DIRECTIONS FOR GROUP PROJECT
Chapter 2 Review Extended Application: Power Functions
80. Area A rectangular field is to have a perimeter of 6000 ft
a.Write the area,A, of the field as a function of the width,w.
b.Find the domain of the function in part a.
c.Use a graphing calculator to sketch the graph of the function in part a.
d.Describe what the graph found in part c tells you about how the area of the field varies with the width.
500 m2.
EU31 EU31 EU31
EU31 EU31 EU31
EU31 EU31 EU31
1855 1875 1895 1915 1935 1955 1975 1995 0
20 40 60 80 100
Share of global energy consumption (percent)
Year Wood Coal Oil
Natural gas Hydropower Nuclear
a.In what year were the percent of wood and coal use equal?
What was the percent of each used in that year?
b.In what year were the percent of oil and coal use equal?
What was the percent of each used in that year?
YOUR TURN ANSWERS 1.
2. (a) 2x214xh12h223x23h24(b) 1 and 1/2 12`, 222 <12, `2, 10, `2
Quadratic Functions; Translation and Reflection
How much should a company charge for its seminars? When Power and Money, Inc., charges $600 for a seminar on management techniques, it attracts 1000 people. For each $20 decrease in the fee, an additional 100 people will attend the seminar.The managers wonder how much to charge for the seminar to maximize their revenue.
2.2
APPLY IT
In Example 6 in this section we will see how knowledge of quadratic functionswill help provide an answer to the question above.
A linear function is defined by
for real numbers aand b. In a quadratic functionthe independent variable is squared. A quadratic function is an especially good model for many situations with a maximum or a minimum function value. Quadratic functions also may be used to describe supply and demand curves; cost, revenue, and profit; as well as other quantities. Next to linear func- tions, they are the simplest type of function, and well worth studying thoroughly.
f1x25ax1b,
FOR REVIEW
In this section you will need to know how to solve a quadratic equation by factoring and by the quadratic formula, which are covered in Sections R.2 and R.4.
Factoring is usually easiest;
when a polynomial is set equal to zero and factored, then a solu- tion is found by setting any one factor equal to zero. But factor- ing is not always possible. The quadratic formula will provide the solution to anyquadratic equation.
Quadratic Function
Aquadratic functionis defined by
where a, b, and care real numbers, with a20.
f1x2 5ax21bx1c,
The simplest quadratic function has with and This function describes situations where the dependent variable yis proportional to the squareof the independent variable x. The graph of this function is shown in Figure 14. This graph is called a parabola. Every quadratic function has a parabola as its graph. The lowest (or highest) point on a parabola is the vertexof the parabola. The vertex of the parabola in Figure 14 is
If the graph in Figure 14 were folded in half along the y-axis, the two halves of the parabola would match exactly. This means that the graph of a quadratic function is symmet- ricwith respect to a vertical line through the vertex; this line is the axisof the parabola.
There are many real-world instances of parabolas. For example, cross sections of spot- light reflectors or radar dishes form parabolas. Also, a projectile thrown in the air follows a parabolic path. For such applications, we need to study more complicated quadratic func- tions than as in the next several examples.
Graphing a Quadratic Function Graph
SOLUTION Each value of ywill be 4 less than the corresponding value of yin The graph of has the same shape as that of but is 4 units lower. See Figure 15. The vertex of the parabola (on this parabola, the lowestpoint) is at The x-intercepts can be found by letting to get
from which and are the x-intercepts. The axis of the parabola is the vertical line
Example 1 suggests that the effect of cin is to lower the graph ifcis negative and to raise the graph ifcis positive. This is true for any function; the movement up or down is referred to as avertical translationof the function.
Graphing Quadratic Functions
Graph with and
SOLUTION Figure 16 shows all four functions plotted on the same axes. We see that since ais negative, the graph opens downward. When ais between 1 and 1 (that is, when the graph is wider than the original graph, because the values of yare smaller in magnitude. On the other hand, when ais greater than 1 or less than 1, the graph is steeper.
Example 2 shows that the sign of ain determines whether the parabola opens upward or downward. Multiplying by a negative number flips the graph of f upside down. This is called a vertical reflectionof the graph. The magnitude of adeter- mines how steeply the graph increases or decreases.
