Calculus

1143 Pages·2005·56.044 MB·English

by  Ron Larson, Robert P. Hostetler, Bruce H. Edwards

2 CHAPTERP Preparation for Calculus
Section P.1 Graphs and Models
• Sketch the graph of an equation.
• Find the intercepts of a graph.
RENÉDESCARTES(1596–1650)
• Test a graph for symmetry with respect to an axis and the origin.
Descartes made many contributions to
• Find the points of intersection of two graphs.
philosophy, science, and mathematics. The idea
• Interpret mathematical models for real-life data.
of representing points in the plane by pairs of
real numbers and representing curves in the
plane by equations was described by Descartes The Graph of an Equation
.in his book La Géométrie, published in 1637.
In 1637 the French mathematician René Descartes revolutionized the study of mathe-
matics by joining its two major fields—algebra and geometry. With Descartes’s
MathBio coordinate plane, geometric concepts could be formulated analytically and algebraic
concepts could be viewed graphically. The power of this approach is such that within
a century,much of calculus had been developed.
The same approach can be followed in your study of calculus. That is,by viewing
calculus from multiple perspectives—graphically, analytically, and numerically—
you will increase your understanding of core concepts.
Consider the equation 3x(cid:3)y(cid:1)7. The point (cid:1)2, 1(cid:2) is a solution point of the
equation because the equation is satisfied (is true) when 2 is substituted for xand 1 is
substituted for y.This equation has many other solutions,such as (cid:1)1, 4(cid:2)and (cid:1)0, 7(cid:2).To
find other solutions systematically,solve the original equation for y.
y(cid:1)7(cid:2)3x Analytic approach
Then construct a table of valuesby substituting several values of x.
y
8 x 0 1 2 3 4
(0, 7)
Numerical approach
6 y 7 4 1 (cid:2)2 (cid:2)5
(1, 4)
4 3x + y = 7
2
(2, 1)
x From the table,you can see that (cid:1)0, 7(cid:2),(cid:1)1, 4(cid:2),(cid:1)2, 1(cid:2),(cid:1)3, (cid:2)2(cid:2),and (cid:1)4, (cid:2)5(cid:2)are solutions
2 4 6 8
−2 (3, −2) of the original equation 3x(cid:3)y(cid:1)7.Like many equations,this equation has an infinite
−4 number of solutions. The set of all solution points is the graph of the equation, as
(4, −5)
−6 shown in Figure P.1.
Graphical approach: 3x(cid:3)y(cid:1)7 NOTE Even though we refer to the sketch shown in Figure P.1 as the graph of 3x(cid:3)y(cid:1)7,
Figure P.1 it really represents only a portionof the graph. The entire graph would extend beyond the page.
In this course, you will study many sketching techniques. The simplest is point
plotting—that is,you plot points until the basic shape of the graph seems apparent.
y EXAMPLE 1 Sketching a Graph by Point Plotting
7
6 Sketch the graph of y(cid:1)x2(cid:2)2.
5
4 y = x2 − 2 Solution First construct a table of values. Then plot the points shown in the table.
3
2
1 x (cid:2)2 (cid:2)1 0 1 2 3
x
−4 −3−2 2 3 4 y 2 (cid:2)1 (cid:2)2 (cid:2)1 2 7
Th...e parabola y(cid:1)x2(cid:2)2 Finally,connect the points with a smoothcurve,as shown in Figure P.2. This graph is
Figure P.2 a parabola.It is one of the conics you will study in Chapter 10.
Editable Graph Try It Exploration A
SECTIONP.1 Graphs and Models 3
One disadvantage of point plotting is that to get a good idea about the shape of a
graph, you may need to plot many points. With only a few points, you could badly
misrepresent the graph. For instance,suppose that to sketch the graph of
y(cid:1) 1x(cid:1)39(cid:2)10x2(cid:3)x4(cid:2)
30
you plotted only five points:(cid:1)(cid:2)3, (cid:2)3(cid:2),(cid:1)(cid:2)1, (cid:2)1(cid:2),(cid:1)0, 0(cid:2),(cid:1)1, 1(cid:2),and (cid:1)3, 3(cid:2),as shown
in Figure P.3(a). From these five points, you might conclude that the graph is a line.
This, however, is not correct. By plotting several more points, you can see that the
graph is more complicated,as shown in Figure P.3(b).
y
y y = 1 x(39 − 10x2 + x4)
30
3 (3, 3)
3
2
2
1 (1, 1)
(0, 0) 1
x
−3 −2 −1 1 2 3 x
(−1, −1) −1 Plotting only a −3 −2 −1 1 2 3
−1
few points can
−2 misrepresent a
−2
graph.
