6 ⬇ 2.15 The x-intercepts are approximately - 0.15 and 2.15
99. On one set of coordinate axes, graph the family of parabolas
100. On one set of coordinate axes, graph the family of parabolas f1x2 =x2+bx +1 for b= -4, b= 0, and b=4. Describe the general characteristics of this family.
101. State the circumstances that cause the graph of a quadratic function f1x2 =ax2+ bx+c to have no x-intercepts.
102. Why does the graph of a quadratic function open up if a7 0 and down if a6 0?
103. Can a quadratic function have a range of (-⬁, ⬁)? Justify your answer.
104. What are the possibilities for the number of times the graphs of two different quadratic functions intersect?
‘Are You Prepared?’ Answers
1. (0, -9), (-3, 0), (3, 0) 2. e-4, 1
2f 3. 25
4 4. right; 4
SECTION 3.4 Build Quadratic Models from Verbal Descriptions and from Data 159
Now Workthe ‘Are You Prepared?’ problems on page 164.
OBJECTIVES 1 Build Quadratic Models from Verbal Descriptions (p. 159) 2 Build Quadratic Models from Data (p. 163)
• Problem Solving (Appendix A, Section A.8, pp. A63 –A70)
• Linear Models: Building Linear Functions from Data (Section 3.2, pp. 140–143)
PREPARING FOR THIS SECTION Before getting started, review the following:
3.4 Build Quadratic Models from Verbal Descriptions and from Data
In this section we will first discuss models in the form of a quadratic function when a verbal description of the problem is given. We end the section by fitting a quadratic function to data, which is another form of modeling.
When a mathematical model is in the form of a quadratic function, the properties of the graph of the quadratic function can provide important information about the model. In particular, we can use the quadratic function to determine the maximum or minimum value of the function. The fact that the graph of a quadratic function has a maximum or minimum value enables us to answer questions involving optimization, that is, finding the maximum or minimum values in models.
Build Quadratic Models from Verbal Descriptions
In economics, revenue R, in dollars, is defined as the amount of money received from the sale of an item and is equal to the unit selling price p, in dollars, of the item times the number x of units actually sold. That is,
R =xp
The Law of Demand states that p and x are related: As one increases, the other decreases. The equation that relates p and x is called the demand equation. When the demand equation is linear, the revenue model is a quadratic function.
Maximizing Revenue
The marketing department at Texas Instruments has found that, when certain calculators are sold at a price of p dollars per unit, the number x of calculators sold is given by the demand equation
x= 21,000- 150p
(a) Find a model that expresses the revenue R as a function of the price p.
(b) What is the domain of R?
(c) What unit price should be used to maximize revenue?
(d) If this price is charged, what is the maximum revenue?
(e) How many units are sold at this price?
(f) Graph R.
(g) What price should Texas Instruments charge to collect at least $675,000 in revenue?
(a) The revenue R is R= xp, where x =21,000 -150p.
R= xp= (21,000- 150p)p= -150p2 +21,000p The Model
(b) Because x represents the number of calculators sold, we have xÚ 0, so 21,000-150pÚ 0. Solving this linear inequality, we find that p …140. In addition, Texas Instruments will only charge a positive price for the calculator, so p7 0. Combining these inequalities, the domain of R is 5 p 兩 0 6p …140 6 .
1
E X A M P L E 1
Solution
(c) The function R is a quadratic function with a= -150, b= 21,000, and c = 0.
Because a6 0, the vertex is the highest point on the parabola. The revenue R is a maximum when the price p is
p= - b
2a = - 21,000
2(-150) = -21,000
-300 = +70.00 c
a= -150, b= 21,000 (d) The maximum revenue R is
R(70) = -150(70)2+ 21,000(70) = +735,000
(e) The number of calculators sold is given by the demand equation x =21,000 -150p. At a price of p= +70,
x= 21,000- 150(70) =10,500 calculators are sold.
(f) To graph R, plot the intercept (140, 0) and the vertex (70, 735,000). See Figure 24 for the graph.
Figure 24
(70, 735,000)
Price p per calculator (dollars) 800,000
700,000 600,000 500,000 400,000 200,000 100,000 300,000
14 28 42 56 70 84 98 112 126 140
0 p
R
Revenue (dollars)
Figure 25
0 14 28 42 56
Price p per calculator (dollars) 70 84 98 112 126 140 800,000
700,000 600,000 500,000
Revenue (dollars)
400,000 200,000 100,000 300,000
p R
(50, 675,000) (70, 735,000)
(90, 675,000)
(g) Graph R= 675,000 and R1p2 = -150p2+ 21,000p on the same Cartesian plane. See Figure 25. We find where the graphs intersect by solving
675,000= -150p2 +21,000p
150p2- 21,000p+ 675,000=0 Add 150p2-21,000p to both sides.
p2- 140p +4500= 0 Divide both sides by 150.
1p- 502 1p-902 = 0 Factor.
p= 50 or p=90 Use the Zero-Product Property.