Graphing Quadratic Functions
Graph for 0, and
SOLUTION Figure 17 shows all three functions on the same axes. Notice that since the number is subtracted beforethe squaring occurs, the graph does not move up or down but instead moves left or right. Evaluating at gives the same result as evaluating at Therefore, when we subtract the positive number 3 from x, the graph shifts 3 units to the right, so the vertex is at f1x25x2 x50. f1x25 1x23123, 02 2.xSimilarly, when we subtract53
24.
h53, y5 1x2h22
f1axx221bx1c a5 20.5),
a5 24.
a5 22, a5 21,
a5 20.5, y5ax2
ax21bx1c x50. x52 x5 22
05x224,
y50 y5x2 10, 242.
y5x224 y5x2.
y5x224.
y5x2, 10, 02.
c50.
b50, a51, f1x2 5x2,
EXAMPLE 1 FIGURE 14
y = x2
–4 –3 –2 –10 1 2 3 4 x y
2 4 6 8 10 12 14
FIGURE 15
EXAMPLE 2
EXAMPLE 3 FIGURE 16
–4 –3 –2 –1 –4
2 3
1 4
0 x
y
4 8 16
12 y = x2 – 4
–4 –2 0 2 4 x
y
–2 –4 –6 –8 –10 –12 –14 y = –0.5x2
y = –2x2 y = –4x2 y = –x2
Method 2 The Quadratic Formula Method 1 Completing the Square
2.2 Quadratic Functions;Translation and Reflection 59 the negative number from x—in other words, when the function becomes
graph shifts to the left 4 units.
The left or right shift of the graph illustrated in Figure 17 is called a horizontal trans- lationof the function.
If a quadratic equation is given in the form we can identify the transla- tions and any vertical reflection by rewriting it in the form
In this form, we can identify the vertex as A quadratic equation not given in this form can be converted by a process called completing the square.The next example illus- trates the process.
Graphing a Quadratic Function Graph
SOLUTION To begin, factor from the x-terms so the coefficient of is 1:
Next, we make the expression inside the parentheses a perfect square by adding the square of one-half of the coefficient of x, which is . Since there is a factor of 3 out- side the parentheses, we are actually adding . To make sure that the value of the function is not changed, we must also add to the function. Actually, we are simply adding , and not changing the function. To summarize our steps,
Factor out
Add and subtract times the coefficient of
. Factor and combine terms.
The function is now in the form ya(x h)2k. Since h and k , the graph is the graph of the parabola y x2 translated unit to the left and units upward. This puts the vertex at ( , ). Since a 3 is negative, the graph will be flipped upside down. The 3 will cause the parabola to be stretched vertically by a factor of 3.
These results are shown in Figure 18.
Instead of completing the square to find the vertex of the graph of a quadratic function given in the form we can develop a formula for the vertex. By the qua-
dratic formula, if where then
Notice that this is the same as x5 2b
2a 6 "b224ac 2a 52b
2a 6Q,
x5 2b6"b224ac
2a .
a20, ax21bx1c50,
y5ax21bx1c, 4/3
21/3
4/3
1/3
4/3
21/3
5 23ax1 1 3b214
3
x22. 112
23 5 23ax212
3x1 1
9b 1113a1 9b
23.
y5 23ax212 3xb 11 23.1192 13.119250
31.1221.19322.1912
322519
y5 23ax21 2 3xb 11.
x2 23
y5 23x222x11.
1h, k2. y5a1x2h221k.
ax21bx1c, 1x1422— the
f1x25 24
EXAMPLE 4 FIGURE 17
–5 –4 –3 –2 –1 0 1
2 3 4 5 –2
2 4 8 10 12
x y
y = (x + 4)2 y = (x – 3)2 y = x2
0 x
y
y = –3x2 – 2x + 1 2 1 –1 –2
2
–1 1
–2 1_ 3
4_
( ) – ,3
FIGURE 18
Another situation that may arise is the absence of any x-intercepts, as in the next example.
Graphing a Quadratic Function Graph
SOLUTION This does not appear to factor, so we’ll try the quadratic formula.
