In Problems 95–100, use linear regression to construct linear models of the form
95.COLLEGE TUITIONFind a linear model for the public col- lege tuition data given in Table 5, where x is years after 1999 and y is the average tuition in dollars. Round a and b to the nearest dollar. Use your model to predict the average public college tuition in 2008, and to estimate when the average public college tuition will reach $15,000 per year.
Table 5 Average Annual Tuition at Public and Private Colleges
School Year
Ending Public Private
1999 $7,107 $19,368
2000 $7,310 $20,186
2001 $7,586 $21,368
2002 $8,022 $22,413
2003 $8,502 $23,340
2004 $9,249 $24,636
2005 $9,877 $26,025
Source: www.infoplease.com yaxb.
VTakeoff ground speed at altitude A for the same AAltitude above sea level
VsTakeoff
96.COLLEGE TUITIONFind a linear model for the private col- lege tuition data given in Table 5, where x is years after 1999 and y is the average tuition in dollars. Round a and b to the nearest dollar. Use your model to predict the average private college tu- ition in 2008, and to estimate when the average private college tuition will reach $35,000 per year.
97.OLYMPIC GAMESFind a linear model for the men’s 100-meter freestyle data given in Table 6 where x is years since 1968 and y is winning time (in seconds). Do the same for the women’s 100-meter freestyle data (round to three decimal places). Do these models indicate that the women will eventu- ally catch up with the men? If so, when? Do you think this will actually occur?
Table 6 Winning Times in Olympic Swimming Events
200-Meter Backstroke 100-Meter Freestyle
Men Women
Men Women (minutes: (minutes:
(seconds) (seconds) seconds) seconds)
1968 52.20 60.00 2:09.60 2:24.80
1972 51.22 58.59 2:02.82 2:19.19
1976 49.99 55.65 1:59.19 2:13.43
1980 50.40 54.79 2:01.93 2:11.77
1984 49.80 55.92 2:00.23 2:12.38
1988 48.63 54.93 1:59.37 2:09.29
1992 49.02 54.65 1:58.47 2:07.06
1996 48.74 54.50 1.58.54 2:07.83
2000 48.30 53.83 1:56.76 2:08.16
2004 48.17 53.84 1:54.95 2:09.19
Source: www.infoplease.com
98.OLYMPIC GAMESFind a linear model for the men’s 200-meter backstroke data given in Table 6, where x is years since 1968 and y is winning time (in seconds). Do the same for the women’s 200-meter backstroke data (round to three decimal places). Do these models indicate that the women will eventually catch up with the men? If so, when? Do you think this will actually occur?
yaxb yaxb
99. SUPPLY AND DEMANDTable 7 contains price–supply data and price–demand data for corn. Find a linear model for the price–supply data where x is price (in dol- lars) and y is supply (in billions of bushels). Do the same for the price–demand data. Find the equilibrium price for corn.
Table 7 Supply and Demand for U.S. Corn
Price Supply Price Demand
$/bu (billion bu) $/bu (billion bu)
2.15 6.29 2.07 9.78
2.29 7.27 2.15 9.35
2.36 7.53 2.22 8.47
2.48 7.93 2.34 8.12
2.47 8.12 2.39 7.76
2.55 8.24 2.47 6.98
2.71 9.23 2.59 5.57
Source: www.usda.gov/nass/pubs/histdata.htm
100. SUPPLY AND DEMANDTable 8 contains price–supply data and price–demand data for soybeans. Find a linear model for the price–supply data where x is supply (in billions of bushels) and y is price (in dollars). Do the same for the price–demand data. Find the equilibrium price for soybeans.
Table 8 Supply and Demand for U.S. Soybeans
Price Supply Price Demand
$/bu (billion bu) $/bu (billion bu)
5.15 1.55 4.93 2.60
5.79 1.86 5.48 2.40
5.88 1.94 5.71 2.18
6.07 2.08 6.07 2.05
6.15 2.15 6.40 1.95
6.25 2.27 6.66 1.85
6.65 2.53 7.25 1.67
Source: www.usda.gov/nass/pubs/histdata.htm yaxb
yaxb
2-3 Quadratic Functions
Z Defining Quadratic Functions
Z The Vertex Form of a Quadratic Function Z Completing the Square
Z Finding the Equation of a Parabola Z Modeling with Quadratic Functions
The graph of the squaring function is shown in Figure 1. Notice that h is an even function; that is, the graph of h is symmetrical with respect to the y axis.