(−3, −3) −3
−3
(a) (b)
Figure P.3
EXPLORATION TECHNOLOGY Technology has made sketching of graphs easier. Even with
technology,however,it is possible to misrepresent a graph badly. For instance,each
Comparing Graphical and Analytic
of the graphing utility screens in Figure P.4 shows a portion of the graph of
Approaches Use a graphing utility
to graph each equation. In each case, y(cid:1)x3(cid:2)x2(cid:2)25.
find a viewing window that shows the
From the screen on the left, you might assume that the graph is a line. From the
important characteristics of the graph.
screen on the right, however, you can see that the graph is not a line. So, whether
a. y(cid:1)x3(cid:2)3x2(cid:3)2x(cid:3)5 you are sketching a graph by hand or using a graphing utility,you must realize that
b. y(cid:1)x3(cid:2)3x2(cid:3)2x(cid:3)25 different “viewing windows” can produce very different views of a graph. In
c. y(cid:1)(cid:2)x3(cid:2)3x2(cid:3)20x(cid:3)5 choosing a viewing window,your goal is to show a view of the graph that fits well
d. y(cid:1)3x3(cid:2)40x2(cid:3)50x(cid:2)45 in the context of the problem.
e. y(cid:1)(cid:2)(cid:1)x(cid:3)12(cid:2)3
10 5
f. y(cid:1)(cid:1)x(cid:2)2(cid:2)(cid:1)x(cid:2)4(cid:2)(cid:1)x(cid:2)6(cid:2)
−5 5
A purely graphical approach to this
problem would involve a simple
“guess,check,and revise”strategy. −10 10
What types of things do you think an
analytic approach might involve? For
instance,does the graph have symme-
try? Does the graph have turns? If so,
where are they? −10 −35
As you proceed through Chapters Graphing utility screens of y(cid:1)x3(cid:2)x2(cid:2)25
1,2,and 3 of this text,you will study Figure P.4
many new analytic tools that will
help you analyze graphs of equations
such as these. NOTE In this text,the term graphingutilitymeans either a graphing calculator or computer
graphing software such as Maple,Mathematica,Derive,Mathcad,or the TI-89.
4 CHAPTERP Preparation for Calculus
Intercepts of a Graph
Two types of solution points that are especially useful in graphing an equation are those
having zero as their x- or y-coordinate. Such points are called interceptsbecause they
are the points at which the graph intersects the x- or y-axis. The point (cid:1)a, 0(cid:2) is an
x-interceptof the graph of an equation if it is a solution point of the equation. To find
the x-intercepts of a graph,let ybe zero and solve the equation for x.The point (cid:1)0, b(cid:2)
is a y-interceptof the graph of an equation if it is a solution point of the equation. To
find the y-intercepts of a graph,let xbe zero and solve the equation for y.
NOTE Some texts denote the x-intercept as the x-coordinate of the point (cid:1)a, 0(cid:2)rather than the
point itself. Unless it is necessary to make a distinction,we will use the term interceptto mean
either the point or the coordinate.
It is possible for a graph to have no intercepts, or it might have several. For
instance,consider the four graphs shown in Figure P.5.
y y y y
x x x x
No x-intercepts Three x-intercepts One x-intercept No intercepts
One y-intercept One y-intercept Two y-intercepts
Figure P.5
EXAMPLE 2 Finding x- and y-intercepts
Find the x-and y-intercepts of the graph of y(cid:1)x3(cid:2)4x.
y Solution To find the x-intercepts,let ybe zero and solve for x.
y = x3 − 4x 4 x3(cid:2)4x(cid:1)0 Let ybe zero.
3 x(cid:1)x(cid:2)2(cid:2)(cid:1)x(cid:3)2(cid:2)(cid:1)0 Factor.
x(cid:1)0, 2, or (cid:2)2 Solve for x.
(−2, 0) (0, 0) (2, 0) Because this equation has three solutions, you can conclude that the graph has three
x x-intercepts:
−4 −3 −1 1 3 4
−1 (cid:1)0, 0(cid:2),(cid:1)2, 0(cid:2), and (cid:1)(cid:2)2, 0(cid:2). x-intercepts
−2
To find the y-intercepts,let xbe zero. Doing this produces y(cid:1)0.So,the y-intercept is
−3
(cid:1)0, 0(cid:2). y-intercept
In.....tercepts of a graph
Figure P.6 (See Figure P.6.)