As soon as we see the negative value under the square root sign, we know the solutions are complex numbers. Therefore, there are no x-intercepts. Nevertheless, the vertex is still at
Substituting this into the equation gives
The y-intercept is at which is 2 units to the right of the parabola’s axis Using the symmetry of the figure, we can also plot the mirror image of this point on the opposite side of the parabola’s axis: at (2 units to the left of the axis), yis also equal to 6. Plot- ting the vertex, the y-intercept, and the point gives the graph in Figure 19.
We now return to the question with which we started this section.
Management Science
When Power and Money, Inc., charges $600 for a seminar on management techniques, it attracts 1000 people. For each $20 decrease in the fee, an additional 100 people will attend the seminar. The managers are wondering how much to charge for the seminar to maximize their revenue.
SOLUTION Let xbe the number of $20 decreases in the price. Then the price charged per person will be
and the number of people in the seminar will be
Number of people510001100x.
Price per person5600220x, 124, 62
x5 24
x5 22.
10, 62,
y5 122221412221652.
x5 2b 2a 524
2 5 22.
5 246"4224112 162
2112 5 246"28
2
a51, b54, c56 x5 2b6"b224ac
2a y5x214x16.
YOUR TURN 1 For the func- tion
(a)complete the square, (b)find the y-intercept, (c)find the x-intercepts, (d)find the vertex, and (e)sketch the graph.
y52x226x21,
EXAMPLE 5
FIGURE 19
0 x
y
y = x2 + 4x + 6 (–2, 2)2
(–4, 6) (0, 6) 4 6 8
EXAMPLE 6
where Since a parabola is symmetric with respect to its axis, the vertex is halfway between its two roots. Halfway between and is Once we have the x-coordinate of the vertex, we can easily find the y-coordinate by substituting the x-coordinate into the original equation. For the function in this example, use the quadratic formula to verify that the x-intercepts are at and and the vertex is at The y-intercept (where ) is 1.
The graph is in Figure 18.
x50 121/3, 4/32.
x51/3,
x5 21
x5 2b/12a2.
x5 2Qb/512a"2b222Q4ac/12a2. x 5 2b/12a21Q
Graph of the Quadratic Function
The graph of the quadratic function has its vertex at
The graph opens upward if a.0and downward if a,0.
a2b 2a, fa2b
2abb. f1x2 5ax21bx1c
APPLY IT
TRY YOUR TURN 1
2.2 Quadratic Functions;Translation and Reflection 61 The total revenue, is given by the product of the price and the number of people attending, or
We see by the negative in the -term that this defines a parabola opening downward, so the maximum revenue is at the vertex. The x-coordinate of the vertex is
The y-coordinate is then
Therefore, the maximum revenue is $800,000, which is achieved by charging
per person. TRY YOUR TURN 2
Notice in this last example that the maximum revenue was achieved by charging less than the current price of $600, which was more than made up for by the increase in sales.
This is typical of many applications. Mathematics is a powerful tool for solving such prob- lems, since the answer is not always what one might have guessed intuitively.
To solve problems such as Example 6, notice the following:
1.The key step after reading and understanding the problem is identifying a useful variable.
2. Revenue is always price times the number sold.
3.The expressions for the price and for the number of people are both linear functions of x.
4.We know the constant term in each linear function because we know what happens when
5.We know how much both the price and the number of people change each time x increases by 1, which gives us the slope of each linear function.
6.The maximum or minimum of a quadratic function occurs at its vertex.
The concept of maximizing or minimizing a function is important in calculus, as we shall see in future chapters.
In the next example, we show how the calculation of profit can involve a quadratic function.
Profit
A deli owner has found that his revenue from producing xpounds of vegetable cream cheese is given by while the cost in dollars is given by
(a) Find the minimum break-even quantity.
C1x255x1100.
R1x25 2x2130x, x50.
6002201102 5$400 600220x5
5800,000.
y5600,000140,00011022200011022 x5 2b
2a 5 240,000 21220002 510.
x2
5600,000140,000x22000x2. R1x25 1600220x2 110001100x2 R1x2,
YOUR TURN 2 Solve Exam- ple 6 with the following changes: a
$1650 price attracts 900 people, and each $40 decrease in the price attracts an additional 80 people.
EXAMPLE 7
FIGURE 20
5 15 25 35 45 55
20 60 100 140 180 220 260
x y
C(x) = 5x + 100
R(x) = –x2 + 30x
SOLUTION Notice from the graph in Figure 20 that the revenue function is a parabola opening downward and the cost function is a linear function that crosses the revenue function at two points. To find the minimum break-even quantity, we find where the two functions are equal.
