right and passes through (0, 5).
(b) First, solve the equation for y: y 2x 7. Here the slope is m 2 and
the y-intercept is b 7. The line falls from left to right and passes through
(0, 7). ■
EXAMPLE 4
Show that the graph of 2y 5x 2 is a straight line. Find its slope, and graph the
line.
SOLUTION We begin by solving the equation for y:
Add 5x to both sides: 2y 5x 2
Divide both sides by 2: y 2.5x 1.
The equation now has the form in the preceding box, with m 2.5 and b 1.
Therefore, its graph is the line with slope 2.5 and y-intercept 1.
Since the y-intercept is 1, the point (0, 1) is on the graph. To find another
point on the line, choose a value for x, say, x 2, and compute the corresponding
value of y:
y 2.5x 1 2.5(2) 1 6.
Hence, (2, 6) is on the line. Plotting the line through (0, 1) and (2, 6) produces
Figure 1–36. ■
EXAMPLE 5
Describe and sketch the graph of the equation y 3.
SOLUTION We can write y 3 as y 0x 3. So its graph is a line with slope
0, which means that the line is horizontal, and y-intercept 3, which means that the
line crosses the y-axis at 3. This is sufficient information to obtain the graph in
Figure 1–37. ■
Example 5 is an illustration of this fact.
SECTION 1.4 Lines 57
Slope-Intercept
Form
The line with slope m and y-intercept b is the graph of the equation
y mx b.
4
6
2
2
y
x
Figure 1–36
y
1
−3
−1
2
x
Figure 1–37
Because slope is not defined for vertical lines, the equations of such lines
have a different form from those examined above.
EXAMPLE 6
We can easily list some points on the line in Figure 1–38: (2, 0), (2, 1), (2, 1),
(2, 1.5), and so on. Every point on this line has first coordinate 2. So every point
satisfies x 0y 2. Hence, the line is the graph of x 2. ■
Example 6 illustrates these facts:
POINT-SLOPE FORM
Suppose the line L passes through the point (x
1
, y
1
) and has slope m. Let (x, y) be
any other point on L. Using the points (x
1
, y
1
) and (x, y) to compute the slope m of
L (see Figure 1–39), we have
y
x
y
x
1
1
slope of L
y
x
y
x
1
1
m
Multiply both sides by x x
1
: y y
1
m(x x
1
).
Thus, the coordinates of every point on L satisfy the equation
y y
1
m(x x
1
),
and we have this fact:
58 CHAPTER 1 Basics
Horizontal
Lines
The horizontal line with y-intercept b is the graph of the equation
y b.
y
1
1
−1
3
x
Figure 1–38
Vertical
Lines
The vertical line with x-intercept c is the graph of the equation
x c.
The slope of this line is undefined.
y
(x, y)
(x
1
, y
1
)
x
Figure 1–39
Point-Slope
Form
The line with slope m through the point (x
1
, y
1
) is the graph of the equation
y y
1
m(x x
1
).
EXAMPLE 7
The tangent line to the graph of y 2x
2
8 at the point (1, 6) is shown in
Figure 1–40. In calculus, it is shown that this line passes through (1, 6) and has
slope 4. Find the equation of the tangent line.
SOLUTION Substitute 4 for m and (1, 6) for (x
1
, y
1
) in the point-slope equation:
y y
1
m(x x
1
)
y (6) 4(x 1) [point-slope form]
y 6 4x 4
y 4x 10 [slope-intercept form] ■
EXAMPLE 8
The annual out-of-pocket spending (per person) on doctors and clinical services
was approximately $105 in 1997. According to projections from the Health Care
Financing Committee, this cost is expected to rise linearly to $196 in 2010, as
indicated in Figure 1–41.
(a) Find an equation that gives the out-of-pocket cost y in year x.
(b) Use this equation to estimate the out-of-pocket costs in 2006 and 2009.
Figure 1–41
SOLUTION
(a) Let x 0 correspond to 1997, so that x 13 corresponds to 2010. Then the
given information can be represented by the points (0, 105) and (13, 196). We
must find the equation of the line through these points. Its slope is
19
1
6
3
1
0
05
9
1
1
3
7.