Editable Graph Try It Exploration A Video Video
TECHNOLOGY Example 2 uses an analytic approach to finding intercepts.
When an analytic approach is not possible, you can use a graphical approach by
finding the points at which the graph intersects the axes. Use a graphing utility to
approximate the intercepts.
SECTIONP.1 Graphs and Models 5
y Symmetry of a Graph
Knowing the symmetry of a graph beforeattempting to sketch it is useful because you
need only half as many points to sketch the graph. The following three types of
(−x, y) (x, y) symmetry can be used to help sketch the graphs of equations (see Figure P.7).
x 1. A graph issymmetric with respect to the y-axisif,whenever (cid:1)x, y(cid:2)is a point on
the graph, (cid:1)(cid:2)x, y(cid:2) is also a point on the graph. This means that the portion of
y-axis
symmetry the graph to the left of the y-axis is a mirror image of the portion to the right of the
y-axis.
2. A graph issymmetric with respect to the x-axisif,whenever (cid:1)x, y(cid:2)is a point on
y the graph,(cid:1)x, (cid:2)y(cid:2)is also a point on the graph. This means that the portion of the
graph above the x-axis is a mirror image of the portion below the x-axis.
3. A graph is symmetric with respect to the originif,whenever (cid:1)x, y(cid:2)is a point on
the graph, (cid:1)(cid:2)x, (cid:2)y(cid:2) is also a point on the graph. This means that the graph is
(x, y)
unchanged by a rotation of 180(cid:4)about the origin.
x
x-axis (x, −y)
Tests for Symmetry
symmetry
1. The graph of an equation in xand yis symmetric with respect to the y-axis if
replacing xby (cid:2)xyields an equivalent equation.
y 2. The graph of an equation in xand yis symmetric with respect to the x-axis if
replacing yby (cid:2)yyields an equivalent equation.
3. The graph of an equation in xand yis symmetric with respect to the origin if
replacing xby (cid:2)xand yby (cid:2)yyields an equivalent equation.
(x, y)
x
(−x, −y) The graph of a polynomial has symmetry with respect to the y-axis if each term
Origin
has an even exponent (or is a constant). For instance,the graph of
symmetry
y(cid:1)2x4(cid:2)x2(cid:3)2 y-axis symmetry
Figure P.7 has symmetry with respect to the y-axis. Similarly, the graph of a polynomial has
symmetry with respect to the origin if each term has an odd exponent,as illustrated in
Example 3.
EXAMPLE 3 Testing for Origin Symmetry
Show that the graph of
y y = 2x3 − x y(cid:1)2x3(cid:2)x
2 is symmetric with respect to the origin.
1 (1, 1) Solution
y(cid:1)2x3(cid:2)x Write original equation.
x
−2 −1 1 2 (cid:2)y(cid:1)2(cid:1)(cid:2)x(cid:2)3(cid:2)(cid:1)(cid:2)x(cid:2) Replace xby (cid:2)xand yby (cid:2)y.
(−1, −1) −1 (cid:2)y(cid:1)(cid:2)2x3(cid:3)x Simplify.
y(cid:1)2x3(cid:2)x Equivalent equation
−2
Because the replacements yield an equivalent equation, you can conclude that the
..... Origin symmetry graph of y(cid:1)2x3(cid:2)xis symmetric with respect to the origin,as shown in Figure P.8.
Figure P.8
Editable Graph Try It Exploration A Video Video
6 CHAPTERP Preparation for Calculus
EXAMPLE 4 Using Intercepts and Symmetry to Sketch a Graph
Sketch the graph of x(cid:2)y2(cid:1)1.
y Solution The graph is symmetric with respect to the x-axis because replacing yby
x − y2 = 1 (5, 2) (cid:2)yyields an equivalent equation.
2
(2, 1) x(cid:2)y2(cid:1)1 Write original equation.
1 x(cid:2)(cid:1)(cid:2)y(cid:2)2(cid:1)1 Replace yby (cid:2)y.
(1, 0)
x x(cid:2)y2(cid:1)1 Equivalent equation
2 3 4 5
−1 This means that the portion of the graph below the x-axis is a mirror image of the
x-intercept portion above the x-axis. To sketch the graph,first plot the x-intercept and the points
−2
..... above the x-axis. Then reflect in the x-axis to obtain the entire graph, as shown in
Figure P.9 Figure P.9.