Subtract from both sides.
Factor.
The two graphs cross when and The minimum break-even point is at The deli owner must sell at least 5 lb of cream cheese to break even.
x55. x55 x520.
5 1x252 1x2202
2x2130x 05x2225x1100
2x2130x55x1100 R1x2 5C1x2
(b)Find the maximum revenue.
SOLUTION By factoring the revenue function,
we can see that it has two roots, and The maximum is at the vertex, which has a value of xhalfway between the two roots, or (Alternatively, we
could use the formula The maximum revenue is
or $225.
(c)Find the maximum profit.
SOLUTION The profit is the difference between the revenue and the cost, or
This is just the negative of the expression factored in part (a), where we found the roots to be and The value ofxat the vertex is halfway between these two roots, or
(Alternatively, we could use the formula The value of the function here is
It is clear that this is a maximum, not only from Figure 20, but also because the profit function is a quadratic with a negative -term. A maximum profit of
$56.25 is achieved by selling 12.5 lb of cream cheese. TRY YOUR TURN 3
Below and on the next page, we provide guidelines for sketching graphs that involve translations and reflections.
x2 100556.25.
P112.525 212.52125112.52 2 12.5.)
225/1222 5 x5 2b/12a2 5 x5 151202/2512.5.
x520.
x55
5 2x2125x2100.
5 12x2130x2 2 15x11002 P1x25R1x22C1x2
225, 3011525 21521
R1152 5 x5 2b/12a2 5230/1222515.) x515.
x530.
x50
x12x1302, R1x2 5 2x2130x 5
YOUR TURN 3 Suppose the revenue in dollars is given by
and the cost is
given by Find
(a)the minimum break-even quan- tity, (b)the maximum revenue, and (c)the maximum profit.
C1x2 58x1192.
R1x2 5 2x2140x
Translations and Reflections of Functions
Let fbe any function, and let hand kbe positive constants (Figure 21).
The graph of y5f1x2 1kis the graph of y5f1x2translated upward by an amount k(Figure 22).
The graph of is the graph of translated downward by an amount k(Figure 23).
The graph of is the graph of translated to the right by an amount h(Figure 24).
The graph of y5f1x1h2is the graph of y5f1x2translated to the left by an amount h(Figure 25).
y5f1x2 y5f1x2h2
y5f1x2 y5f1x2 2k
x y
y = f(x)
x y
k
y = f(x) + k
x h
y
y = f(x + h)
x y
k
y = f(x) – k
x h
y
y = f(x – h)
FIGURE 21
FIGURE 23 FIGURE 24 FIGURE 25
FIGURE 22
2.2 Quadratic Functions;Translation and Reflection 63
Notice in Figure 27 another type of reflection, known as a horizontal reflection. Multi- plying xor by a constant a, to get or does not change the general appearance of the graph, except to compress or stretch it. When ais negative, it also causes a reflection, as shown in the last two figures in the summary for Also see Exercises 39–42 in this section.
Translations and Reflections of a Graph Graph
SOLUTION Begin with the simplest possible function, then add each variation in turn.
Start with the graph of As Figure 28 reveals, this is just one-half of the graph of lying on its side.
Now add another component of the original function, the negative in front of thex, giv- ing This is a horizontal reflection of the graph, as shown in Figure 29. Next, include the 4 under the square root sign. To get into the form
or we need to factor out the negative:
Now the 4 is subtracted, so this function is a translation to the right of the function by 4 units, as Figure 30 indicates.
f1x2 5"2x
"42x5"21x242.
f1x1h2,
f1xf12xh225"2x. f1x25"x42x
f1x2 5x2 f1x2 5"x.
f1x2 5 2"42x13.
a5 21.
y5a.f1x2 y5f1ax2
f1x2
Translations and Reflections of Functions
The graph of is the graph of reflected vertically across the x-axis, that is, turned upside down (Figure 26).
The graph of yy5 25f1f12xx22is the graph of yy55f1f1xx22reflected horizontally across the y-axis, that is, its mirror image (Figure 27).