Also, the lowest point on the graph is (0, 0). Let’s explore the effect of applying a sequence of basic transformations to the graph of h. (A brief review of Section 1-4 would be helpful at this point.)
h(x)x2
ZFigure 1 Squaring function h(x)x2.
ZZZEXPLORE-DISCUSS 1
Indicate how the graph of each function is related to the graph of
Discuss the symmetry of the graphs and find the highest or lowest point, whichever exists, on each graph.
(A) (B) (C)
(D) n(x) 3(x1)21 3x26x4
m(x) (x4)28 x28x8
g(x)0.5(x2)230.5x22x5 f (x)(x3)27x26x2
h(x)x2.
h(x)
5
5 5 x
Z Defining Quadratic Functions
Graphing the functions in Explore-Discuss 1 produces figures similar in shape to the graph of the squaring function in Figure 1. These figures are called parabolas. The functions that produced these parabolas are examples of the important class of quad- ratic functions, which we will now define.
Z DEFINITION 1 Quadratic Functions
If a, b, and c are real numbers with then the function
is called a quadratic function and its graph is called a parabola.* This is known as the general form of a quadratic function.
f(x)ax2bxc a0,
*A more general definition of a parabola that is independent of any coordinate system is given in Section 8-1.
Because the expression represents a real number no matter what number we substitute for x,
the domain of a quadratic function is the set of all real numbers.
We will discuss methods for determining the range of a quadratic function later in this section. Typical graphs of quadratic functions are illustrated in Figure 2.
ax2bxc
*In Problem 63 of Exercises 2-3, you will be asked to show that any function of this form fits the definition of quadratic function in Definition 1.
ZFigure 2 Graphs of quadratic functions.
10 10
10
10
10 10
10
10
10 10
10
10
Z The Vertex Form of a Quadratic Function
We will begin our detailed study of quadratic functions by examining some in a spe- cial form, which we will call the vertex form:*
We’ll see where the name comes from in a bit. For now, refer to Explore-Discuss 1.
Any function of this form is a transformation of the basic squaring function so we can use transformations to analyze the graph.
g(x)x2, f(x)a(xh)2k
EXAMPLE 1 The Graph of a Quadratic Function
Use transformations of to graph the function Use
your graph to determine the graphical significance of the constants 2, 3, and 4 in this function.
SOLUTION
Multiplying by 2 vertically stretches the graph by a factor of 2. Subtracting 3 inside the square moves the graph 3 units to the right. Adding 4 outside the square moves the graph 4 units up. The graph of f is shown in Figure 3, along with the graph of g(x)x2.
f(x)2(x3)24.
g(x)x2
(a) f(x)x2 9 (b) g(x)2x215x30 (c) h(x) 0.3x2x4
The lowest point on the graph of f is (3, 4), so and determine the key point where the graph changes direction. The constant affects the width of the parabola. Our results are checked by graphing f and with a graphing
calculator (Fig. 4).
g(x)x2 a2
k4 h3
ZFigure 3
x y
5 5
10
5
yx2 y 2(x 3)2 4
(3, 4)
10 10
10
10
ZFigure 4
MATCHED PROBLEM 1
Use transformations of to graph the function
Use your graph to determine the significance of the constants and 5 in this function.
12, 2,
f(x) 12(x2)25.
g(x)x2
ZZZEXPLORE-DISCUSS 2
Explore the effect of changing the constants a, h, and k on the graph of (A) Let and Graph function f for and 3 simulta- neously in the same viewing window. Explain the effect of changing k on the graph of f.
(B) Let and Graph function f for and 5 simulta- neously in the same viewing window. Explain the effect of changing h on the graph of f.
(C) Let and Graph function f for and 3 simul- taneously in the same viewing window. Then graph function f for
and simultaneously in the same viewing window. Explain the effect of changing a on the graph of f.
(D) Can all quadratic functions of the form be rewritten as a(xh)2k?
yax2bxc 0.25
a1, 1, a0.25, 1,
k 2.
h5
h 4, 0, k2.
a1
k 4, 0, h5.
a1 f(x)a(xh)2k.
Explore-Discuss 2 will help to clarify the significance of the constants a, h, and k in the form f(x)a(xh)2k.
