Math 27 - Section 11.3 - Part III - Page 1
Section 11.3 - Quadratic Functions and Their Graphs - Part III
I.
Applications
A.
To find the maximum/minimum value for any quadratic function, we need to find the
vertex. So if the equation is in General Form, we first find the axis of symmetry by
B
evaluating the expression x = −
. Then to find the actual max/min, we substitute
2A
this value into the original equation to find y. So the x-value will tell us how many or
how much time, to make to get us the max/min y-value.
B.
Examples - Solve each of the following.
1.
A person standing close to the edge on the top of a 200-foot building throws a
baseball vertically upward. The quadratic function
s(t) = -16t2 + 64t + 200
models the ball's height above the ground, s(t), in feet, t seconds after it was
thrown. (Page 781, #60)
a.
After how many seconds does the ball reach its maximum height? What is
the maximum height?
Any time we see the word "maximum" or "minimum" in this class, we have to
think "vertex". So to find the vertex, we need to find the Axis of Symmetry
B
first. So in this case, we have to find −
.
2A
t= −
B
−(64)
−64
=
=
=2
2 A 2( −16) −32
This tells us that the ball will reach its maximum height after 2 seconds. To
find the maximum height, we find to find s(2).
2
s(2) = -16(2) + 64(2) + 200 = -16(4) + 64(2) + 200 = -64 + 128 + 200 = 264
Answer: The ball will reach a maximum height of 264 feet after 2 seconds.
b.
How many seconds does it take until the ball finally hits the ground? Round
your answer to the nearest tenth of a second.
Remember what s(t) represents, the height of the ball above the ground. So
when the ball hits the ground, s(t) = 0.
2
0 = -16t + 64t + 200
2
0 = 2t - 8t - 25
A = 2, B = -8, C = -25
Divide by -8.
Use the Quadratic Formula to solve.
− ( −8 ) ± ( −8)2 − 4(2)( −25)
8 ± 64 + 200 8 ± 264 8 ± 16.2
=
≈
2(2)
4
4
4
8 + 16.2 24.2
8 − 16.2 −8.2
t=
=
= 6.05 ≈ 6.1 OR t =
=
OOPS!
4
4
4
4
t=
=
Answer: The ball will hit the ground after about 6.1 seconds.
© Copyright 2008 by John Fetcho. All rights reserved.
Math 27 - Section 11.3 - Part III - Page 2
c.
Find s(0) and describe what this means.
2
s(0) = -16(0) + 64(0) + 200 = 0 + 0 + 200 = 200
Answer: The initial height was 200 feet.
d.
Use your results from parts (a) through (c) to graph the quadratic function.
Begin the graph with t = 0 and end with the value of t for which the ball hits
the ground.
(2, 264)
200
6.1
2.
Now you try one:
The graph (top left-hand column, page 781) shows U.S. adult wine
consumption, in gallons per person, for selected years from 1980 through
2005.
The function
f(x) = 0.004x2 − 0.094x + 2.6
models U.S. wine consumption, f(x), in gallons per person, x years after 1980.
(Page 781, #57)
a.
According to this function, what was U.S. adult wine consumption in 2005?
Does this overestimate or underestimate the value shown on the graph? By
how much?
Answer: In 2005, U.S. adult wine consumption was 2.75 gallons per person.
This was an underestimate of 0.05 gallons per person.
b.
According to this function, in which year was wine consumption at a
minimum? Round to the nearest year. What does the function give for per
capita wine consumption for that year?
Answer: The minimum per capita wine consumption of 2.048 gallons per
person occurred in 1992.
© Copyright 2008 by John Fetcho. All rights reserved.
Math 27 - Section 11.3 - Part III - Page 3
4.
You have 200 feet of fencing to enclose a rectangular plot that borders on a
river. If you do not fence the side along the river, find the length and width of
the plot that will maximize the area. What is the largest area that can be
enclosed? (page 782, #66)
Let x = The length
Let y = The width
Intermediate Algebra River
y
y
x
The area is what we want to maximize, so that is our objective equation. We
are constrained by the amount of fencing available, so the constraint equation
will come from our equation for the fencing.
Objective Equation
Area = xy
Constraint Equation
Fencing = 200 = 2y + x
So we want to solve the constraint for either x or y, solving for x looks easier
(why?):
200 − 2y = x
Substitute into the objective equation.
Area = (200 − 2y)y
Multiply and put into general form.
2
Area = 200y − 2y
B
2
Area = −2y + 200y
To find the maximum, we need to evaluate −
.
2A
B
−(200) −200
y= −
=
=
= 50
x = 200 − 2(50) = 200 − 100 = 100
2A
2( −2)
−4
Area = (100)(50) = 5,000
Answer: The maximum area of 5,000 square feet will occur when the width is
50 feet and the length is 100 feet.
© Copyright 2008 by John Fetcho. All rights reserved.