Editable Graph Try It Exploration A Exploration B Open Exploration
TECHNOLOGY Graphing utilities are designed so that they most easily graph
equations in which y is a function of x (see Section P.3 for a definition of
function). To graph other types of equations, you need to split the graph into
two or more parts or you need to use a different graphing mode. For instance,
to graph the equation in Example 4,you can split it into two parts.
y (cid:1)(cid:3)x(cid:2)1 Top portion of graph
1
y (cid:1)(cid:2)(cid:3)x(cid:2)1 Bottom portion of graph
2
Points of Intersection
A point of intersection of the graphs of two equations is a point that satisfies both
equations. You can find the points of intersection of two graphs by solving their
equations simultaneously.
y
EXAMPLE 5 Finding Points of Intersection
2
x − y = 1 Find all points of intersection of the graphs of x2(cid:2)y(cid:1)3and x(cid:2)y(cid:1)1.
1 (2, 1)
Solution Begin by sketching the graphs of both equations on the same rectangular
x
−2 −1 1 2 coordinate system, as shown in Figure P.10. Having done this, it appears that the
−1 graphs have two points of intersection. You can find these two points,as follows.
(−1, −2) −2 y(cid:1)x2(cid:2)3 Solve first equation for y.
x2 − y = 3 y(cid:1)x(cid:2)1 Solve second equation for y.
x2(cid:2)3(cid:1)x(cid:2)1 Equate y-values.
Tw.o points of intersection x2(cid:2)x(cid:2)2(cid:1)0 Write in general form.
Figure P.10 (cid:1)x(cid:2)2(cid:2)(cid:1)x(cid:3)1(cid:2)(cid:1)0 Factor.
Editable Graph x(cid:1)2 or (cid:2)1 Solve for x.
The corresponding values of y are obtained by substituting x(cid:1)2 and x(cid:1)(cid:2)1 into
either of the original equations. Doing this produces two points of intersection:
S..TUDY TIP You can check the
points of intersection from Example 5 (cid:1)2, 1(cid:2) and (cid:1)(cid:2)1, (cid:2)2(cid:2). Points of intersection
by substituting into bothof the original
equations or by using the intersect Try It Exploration A
feature of a graphing utility.
SECTIONP.1 Graphs and Models 7
Mathematical Models
Real-life applications of mathematics often use equations as mathematical models.
In developing a mathematical model to represent actual data,you should strive for two
(often conflicting) goals: accuracy and simplicity. That is, you want the model to be
simple enough to be workable, yet accurate enough to produce meaningful results.
Section P.4 explores these goals more completely.
EXAMPLE6 Comparing Two Mathematical Models
The Mauna Loa Observatory in Hawaii The Mauna Loa Observatory in Hawaii records the carbon dioxide concentration y(in
has been measuring the increasing parts per million) in Earth’s atmosphere. The January readings for various years are
. concentration of carbon dioxide in Earth’s shown in Figure P.11. In the July 1990 issue of Scientific American, these data were
atmosphere since 1958. used to predict the carbon dioxide level in Earth’s atmosphere in the year 2035,using
the quadratic model
Video
y(cid:1)316.2(cid:3)0.70t(cid:3)0.018t2 Quadratic model for 1960–1990 data
where t(cid:1)0represents 1960,as shown in Figure P.11(a).
The data shown in Figure P.11(b) represent the years 1980 through 2002 and can
be modeled by
y(cid:1)306.3(cid:3)1.56t Linear model for 1980–2002 data
where t(cid:1)0represents 1960. What was the prediction given in the Scientific American
article in 1990? Given the new data for 1990 through 2002,does this prediction for the
year 2035 seem accurate?
y y
375 375
on) 337605 on) 337605
milli 336505 milli 336505
er 350 er 350
s p 345 s p 345
part 333450 part 333450
O (in 2 333223050 O (in 2 333223050
C C
315 315
t t
5 10 15 20 25 30 35 40 45 5 10 15 20 25 30 35 40 45
Year (0 ↔ 1960) Year (0 ↔ 1960)
(a) (b)
Figure P.11
Solution To answer the first question,substitute t(cid:1)75(for 2035) into the quadratic
model.
y(cid:1)316.2(cid:3)0.70(cid:1)75(cid:2)(cid:3)0.018(cid:1)75(cid:2)2(cid:1)469.95 Quadratic model
So, the prediction in the Scientific American article was that the carbon dioxide
concentration in Earth’s atmosphere would reach about 470 parts per million in the
NOTE The models in Example 6 were
year 2035. Using the linear model for the 1980–2002 data,the prediction for the year
developed using a procedure called least
2035 is
squares regression(see Section 13.9).