Every parabola has a point where the graph reaches a maximum or minimum and changes direction. We will call that point the vertex of the parabola. Finding the vertex is key to many of the things we’ll do with parabolas. Example 1 and Explore-Discuss 2 demonstrate that if a quadratic function is in the form then the vertex is the point (h, k).
Next, notice that the graph of is symmetric about the y axis. As a result, the transformation is symmetric about the vertical line (which runs through the vertex). We will call this vertical line of symmetry the axis, or axis of symmetry of a parabola. If the page containing the graph of f is folded along the line the two halves of the graph would match exactly.
Finally, Explore-Discuss 2 illustrates the significance of the constant a in If a is positive, the graph has a minimum and opens upward.
But if a is negative, the graph will be a vertical reflection of and will have a maximum and open downward. The size of a determines the width of the parabola:
if the graph is narrower than and if it is wider.
These properties of a quadratic function in vertex form are summarized next.
a 6 1, h(x)x2,
a 7 1,
h(x)x2 f(x)a(xh)2k.
x3,
x3 f(x)2(x3)24
h(x)x2
f(x)a(xh)2k,
Z PROPERTIES OF A QUADRATIC FUNCTION AND ITS GRAPH Given a quadratic function in vertex form
we summarize general properties as follows:
1.The graph of f is a parabola:
a0 f(x)a(xh)2k
x f(x)
k
h Axis xh
Vertex (h, k)
Min f(x)
a 0 Opens upward
x f(x)
k
h Axis xh
Vertex (h, k)
Max f(x)
a 0 Opens downward
2.Vertex: (h, k) (parabola rises on one side of the vertex and falls on the other).
3.Axis (of symmetry): (parallel to y axis).
4. is the minimum if and the maximum if
5.Domain: all real numbers; range: if or if 6.The graph of f is the graph of translated horizontally h units
and vertically k units.
g(x)ax2
a 7 0.
[ k, ) a 6 0
(, k ]
a 6 0.
a 7 0 f(h)k
xh
ZFigure 5
EXAMPLE 2 Analyzing a Quadratic Function
For the following quadratic function, analyze the graph, and check your results with a graphing calculator:
SOLUTION
We can rewrite the function as Comparing this equa-
tion to we see that and Therefore,
the vertex is the axis of symmetry is the maximum value is and the range is The function f is increasing on
and decreasing on The graph of f is the graph of shifted to the left one unit and upward five and one-half units. To check these results, we graph f and g simultaneously in the same viewing window, use the MAXIMUM command to locate the vertex, and add the graph of the axis of symmetry (Fig. 5).
g(x) 0.5x2 [1, ).
(, 1 ] (, 5.5 ] .
f(1)5.5,
x 1,
(1, 5.5),
k5.5.
a 0.5, h 1,
ya(xh)2k,
f(x) 0.5 [ x(1) ]25.5.
f(x) 0.5(x1)25.5
MATCHED PROBLEM 2
For the following quadratic function, analyze the graph, and check your results with a graphing calculator:
f(x) (x1.5)21.25
ZZZCAUTIONZZZ
Be careful with the sign when finding the first coordinate of a vertex. The generic vertex form has in it, so when we have the first coordinate of the vertex is actually negative 1.
(x1)2, (xh)2
Z Completing the Square
Now that we can recognize the properties of a quadratic function in vertex form, the obvious question is “What if a quadratic function is not in vertex form?”
More often than not, the quadratic functions we encounter will be in the form The method of completing the square can be used to find the vertex form of such a quadratic function. We’ll also find this process useful in solv- ing equations later in the chapter.
f(x)ax2bxc.
6 6
6
6
Given the quadratic expression
what number should be added to this expression to make it a perfect square? To find out, consider the square of the following expression:
is the square of one-half the coefficient of x.
We see that the third term on the right side of the equation is the square of one-half the coefficient of x in the second term on the right; that is, is the square of This observation leads to the following rule:
1 2(2m).
m2
m2
(xm)2x22mxm2
x2bx ZZZEXPLORE-DISCUSS 3
Replace ? in each of the following with a number that makes the equation valid.
(A) (B)
(C) (D)
Replace ? in each of the following with a number that makes the expression a perfect square of the form
(E) (F)
(G) x2bx?
x212x? x210x?