Now we use the slope 7 and one of the points (0, 105) or (13, 196) to find the
equation of the line. It doesn’t matter which point, since both lead to the same
equation.
y y
1
m(x x
1
) y y
1
m(x x
1
)
y 105 7(x 0) y 196 7(x 13)
y 7x 105 y 196 7x 91
y 7x 105.
(b) Since 2006 corresponds to x 9, the projected out-of-pocket costs in 2006 are
y 7x 105 7
9 105 $168.
The costs in 2009 (x 12) are
y 7x 105 7
12 105 $189. ■
100
120
140
160
180
200
1997
$105
$196
2010
y
x
SECTION 1.4 Lines 59
4
2
−4
−2
−6
−8
21
0
−1−23
y
x
Figure 1–40
GENERAL FORM
By rearranging terms, if necessary, the equation of any line can be written in the
general form Ax By C for some constants A, B, and C. For instance,
This equation can be written as
y 4x 10 4x 1y 10
y 196 7(x 13) 7x 1y 105
y 30x 1y 3
x 21x 0y 2
In summary:
EXAMPLE 9
Graph 3x 2y 6.
SOLUTION From the preceding box, we know that the graph is a line. It is easily
graphed by finding its intercepts.
y-intercept x-intercept
Set x 0 and solve for y. Set y 0 and solve for x.
3x 2y 63x 2y 6
3
0 2y 63x 2
0 6
2y 63x 6
y 3 x 2
The y-intercept is 3 and the x-intercept is 2, which leads to the graph in
Figure 1–42. ■
PARALLEL AND PERPENDICULAR LINES
The slope of a line measures how steeply it rises or falls. Since parallel lines rise
or fall equally steeply, the following fact should be plausible (see Exercises 93–94
for a proof).
60 CHAPTER 1 Basics
General
Form
Every line is the graph of an equation of the form
Ax By C,
where A and B are not both zero.
3
2
y
x
Figure 1–42
Parallel
Lines
Two nonvertical lines are parallel exactly when they have the same slope.
EXAMPLE 10
Find the equation of the line L through (2, 1) that is parallel to the line M whose
equation is 3x 2y 6 0.
SOLUTION First find the slope of M by rewriting its equation in slope-
intercept form:
3x 2y 6 0
2y 3x 6
y
3
2
x 3.
Therefore M has slope 3/2. The parallel line L must have the same slope, 3/2.
Since (2, 1) is on L, we can use the point-slope form to find its equation:
y y
1
m(x x
1
)
y (1)
3
2
(x 2) [point-slope form]
y 1
3
2
x 3
y
3
2
x 4 [slope-intercept form] ■
Two lines that meet in a right angle (90° angle) are said to be perpendicular.
As you might suspect, there is a close relationship between the slopes of two
perpendicular lines.
A proof of this fact is outlined in Exercise 96.
EXAMPLE 11
In Figure 1–43, the line L through (0, 2) and (1, 5) appears to be perpendicular
to the line M through (3, 2) and (3, 4). Verify algebraically that the lines are
perpendicular.
SOLUTION Compute the slope of the lines:
Slope L
5
1
2
0
3 and slope M
3
4
(
(
3
2
)
)
6
2
1
3
.
Since 3(1/3) 1, the lines L and M are perpendicular. ■
SECTION 1.4 Lines 61
Perpendicular
Lines
Two nonvertical lines, with slopes m
1
and m
2
, are perpendicular exactly
when the product of their slopes is 1, that is,
m
1
m
2
1, or equivalently, m
1
m
1
2
.
y
M
L
x
Figure 1–43
EXAMPLE 12
Find the equation of the perpendicular bisector of the line segment with endpoints
(5, 4) and (7, 2).
SOLUTION The perpendicular bisector M goes through the midpoint of the
line segment from (5, 4) and (7, 2). The midpoint formula (page 43) shows
that this midpoint is
x
1
2
x
2
,
y
1
2
y
2
5
2
7
,
4
2
2
(1, 1).
The line L through (5, 4) and (7, 2) has slope
y
x
2
2
y
x
1
1
2
7
(
(
4
5
)
)
1
6
2
1
2
.
Since M is perpendicular to L, we have (slope M)(slope L) 1, so that
Slope M
slo
p
1
e L
1
/
1
2
2.