The quadratic and linear models have a y(cid:1)306.3(cid:3)1.56(cid:1)75(cid:2)(cid:1)423.3. Linear model
correlation given by r2(cid:1)0.997and
.. r2(cid:1)0.996,respectively. The closer r2 So,based on the linear model for 1980–2002,it appears that the 1990 prediction was
is to 1,the “better”the model. too high.
Try It Exploration A
8 CHAPTERP Preparation for Calculus
Exercises for Section P.1
The symbol indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.
Click on to view the complete solution of the exercise.
Click on to print an enlarged copy of the graph.
In Exercises 1–4, match the equation with its graph. [Graphs In Exercises 19–26,find any intercepts.
are labeled (a),(b),(c),and (d).]
19. y(cid:2)x2(cid:4)x(cid:3)2 20. y2(cid:2)x3(cid:3)4x
(a) y (b) y 21. y(cid:2)x2(cid:3)25(cid:3)x2 22. y(cid:2)(cid:4)x(cid:3)1(cid:5)(cid:3)x2(cid:4)1
(cid:4) (cid:5)
5 3 2(cid:3)(cid:3)x x2(cid:4)3x
3 4 23. y(cid:2) x 24. y(cid:2)(cid:4)3x(cid:4)1(cid:5)2
3
2 25. x2y(cid:3)x2(cid:4)4y(cid:2)0 26. y(cid:2)2x(cid:3)(cid:3)x2(cid:4)1
1 1
x x In Exercises 27–38,test for symmetry with respect to each axis
−1 1 1 2 3 4
and to the origin.
(c) y (d) y 27. y(cid:2)x2(cid:3)2 28. y(cid:2)x2(cid:3)x
2 4 29. y2(cid:2)x3(cid:3)4x 30. y(cid:2)x3(cid:4)x
1 2 31. xy(cid:2)4 32. xy2(cid:2)(cid:3)10
x
−2 1 2 x 33. y(cid:2)4(cid:3)(cid:3)x(cid:4)3 34. xy(cid:3)(cid:3)4(cid:3)x2(cid:2)0
−1 −2 2
x x2
−2 −2 35. y(cid:2) 36. y(cid:2)
x2(cid:4)1 x2(cid:4)1
(cid:2) (cid:2) (cid:2) (cid:2)
1. y(cid:2)(cid:3)1x(cid:4)2 2. y(cid:2)(cid:3)9(cid:3)x2 37. y(cid:2) x3(cid:4)x 38. y (cid:3)x(cid:2)3
2
3. y(cid:2)4(cid:3)x2 4. y(cid:2)x3(cid:3)x
In Exercises 39–56, sketch the graph of the equation. Identify
any intercepts and test for symmetry.
In Exercises 5–14, sketch the graph of the equation by point
plotting. 39. y(cid:2)(cid:3)3x(cid:4)2 40. y(cid:2)(cid:3)1x(cid:4)2
2
5. y(cid:2)32x(cid:4)1 6. y(cid:2)6(cid:3)2x 41. y(cid:2)12x(cid:3)4 42. y(cid:2)23x(cid:4)1
7. y(cid:2)4(cid:3)x2 8. y(cid:2)(cid:4)x(cid:3)3(cid:5)2 43. y(cid:2)1(cid:3)x2 44. y(cid:2)x2(cid:4)3
9. y(cid:2)(cid:2)x(cid:4)2(cid:2) 10. y(cid:2)(cid:2)x(cid:2)(cid:3)1 45. y(cid:2)(cid:4)x(cid:4)3(cid:5)2 46. y(cid:2)2x2(cid:4)x
11. y(cid:2)(cid:3)x(cid:3)4 12. y(cid:2)(cid:3)x(cid:4)2 47. y(cid:2)x3(cid:4)2 48. y(cid:2)x3(cid:3)4x
2 1 49. y(cid:2)x(cid:3)x(cid:4)2 50. y(cid:2)(cid:3)9(cid:3)x2
13. y(cid:2) 14. y(cid:2)
x x(cid:3)1 51. x(cid:2)y3 52. x(cid:2)y2(cid:3)4
1 10
53. y(cid:2) 54. y(cid:2)
In Exercises 15 and 16,describe the viewing window that yields x x2(cid:4)1
the figure. (cid:2) (cid:2) (cid:2) (cid:2)
55. y(cid:2)6(cid:3) x 56. y(cid:2) 6(cid:3)x
(cid:2) (cid:2) (cid:2) (cid:2)
15. y(cid:2)x3(cid:3)3x2(cid:4)4 16. y(cid:2) x (cid:4) x(cid:3)10
In Exercises 57–60,use a graphing utility to graph the equation.