(xh)2.
(x4)2x28x? (x3)2x26x?
(x2)2x24x? (x1)2x22x?
Z COMPLETING THE SQUARE
To complete the square of the quadratic expression
Leading coefficient 1
add the square of one-half the coefficient of x; that is, add or
The resulting expression can be factored as a perfect square:
x2bxab
2b2ax b 2b2 b2 ab 4
2b2 x2bx
{ {
EXAMPLE 3 Completing the Square
Complete the square for each of the following:
(A) (B) (C)
SOLUTIONS
(A)
Add that is, 9.
(B)
Add that is,
(C)
Add that is,
Note: In each case, the quadratic expression ends up factoring as (x half the coef-
ficient of the x term).
MATCHED PROBLEM 3
Complete the square for each of the following:
(A) (B) (C)
x22
3x x27
4x x28x
9 64. a1
23 4 b2;
x2 3 4x 9
64ax3 8b2 x2 3
4x
4 25. a1
24 5b2;
x2 4 5x 4
25ax2 5b2 x2 4
5x
c1 2(6)d2;
x26x9(x3)2 x26x
x23 4x x24
5x x26x
It is important to note that the rule for completing the square applies to only quadratic expressions in which the coefficient of is 1. We’ll see how to overcome this limitation later.
Z Finding the Equation of a Parabola
x2
EXAMPLE 4 Finding the Vertex Form of a Parabola
Find the vertex form of f (x)x28x4,then write the vertex and axis.
SOLUTION
We will separate with parentheses, then use completing the square to factor part of f as a perfect square.
Group together.
Find the number needed to complete the square.
add 16, then subtract it at the end.
Factor parentheses; combine like terms.
fis now in vertex form.
The vertex form is The vertex is and the axis is
MATCHED PROBLEM 4
Find the vertex form of then write the vertex and axis.
When the coefficient of x2is not 1, the procedure is just a bit more complicated.
g(x)x210x1, x4.
(4, 12) f (x)(x4)212.
(x4)212
(x28x16)416
c1
2(8)d216;
(x28x?)4 (x28x)4
x28x
f (x)x28x4 x28x
EXAMPLE 5 Finding the Vertex Form of a Parabola
Find the vertex form of Find the vertex and axis, then describe the graph verbally and check your answer with a graphing calculator.
SOLUTION
We need a coefficient of 1 on the so after grouping the first two terms, we’ll fac- tor out
Factor out of the first two terms.
add this number inside the parentheses.
Factor the parentheses; combine like terms.
fis in vertex form.
The vertex is and the axis is Because is negative, the parabola opens downward and has a maximum value of The function f is increasing on and decreasing on The range of f is The graph is shown in
Figure 6.
(, 13] . [43, ).
(, 43]
1 3.
a 3
x43. (43, 13)
3ax 4 3b2 1
3 3ax2 8
3x16
9b5 16 3 3ax2 8
3x16
9b5?
a1 28
3 b216 9;
3ax2 8 3xb5
3
f (x)(3x28x)5 3.
x2,
f(x) 3x28x5.
Because of the factor, we actually subtracted so we also add 16
3.
16 3, 3
1 1
5
5
ZFigure 6
MATCHED PROBLEM 5
Repeat Example 5 for
g(x) 2x27x3.
ZZZCAUTIONZZZ
When completing the square on a quadratic function with the num- ber that you add or subtract at the end will be different from the number you added inside the parentheses.
a1,
A key observation based on Examples 4 and 5 will help us to find the vertex of a parabola quickly, without completing the square. In both Example 4 and 5, the first coordinate of the vertex worked out to be where the function f was written in the form This provides a simple way to find the vertex of a parabola in that form. (For a proof of this fact, see Problem 64 in Exercises 2-3.)
f(x)ax2bxc.
2ab,
Z FINDING THE VERTEX OF A PARABOLA
When a quadratic function is written in the form the first coordinate of the vertex can be found using the formula
The second coordinate can then be found by evaluating f at the first coordinate.
x b
2a
f(x)ax2bxc,
EXAMPLE 6 Finding the Vertex of a Parabola
Find the vertex of the parabola
SOLUTION
The first coordinate is given by
The second coordinate is f (3):
The vertex is (3, 43).
f (3) 5(3)230(3)243
x b
2a
30 103 f (x) 5x230x2.