Quadratic Functions:
Problem Solving
(Section 11.8)
We will look at some situations where quadratic functions are used for problem solving.
Maximum and Minimum Problems
For any quadratic function, the value of the function at the vertex is either a maximum or a minimum.
Problems that ask us to find the maximum or minimum of a quantity can be solved by finding the
coordinates of the vertex.
Vertex is at the minimum
Vertex is at the maximum
Example 1: Newborn calves. The number of pound of milk per day recommended for a calf that is x
weeks old can be approximated by the function p(x) = -0.2x2 + 1.3x + 6.2. When is the calf’s milk
consumption greatest and how much milk doe sit consume at that time?
Solution:
Step 1: Summarize the information
We are given a quadratic function that describes the milk consumption of a calf.
The x in the function represents the calf’s age in weeks.
The y of the function represents how many gallons of milk the calf consumes.
We are asked to find the x and the y values for the maximum consumption. This means that we
need to find the vertex of the parabola.
Step 2: Plan
Find the vertex of the parabola
Step 3: Carry out the plan
The function we are given is: p(x) = -0.2x2 + 1.3x + 6.2
The x-value of the vertex is: 
b
1.3

 3.25
2a
2(0.2)
Find the y-value of the vertex by plugging x = 3.25 into the quadratic expression:
-0.2(3.25)2 + 1.3(3.25) + 6.2 = 8.3125
Step 4: State the solution
The maximum milk consumption occurs at 3.25 weeks.
The maximum amount of milk consumed is 8.3125 pounds.
Step 5: Check
We can check if this is the maximum by picking values of x smaller and larger than 3.25 and
evaluating the function at these points.
x
3
3.25
3.5
p(x)
8.3
8.3125
8.3
Since the value at x = 3.25 is the biggest, that must be the maximum.
If we make the graph, we can also see that (3.25, 8.3125)
is the vertex, so 8.3125 is the maximum value.
Activity 1: Swimming area: a lifeguard has 100 m of roped-together flotation devices with which to
cordon off a rectangular swimming area at Lakeside Beach. If the shoreline forms on side of the
rectangle, what dimensions will maximize the size of the area for swimming?
L
Solution:
Step 1: Summarize the information
W
water
L
Step 2: Plan
Step 3: Carry out the plan
Equations in terms of Length and Width:
The function for area in terms of W:
The x-value of the vertex is:
Find the y-value of the vertex:
Step 4: State the solution
The width that maximizes area is:
The length that maximizes the area is:
The maximum area is:
Step 5: Check
beach
W
Fitting Quadratic Functions to Data
We can find the quadratic function that goes with a set of data points if we know at least three data
points.
Example 2: The decline of teen smoking: Deadly and less fashionable than ever, teen smoking has
declined since 1997. According to the center for Disease Control and Prevention, the percentage of highschool students who reported having smoked a cigarette in the preceding 30 days increased from 30.5%
in 1993 to 36.4% in 1997, but declined since then to 21.9% in 2003.
Years after
1993
0
4
10
Percentage of students who
smoked in last 30 days
30.5
36.4
21.9
The rise and fall in percentage of teen smokers suggests that the situation can be modeled by a
quadratic function.
a) Use the data points (0, 30.5), (4, 36.4), and (10,21.9) to find a quadratic function that fits the
data.
b) Use this function to estimate the percentage of high-school students in 2005 who smoked a
cigarette in the preceding 30 days.
Solution:
Step 1: Start with the general form of a quadratic function: f ( x)  ax 2  bx  c .
Here x represent the years after 1993 and f(x) represents the percentage of smoking teens.
Step 2: Plan
We need to find the values of a, b and c.
We do this by plugging in the given points.
Step 3: Carry out the plan
Point (0, 30.5) means x = 0, and f(0) = 30.5:
30.5 = a(0)2 + b(0) + c, so c = 30.5
So far we know: f ( x)  ax 2  bx  30.5
Point (4, 36.4) means x = 4, and f(4) = 36.4:
36.4 = a (4)2 + b(4) + 30.5
36.4 = 16a + 4b + 30.5 or
5.9 = 16a + 4b
Point (10, 21.9) means x = 10, and f(10) = 21.9: 21.9 = a (10)2 + b(10) + 30.5
21.9 = 100a + 10b + 30.5 or
-8.6 = 100a + 10b
To find a and b, we need to solve the equations:
5.9 = 16a + 4b
8.6 = 100a + 10b
Multiply all the terms of the first equation by -2.5 and add the equations:
-14.75 = -40a – 10b
-8.6
= 100a + 10b
-23.35 = 60a
so
a = - 0.39
To find b, plug the value of a in the first equation:
5.9 = 16(-0.39) + 4b
5.9 = -6.23 + 4b
12.13 = 4b
so
b = 3.03
Step 4: State the solution
Since we have our values for a, b, and c, we can write the quadratic function as:
f ( x)  0.39 x 2  3.03x  30.5
Step 5: Check
To check that this is correct, plug in the given values of x: 0, 4, and 10 and check that we get the given
values of y.
x = 0;
f (0)  0.39(0) 2  3.03(0)  30.5  30.5
x = 4;
f (4)  0.39(4) 2  3.03(4)  30.5  36.4
x = 10; f (10)  0.39(10) 2  3.03(10)  30.5  21.9
Part b: estimate the percentage of high-school students in 2005 who smoked a cigarette
2005 is 12 years after 2003, so x = 12. We plug in x = 12 into our function to find the percentage.
f (12)  0.39(12) 2  3.03(12)  30.5  10.84%
Activity 2: Find a quadratic function that fits the set of data: (0, 4), (-1, 6), (-2, 16)
Solution:
Step 1: The general form of a quadratic function is:
Step 2: Plan
Step 3: Carry out the plan
Plug in x = 0
Find c:
Rewrite the function:
Plug in x = -1
Plug in x = -2
Solve for a, b:
Re-write the function:
Step 4: Check