Identify any intercepts and test for symmetry.
57. y2(cid:3)x(cid:2)9 58. x2(cid:4)4y2(cid:2)4
59. x(cid:4)3y2(cid:2)6 60. 3x(cid:3)4y2(cid:2)8
In Exercises 61–68,find the points of intersection of the graphs
In Exercises 17 and 18, use a graphing utility to graph the of the equations.
equation. Move the cursor along the curve to approximate the
unknown coordinate of each solution point accurate to two 61. x(cid:4)y(cid:2)2 62. 2x(cid:3)3y(cid:2)13
decimal places. 2x(cid:3)y(cid:2)1 5x(cid:4)3y(cid:2) 1
17. y(cid:2)(cid:3)5(cid:3)x (a) (cid:4)2, y(cid:5) (b) (cid:4)x, 3(cid:5) 63. x2(cid:4)y(cid:2)6 64. x(cid:2)3(cid:3)y2
18. y(cid:2)x5(cid:3)5x (a) (cid:4)(cid:3)0.5, y(cid:5) (b) (cid:4)x, (cid:3)4(cid:5) x(cid:4)y(cid:2)4 y(cid:2)x(cid:3)1
SECTIONP.1 Graphs and Models 9
65. x2(cid:4)y2(cid:2)5 66. x2(cid:4)y2(cid:2)25 where xis the diameter of the wire in mils (0.001 in.). Use a
x(cid:3)y(cid:2)1 2x(cid:4)y(cid:2)10 graphing utility to graph the model. If the diameter of the wire
is doubled,the resistance is changed by about what factor?
67. y(cid:2)x3 68. y(cid:2)x3(cid:3)4x
y(cid:2)x y(cid:2)(cid:3)(cid:4)x(cid:4)2(cid:5)
Writing About Concepts
In Exercises 69–72, use a graphing utility to find the points In Exercises 77 and 78,write an equation whose graph has
of intersection of the graphs. Check your results analytically. the indicated property. (There may be more than one
correct answer.)
69. y(cid:2)x3(cid:3)2x2(cid:4)x(cid:3)1 70. y(cid:2)x4(cid:3)2x2(cid:4)1
y(cid:2)(cid:3)x2(cid:4)3x(cid:3)1 y(cid:2)1(cid:3)x2 77. The graph has intercepts at x(cid:2)(cid:3)2,x(cid:2)4,and x(cid:2)6.
71. y(cid:2)(cid:3)x(cid:4)6 72. y(cid:2)(cid:3)(cid:2)2x(cid:3)3(cid:2)(cid:4)6 78. The graph has intercepts at x(cid:2)(cid:3)52,x(cid:2)2,and x(cid:2)32.
y(cid:2)(cid:3)(cid:3)x2(cid:3)4x y(cid:2)6(cid:3)x 79. Each table shows solution points for one of the following
equations.
73. Modeling Data The table shows the Consumer Price Index
(i) y(cid:2)kx(cid:4)5 (ii) y(cid:2)x2(cid:4)k
(CPI) for selected years. (Source: Bureau of Labor Statistics)
(iii) y(cid:2)kx3(cid:8)2 (iv) xy(cid:2)k
Year 1970 1975 1980 1985 1990 1995 2000
Match each equation with the correct table and find k.
CPI 38.8 53.8 82.4 107.6 130.7 152.4 172.2 Explain your reasoning.
(a) (b)
x 1 4 9 x 1 4 9
(a) Use the regression capabilities of a graphing utility to find
a mathematical model of the form y(cid:2)at2(cid:4)bt(cid:4)cfor the y 3 24 81 y 7 13 23
data. In the model,yrepresents the CPI and trepresents the
year,with t(cid:2)0corresponding to 1970. (c) (d)
x 1 4 9 x 1 4 9
(b) Use a graphing utility to plot the data and graph the model.
Compare the data with the model. y 36 9 4 y (cid:3)9 6 71
(c) Use the model to predict the CPI for the year 2010.
80. (a) Prove that if a graph is symmetric with respect to the
74. Modeling Data The table shows the average numbers of acres
x-axis and to the y-axis, then it is symmetric with
per farm in the United States for selected years. (Source:
respect to the origin. Give an example to show that the
U.S. Department of Agriculture)
converse is not true.
(b) Prove that if a graph is symmetric with respect to one
Year 1950 1960 1970 1980 1990 2000
axis and to the origin,then it is symmetric with respect
Acreage 213 297 374 426 460 434 to the other axis.