Thus M is the line through (1, 1) with slope 2, and its equation is
y (1) 2(x 1) [point-slope form]
y 2x 1. [slope-intercept form] ■
RATES OF CHANGE
We have seen that the geometric interpretation of slope is that it measures the
steepness or direction of a line. The next examples show that slope can also be
interpreted as a rate of change.
EXAMPLE 13
According to the Kelley Blue Book, a Ford Focus ZX5 hatchback that is worth
$14,632 today will be worth $10,120 in 3 years (if it is in good condition with
average mileage).
(a) Assuming linear depreciation, find the equation that gives the value y of the
car in year x.
(b) At what rate is the car depreciating?
(c) What will the car be worth in 6 years?
SOLUTION
(a) Linear depreciation means that the value equation is linear. So the equation is
of the form y mx b for some constants m and b. Since the car is worth
$14,632 now (that is, y 14,632 when x 0), we have
y mx b
Let x 0 and y 14,632: 14,632 m
0 b
b 14,632.
62 CHAPTER 1 Basics
So the equation is y mx 14,632. Since the car is worth $10,120 in 3 years
(that is, y 10,120 when x 3), we have
y mx 14,632
Let x 3 and y 10,120: 10,120 m
3 14,632
Subtract 14,632 from both sides: 4512 3m
Divide both sides by 3: m 1504
Therefore, the value equation is y 1504x 14,632.
(b) Consider this table:
SECTION 1.4 Lines 63
Year x 01234
y 1504x 14,632 14,632 13,128 11,624 10,120 8616
You can easily verify that the car depreciates (decreases in value) $1504 each
year (that is, each time x changes by 1). In other words, the value changes at the
rate of 1504 per year. This rate is the slope of the line y 1504x 14,632.
(c) The value of the car after 6 years (x 6) is given by
y 1504x 14,632
Let x 6: y 1504(6) 14,632 5608.
The car is worth $5608 in 6 years. ■
EXAMPLE 14
A factory that makes can openers has fixed costs (for building, fixtures,
machinery, etc.) of $26,000. The variable cost (materials and labor) for making
one can opener is $2.75.
(a) Find the cost equation that gives the total cost y of producing x can openers
and sketch its graph.
(b) At what rate does the total cost increase as more can openers are made?
(c) What is the total cost of making 1000 can openers? 20,000? 40,000?
(d) In part (c), what is the average cost per can opener in each case?
SOLUTION
(a) Since each can opener costs $2.75, the variable cost of making x can openers
is 2.75x. The total cost y of making x can openers is
y variable costs fixed costs
y 2.75x 26,000.
The graph of this equation is the line in Figure 1–44.
(b) The cost equation shows that y increases by 2.75 each time x increases by 1.
That is, total cost is increasing at the rate of $2.75 per can opener. This rate of
change is the slope the cost equation line y 2.75x 26,000.
(c) The cost of making 1000 can openers is
y 2.75x 26,000 2.75(1000) 26,000 $28,750.
Similarly, the cost of making 20,000 can openers is
y 2.75(20,000) 26,000 $81,000,
and the cost of 40,000 is
y 2.75(40,000) 26,000 $136,000.
20,000 40,000 60,000
Months
x
y
25,000
50,000
150,000
75,000
175,000
100,000
200,000
125,000
Figure 1–44
(d) The average cost per can opener in each case is the total cost divided by the
number of can openers. So the average cost per can opener is as follows.
For 1000: $28,750/1000 $28.75 per can opener;
For 20,000: $81,000/20,000 $4.05 per can opener;
For 40,000: $136,000/40,000 $3.40 per can opener. ■
Examples 13 and 14 illustrate this fact.
64 CHAPTER 1 Basics
Linear Rate
of Change
The slope m of the line with equation
y mx b
is the rate of change of y with respect to x.
EXERCISES 1.4
1. For which of the line segments in the figure is the slope
(a) largest? (b) smallest?
(c) largest in absolute value? (d) closest to zero?
2. The doorsill of a campus building is 5 feet above ground
level. To allow wheelchair access, the steps in front of the
door are to be replaced by a straight ramp with constant
slope 1/12, as shown in the figure. How long must the ramp
be? [The answer is not 60 feet.]
In Exercises 3–6, find the slope of the line through the given points.