FIGURE 26 FIGURE 27
x y
y = f(–x) x
y y = – f(x)
0 x
y
f(x) = 4 – x 2
4
4
–4 –2 2
–2 –4
0 x
y 4
4
–4 –2 2
2
–2
–4 f(x) = – x
0 x
y 4
4
–4 –2 2
2
–2
–4 f(x) = x
FIGURE 28 FIGURE 29 FIGURE 30
FIGURE 31
EXAMPLE 8
0 x
y 4
4
–4 –2 2
2
–2
–4 f(x) = – 4 – x + 3 (a)
0 x
y 4
4
–4 –2 2
2
–2
–4 f(x) = – 4 – x
5 5
3.1
3.1
f(x) 4 x 3
(b)
The effect of the negative in front of the radical is a vertical reflection, as in Figure 31, which shows the graph of Finally, adding the constant 3 raises the entire graph by 3 units, giving the graph off1x2 52"42x13 in Figure 32(a).
f1x25 2"42x.
2.2 EXERCISES
1.How does the value of aaffect the graph of Discuss the case for and for
2.How does the value of aaffect the graph of if In Exercises 3–8, match the correct graph A–F to the function without using your calculator. Then, if you have a graphing cal- culator, use it to check your answers. Each graph in this group shows xand yin
3. 4.
5. 6.
7.y5 2132x2212 8. y5 21x132212 y5 1x132212 y51x232212
y5 1x2322 y5x223
3210, 104.
a#0?
y5ax2 0#a#1.
a$1 y5ax2?
Given the following graph, sketch by hand the graph of the function described, giving the new coordinates for the three points labeled on the original graph.
210 10
10
210
210 10
10
210
210 10
10
210
210 10
10
210
210 10
10
210
210 10
10
210
Complete the square and determine the vertex for each of the following.
9. 10.
11. 12.
In Exercises 13–24, graph each parabola and give its vertex, axis, x-intercepts, and y-intercept.
13. 14.
15. 16.
17. 18.
19. 20.
3 f1x25 1
2 x216x124 f1x2 52x224x15
f1x25 2x216x26 f1x2 52x218x28
y5 23x226x14 y5 22x2212x216
y5x214x25 y5x215x16
y5 25x228x13 y5 22x218x29
y54x2220x27 y53x219x15
210 10
10
210
210 10
10
210
210 10
10
210
210 10
10
210
210 10
10
210
210 10
10
210
y
x
0 (5, 0)
(–1, 4)
(–3, –2)
If you viewed a graphing calculator image such as Figure 32(b), you might think the function contin- ues to go up and to the right. By realizing that is the vertex of the sideways parabola, we see that this is the rightmost point on the graph. Another approach is to find the domain of fby setting from which we conclude that This demonstrates the importance of knowing the algebraic techniques in order to interpret a graphing calculator image correctly.
x#4.
42x$0,
14, 32
23. 24.
In Exercises 25–30, follow the directions for Exercises 3–8.
25. 26.
27. 28.
29. y5 2"x1224 30.y5 2"x2224
y5"2x2224
y5"2x1224
y5"x2224
y5"x1224
f1x2 5 21
2x22x27 f1x251 2
3x228 3x11
3
(A) (B)
(C) (D)
(E) (F)
(A) (B)
(C) (D)
(E) (F)
31. y5 2f1x2 32.y5f1x222 12 TECHNOLOGY NOTE
2.2 Quadratic Functions;Translation and Reflection 65 Use the ideas in this section to graph each function without a
calculator.
35. 36.
37. 38.
Using the graph of in Figure 21, show the graph of where asatisfies the given condition.
39. 40.
41. 42.
Using the graph of in Figure 21, show the graph of where asatisfies the given condition.
43. 44.
45. 46.
47. If r is an x-intercept of the graph of what is an x-intercept of the graph of each of the following?
a. b.
c.
48. If bis the y-intercept of the graph of what is the y-intercept of the graph of each of the following?
a. b.
c.
APPLICATIONS
B u s i n e s s a n d E c o n o m i c s
Profit In Exercises 49–52, let be the cost to produce x batches of widgets, and let be the revenue in thousands of dollars. For each exercise, (a) graph both functions, (b) find the minimum break-even quantity, (c) find the maximum revenue, and (d) find the maximum profit.
49.
50.
51.
52.