(a) Use the regression capabilities of a graphing utility to find
a mathematical model of the form y(cid:2)at2(cid:4)bt(cid:4)cfor the True or False? In Exercises 81–84, determine whether the
data. In the model, y represents the average acreage and t statement is true or false. If it is false, explain why or give an
represents the year,with t(cid:2)0corresponding to 1950. example that shows it is false.
(b) Use a graphing utility to plot the data and graph the model. 81. If (cid:4)1, (cid:3)2(cid:5)is a point on a graph that is symmetric with respect
Compare the data with the model. to the x-axis,then (cid:4)(cid:3)1, (cid:3)2(cid:5)is also a point on the graph.
(c) Use the model to predict the average number of acres per 82. If (cid:4)1, (cid:3)2(cid:5)is a point on a graph that is symmetric with respect
farm in the United States in the year 2010. to the y-axis,then (cid:4)(cid:3)1, (cid:3)2(cid:5)is also a point on the graph.
75. Break-Even Point Find the sales necessary to break even 83. If b2(cid:3)4ac > 0and a(cid:5)0,then the graph of y(cid:2)ax2(cid:4)bx(cid:4)c
(cid:4)R(cid:2)C(cid:5)if the cost Cof producing xunits is
has two x-intercepts.
C(cid:2)5.5(cid:3)x(cid:4)10,000 Cost equation 84. If b2(cid:3)4ac(cid:2)0and a(cid:5)0,then the graph of y(cid:2)ax2(cid:4)bx(cid:4)c
has only one x-intercept.
and the revenue Rfor selling xunits is
R(cid:2)3.29x. Revenue equation In Exercises 85 and 86, find an equation of the graph that
consists of all points (cid:6)x, y(cid:7) having the given distance from the
76. Copper Wire The resistance y in ohms of 1000 feet of solid
origin. (For a review of the Distance Formula,see Appendix D.)
copper wire at 77(cid:6)Fcan be approximated by the model
85. The distance from the origin is twice the distance from (cid:4)0, 3(cid:5).
10,770
y(cid:2) (cid:3)0.37, 5 ≤ x ≤ 100 86. The distance from the origin is K (cid:4)K(cid:5)1(cid:5) times the distance
x2
from (cid:4)2, 0(cid:5).
10 CHAPTERP Preparation for Calculus
Section P.2 Linear Models and Rates of Change
• Find the slope of a line passing through two points.
• Write the equation of a line with a given point and slope.
• Interpret slope as a ratio or as a rate in a real-life application.
• Sketch the graph of a linear equation in slope-intercept form.
• Write equations of lines that are parallel or perpendicular to a given line.
y
The Slope of a Line
(x ,y )
y 2 2 The slope of a nonvertical line is a measure of the number of units the line rises (or
2
∆y = y − y falls) vertically for each unit of horizontal change from left to right. Consider the two
(x1, y1) 2 1 points (cid:1)x , y (cid:2) and (cid:1)x , y (cid:2) on the line in Figure P.12. As you move from left to right
y 1 1 2 2
1 along this line,a vertical change of
∆x = x − x
2 1
x (cid:3)y(cid:2)y (cid:1)y Change in y
x x 2 1
1 2
units corresponds to a horizontal change of
(cid:3)y(cid:2)y (cid:1)y (cid:2)change in y
(cid:3)x(cid:2)x2 (cid:1)x1 (cid:2)change in x (cid:3)x(cid:2)x (cid:1)x Change in x
2 1 2 1
Figure P.12
units. ((cid:3) is the Greek uppercase letter delta, and the symbols (cid:3)y and (cid:3)x are read
“delta y”and “delta x.”)
Definition of the Slope of a Line
The slopemof the nonvertical line passing through (cid:1)x , y (cid:2)and (cid:1)x , y (cid:2)is
1 1 2 2
(cid:3)y y (cid:1)y
m(cid:2) (cid:2) 2 1, x (cid:4)x .
(cid:3)x x (cid:1)x 1 2
2 1
...
Slope is not defined for vertical lines.
Video Video Video
NOTE When using the formula for slope,note that
y (cid:1)y (cid:1)(cid:1)y (cid:1)y (cid:2) y (cid:1)y
2 1(cid:2) 1 2 (cid:2) 1 2.
x (cid:1)x (cid:1)(cid:1)x (cid:1)x (cid:2) x (cid:1)x
2 1 1 2 1 2
So, it does not matter in which order you subtract as long as you are consistent and both
“subtracted coordinates”come from the same point.