3. (1, 2); (3, 7) 4. (1, 2); (2, 1)
5. (1/4, 0); (3/4, 2) 6. (2
, 1); (2, 9)
Ramp
5
y
A
B
C
D
E
x
In Exercises 7–10, find a number t such that the line passing
through the two given points has slope 2.
7. (0, t); (9, 4) 8. (1, t); (2, 4)
9. (t 1, 5); (6, 3t 7) 10. (t, t); (5, 9)
11. Let L be a nonvertical straight line through the origin.
L intersects the vertical line through (1, 0) at a point P. Show
that the second coordinate of P is the slope of L.
12. On one graph, sketch five line segments, not all meeting at
a single point, whose slopes are five different positive num-
bers. Do this in such a way that the left-hand line has the
largest slope, the second line from the left has the next
largest slope, and so on.
In Exercises 13–16, match the given equation with the line
shown below that most closely resembles its graph.
13. y 3x 2 14. y 3x 2
15. y 3x 2 16. y 3x 2
In Exercises 17–20, find the equation of the line with
y-intercept b and slope m.
17. b 5, m 4 18. b 3, m 7
(d)
x
0
y
x
0
(c)
y
x
0
y
(b)
x
0
y
(a)
19. b 1.5, m 2.3 20. b 4.5, m 2.5
In Exercises 21–24, find the equation of the line.
21.
22.
23.
24.
−5
12345
−4 −3 −2 −1
−1
1
2
3
4
5
−2
−3
−4
−5
x
y
−5
12345
−4 −3 −2 −1
−1
1
2
3
4
5
−2
−3
−4
−5
x
y
−5
12345
−4 −3 −2 −1
−1
1
2
3
4
5
−2
−3
−4
−5
x
y
−5
12345
−4 −3 −2 −1
−1
1
2
3
4
5
−2
−3
−4
−5
x
y
SECTION 1.4 Lines 65
In Exercises 25–28, find the slope and y-intercept of the line
whose equation is given.
25. 2x y 5 0 26. 4x 3y 5
27. 3(x 2) y 7 6(y 4)
28. 2(y 3) (x 6) 4(x 1) 2
In Exercises 29–32, find the equation of the line with slope m
that passes through the given point.
29. m 1; (4, 7) 30. m 2; (2, 1)
31. m 1; (6, 2) 32. m 0; (4, 5)
In Exercises 33–36, find the equation of the line through the
given points.
33. (0, 5) and (3, 2) 34. (4, 3) and (2, 1)
35. (6/5, 3/5) and (1/5, 3) 36. (6, 7) and (6, 15)
In Exercises 37–42, graph the equation. Label all intercepts.
37. 3x 5y 15 38. 2x 3y 12
39. 2y x 2 40. 4x 5y 10
41. 3x 2y 0 42. 2x 6y 0
In Exercises 43–46, determine whether the line through
P and Q is parallel or perpendicular to the line through
R and S or neither.
43. P (2, 5), Q (1, 1) and R (4, 2), S (6, 1).
44. P (0, 3/2), Q (1, 1) and R (2, 7), S (3, 9).
45. P (3, 1/3), Q (1, 1) and R (2, 0), S (4, 2/3).
46. P (3, 3), Q (3, 1) and R (2, 2), S (4, 5).
In Exercises 47–49, determine whether the lines whose
equations are given are parallel, perpendicular, or neither.
47. 2x y 2 0 and 4x 2y 18 0.
48. 3x y 3 0 and 6x 2y 17 0.
49. y 2x 4 and .5x y 3.
50. Do the points (4, 6), (1, 12), and (7, 0) all lie on the
same straight line? [Hint: Use slopes.]
51. Are (9, 6), (1, 2), and (1, 3) the vertices of a right
triangle? [Hint: Use slopes.]
52. Are the points (5, 2) (3, 1), (3, 0), and (5, 3) the
vertices of a parallelogram?
In Exercises 53–56, find the equation of the perpendicular
bisector of the line segment joining the two given points.
53. (1, 3), (3, 7) 54. (3, 6), (7, 2)
55. (2, 3), (4, 7) 56. (6, 2), (6, 7)
In Exercises 57–64, find an equation for the line satisfying the
given conditions.
57. Through (2, 1) with slope 3.
58. y-intercept 7 and slope 1.
59. Through (2, 3) and parallel to 3x 2y 5.
60. Through (1, 2) and perpendicular to y 2x 3.