The graphs intersect at (50, 675,000) and (90, 675,000). Based on the graph in Figure 25, Texas Instruments should charge between $50 and $90 to earn at least
$675,000 in revenue.
Now Work P R O B L E M 3
SECTION 3.4 Build Quadratic Models from Verbal Descriptions and from Data 161
Maximizing the Area Enclosed by a Fence
A farmer has 2000 yards of fence to enclose a rectangular field. What are the dimensions of the rectangle that encloses the most area?
Figure 26 illustrates the situation. The available fence represents the perimeter of the rectangle. If x is the length and w is the width, then
2x +2w =2000 (1)
The area A of the rectangle is
A= xw
To express A in terms of a single variable, solve equation (1) for w and substitute the result in A= xw. Then A involves only the variable x. [You could also solve equation (1) for x and express A in terms of w alone. Try it!]
2x+ 2w= 2000 2w= 2000- 2x
w= 2000-2x
2 = 1000- x Then the area A is
A= xw= x(1000- x) = -x2+ 1000x Now, A is a quadratic function of x.
A(x) = -x2+ 1000x a = -1, b =1000, c =0
Figure 27 shows the graph of A(x) = -x2 +1000x. Since a6 0, the vertex is a maximum point on the graph of A. The maximum value occurs at
x = - b
2a = - 1000 2(-1) = 500 The maximum value of A is
Aa- b
2ab = A(500) = -5002+ 1000(500)= -250,000+ 500,000= 250,000 The largest rectangle that can be enclosed by 2000 yards of fence has an area of 250,000 square yards. Its dimensions are 500 yards by 500 yards.
Now Work P R O B L E M 7
Analyzing the Motion of a Projectile
A projectile is fired from a cliff 500 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 400 feet per second. In physics, it is established that the height h of the projectile above the water can be modeled by
h(x) = -32x2
(400)2 + x+ 500
where x is the horizontal distance of the projectile from the base of the cliff. See Figure 28.
E X A M P L E 2
Solution
E X A M P L E 3
Figure 26
x x
w w
(500, 250,000)
(0, 0) (1000, 0)
A
1000 500 250,000
x Figure 27
1000 2000 3000 4000 500
1000 1500 2500 2000 h (x)
x 45°
5000 Figure 28
(a) Find the maximum height of the projectile.
(b) How far from the base of the cliff will the projectile strike the water?
(a) The height of the projectile is given by a quadratic function.
h(x) = -32x2
(400)2 + x+ 500= -1
5000 x2+ x+ 500
We are looking for the maximum value of h. Since a 6 0, the maximum value is obtained at the vertex, whose x-coordinate is
x= - b
2a = - 1
2a- 1 5000b
= 5000 2 = 2500 The maximum height of the projectile is
h(2500) = -1
5000 (2500)2 +2500+ 500= -1250+2500+ 500= 1750 ft (b) The projectile will strike the water when the height is zero. To find the distance
x traveled, solve the equation h(x) = -1
5000 x2 +x + 500=0 The discriminant of this quadratic equation is
b2 -4ac =12 -4a -1
5000b(500) = 1.4 Then
x= -b{ 2b2 -4ac
2a = -1 { 21.4
2a- 1 5000b
⬇ e-458 5458
Discard the negative solution. The projectile will strike the water at a distance of about 5458 feet from the base of the cliff.
Now Work P R O B L E M 1 1
The Golden Gate Bridge
The Golden Gate Bridge, a suspension bridge, spans the entrance to San Francisco Bay. Its 746-foot-tall towers are 4200 feet apart. The bridge is suspended from two huge cables more than 3 feet in diameter; the 90-foot-wide roadway is 220 feet above the water. The cables are parabolic in shape* and touch the road surface at the center of the bridge. Find the height of the cable above the road at a distance of 1000 feet from the center.
See Figure 29 on the next page. Begin by choosing the placement of the coordinate axes so that the x-axis coincides with the road surface and the origin coincides with the center of the bridge. As a result, the twin towers will be vertical (height 746- 220= 526 feet above the road) and located 2100 feet from the center. Also, the cable, which has the shape of a parabola, will extend from the towers, open up, and have its vertex at (0, 0). The choice of placement of the axes enables us to identify the equation of the parabola as y = ax2, a 7 0. Notice that the points (-2100, 526) and (2100, 526) are on the graph.
Solution
E X A M P L E 4
Solution
* A cable suspended from two towers is in the shape of a catenary, but when a horizontal roadway is suspended from the cable, the cable takes the shape of a parabola.
Seeing the Concept
Graph
h1x2 = -1
5000x2+ x+500 0…x …5500
Use MAXIMUM to find the maximum height of the projectile, and use ROOT or ZERO to find the distance from the base of the cliff to where it strikes the water. Compare your results with those obtained in Example 3.
SECTION 3.4 Build Quadratic Models from Verbal Descriptions and from Data 163
Based on these facts, we can find the value of a in y= ax2. y =ax2
526 =a(2100)2 x=2100, y =526 a = 526