Figure P.13 shows four lines: one has a positive slope, one has a slope of zero,
one has a negative slope,and one has an “undefined”slope. In general,the greater the
absolute value of the slope of a line, the steeper the line is. For instance, in Figure
P.13,the line with a slope of (cid:1)5is steeper than the line with a slope of 1.
5
y y y y
4 m = 1 4 m = 0 4 (0, 4) 4 (3, 4)
1 5 2 m = −5
3 3 3 3 3 m is
(−1, 2) (2, 2) 4
undefined.
2 2 2
(3, 1)
1 1 1 1 (3, 1)
(−2, 0)
x x x x
−2 −1 1 2 3 −2 −1 1 2 3 −1 2 3 4 −1 1 2 4
−1 −1 −1 (1, −1) −1
If mis positive, then the line If mis zero, then the line is If mis negative, then the line If mis undefined, then the line
rises from left to right. horizontal. falls from left to right. is vertical.
Figure P.13
SECTIONP.2 Linear Models and Rates of Change 11
Equations of Lines
EXPLORATION
Any two points on a nonvertical line can be used to calculate its slope. This can be
Investigating Equations of Lines
verified from the similar triangles shown in Figure P.14. (Recall that the ratios of
Use a graphing utility to graph each
corresponding sides of similar triangles are equal.)
of the linear equations. Which point
is common to all seven lines? Which
value in the equation determines the y
slope of each line? (x2*, y2*)
(x , y )
a. y(cid:1)4(cid:2)(cid:1)2(cid:1)x(cid:5)1(cid:2) 2 2
b. y(cid:1)4(cid:2)(cid:1)1(cid:1)x(cid:5)1(cid:2) (x1, y1)
c. y(cid:1)4(cid:2)(cid:1)1(cid:1)x(cid:5)1(cid:2) (x1*, y1*)
2
d. y(cid:1)4(cid:2)0(cid:1)x(cid:5)1(cid:2) x
e. y(cid:1)4(cid:2)12(cid:1)x(cid:5)1(cid:2) m = xy2** −− yx1**=yx2 −− yx1
f. y(cid:1)4(cid:2)1(cid:1)x(cid:5)1(cid:2) 2 1 2 1
g. y(cid:1)4(cid:2)2(cid:1)x(cid:5)1(cid:2) Any two points on a nonvertical line can be
used to determine its slope.
Use your results to write an equation Figure P.14
of a line passing through (cid:1)(cid:1)1, 4(cid:2)
with a slope of m.
You can write an equation of a nonvertical line if you know the slope of the line
and the coordinates of one point on the line. Suppose the slope is mand the point is
(cid:1)x , y (cid:2).If (cid:1)x, y(cid:2)is any other point on the line,then
1 1
y(cid:1)y
1 (cid:2)m.
x(cid:1)x
1
This equation, involving the two variables x and y, can be rewritten in the form
y(cid:1)y (cid:2)m(cid:1)x(cid:1)x (cid:2),which is called the point-slope equation of a line.
1 1
Point-Slope Equation of a Line
An equation of the line with slope mpassing through the point (cid:1)x , y (cid:2)is given
1 1
by
y(cid:1)y (cid:2)m(cid:1)x(cid:1)x (cid:2).
y y = 3x − 5 1 1
1
x EXAMPLE 1 Finding an Equation of a Line
1 3 4
−1 ∆y = 3 Find an equation of the line that has a slope of 3 and passes through the point (cid:1)1, (cid:1)2(cid:2).
−2
∆x = 1
Solution
−3
(1, −2)
−4 y (cid:1)y1(cid:2)m(cid:1)x(cid:1)x1(cid:2) Point-slope form
y(cid:1)(cid:1)(cid:1)2(cid:2)(cid:2)3(cid:1)x(cid:1)1(cid:2) Substitute (cid:1)2for y,1 for x,and 3 for m.
−5 1 1
y (cid:5)2(cid:2)3x(cid:1)3 Simplify.
The line with a slope of 3 passing through y (cid:2)3x(cid:1)5 Solve for y.
. the point (cid:1)1, (cid:1)2(cid:2)
....
Figure P.15 (See Figure P.15.)
Editable Graph
Try It Exploration A Exploration B Exploration C
NOTE Remember that only nonvertical lines have a slope. Consequently,vertical lines cannot
be written in point-slope form. For instance,the equation of the vertical line passing through
the point (cid:1)1, (cid:1)2(cid:2)is x(cid:2)1.