College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
4
Polynomial and
Rational Functions
4.1
Quadratic Functions
and Models
Introduction
A polynomial function is a function that is
defined by a polynomial expression. So a
polynomial function of degree n is a
function of the form:
P(x) = anxn + an–1xn–1 + … +a1x + a0
• We have already studied polynomial functions of
degree 0 and 1. These are functions of the form
P(x) = a0 and P(x) = a1x + a0, respectively, whose
graphs are lines.
Quadratic Functions
In this section, we study polynomial
functions of degree 2. These are called
quadratic functions.
• We see in this section how quadratic
functions model many real-world
phenomena.
Quadratic Functions
A quadratic function is a polynomial
function of degree 2. So a quadratic
function is a function of the form:
f(x) = ax2 + bx + c
(a ≠ 0)
Graphing Quadratic Functions
Using the Standard Form
Quadratic Function
If we take a = 1 and b = c = 0, we get the
simple quadratic function f(x) = x2, whose
graph is the parabola that we drew in
Example 1 of Section 3.2.
• In fact, the graph of any quadratic function
is a parabola.
• It can be obtained from the graph of f(x) = x2
by the transformations given in Section 3.5.
Standard Form of a Quadratic Function
A quadratic function f(x) = ax2 + bx + c
can be expressed in the standard form
f(x) = a(x – h)2 + k
by completing the square.
Standard Form of a Quadratic Function
The graph of f is a parabola with
vertex (h, k).
• The parabola opens upward if a > 0 or
downward if a < 0.
E.g. 1—Standard Form of Quadratic Function
Let f(x) = 2x2 – 12x + 23.
(a) Express f in standard form.
(b) Sketch the graph of f.
E.g. 1—Standard Form
Example (a)
Since the coefficient of x2 is not 1, we must
factor this coefficient from the terms involving
x before we complete the square.
f  x   2 x 2  12 x  23

 2 x

 2 x  6 x  23
2
2

 6 x  9  23  2  9
 2  x  3  5
2
E.g. 1—Standard Form
Example (b)
The standard form tells us that we get
the graph of f by:
1.
2.
3.
4.
Taking the parabola y = x2.
Shifting it to the right 3 units.
Stretching it by a factor of 2.
Moving it upward 5 units.
The vertex of the parabola is at (3, 5)
and the parabola opens upward.
E.g. 1—Standard Form
Example (b)
We sketch the graph after noting that
the y-intercept is f(0) = 23.
Maximum and Minimum Values
of Quadratic Functions
Maximum & Minimum Values of Quadratic Functions
If a quadratic function has vertex (h, k),
then the function has:
• A minimum value at the vertex if it opens upward.
• A maximum value at the vertex if it opens
downward.
Maximum & Minimum Values of Quadratic Functions
For example, the function graphed here
has minimum value 5 when x = 3—since
the vertex (3, 5) is the lowest point on
the graph.
Maximum or Minimum Value of a Quadratic Function
Let f be a quadratic function with standard
form
f(x) = a(x – h)2 + k
The maximum or minimum value of f
occurs at x = h.
Maximum or Minimum Value of a Quadratic Function
If a > 0, the minimum value of f is f(h) = k.
If a < 0, the maximum value of f is f(h) = k.
E.g. 2—Minimum Value of a Quadratic Function
Consider the quadratic function
f(x) = 5x2 – 30x + 49
(a) Express f in standard form.
(b) Sketch the graph of f.
(c) Find the minimum value of f.
E.g. 2—Minimum Value
Example (a)
To express this quadratic function in standard
form, we complete the square.
f ( x )  5 x 2  30 x  49

 5x

 5 x 2  6 x  49
2

 6 x  9  49  5  9
 5  x  3  4
2
E.g. 2—Minimum Value
Example (b)
The graph is a parabola that has
its vertex at (3, 4) and
opens upward.
E.g. 2—Minimum Value
Example (c)
Since the coefficient of x2 is positive,
f has a minimum value.
• The minimum value is f(3) = 4.
E.g. 3—Maximum Value of a Quadratic Function
Consider the quadratic function
f(x) = –x2 + x + 2
(a) Express f in standard form.
(b) Sketch the graph of f.
(c) Find the maximum value of f.
E.g. 3—Maximum Value
Example (a)
To express this quadratic function in standard
form, we complete the square.
y  x 2  x  2

 x

  x2  x  2
2
x
 x 

1 2
2

1
4

9
4
 2   1 41
E.g. 3—Maximum Value
Example (b)
From the standard form, we see that
the graph is a parabola that opens downward
and has vertex (1/2, 9/4).
As an aid to sketching the graph, we find
the intercepts.
• The y-intercept is f(0) = 2.
E.g. 3—Maximum Value
Example (b)
To find the x-intercepts, we set f(x) = 0 and
factor the resulting equation.
x x20
2


 x x 2 0
2
  x  2 x  1  0
• Thus, the x-intercepts are x = 2 and x = –1.
E.g. 3—Maximum Value
Example (b)
The graph of f is sketched here.
E.g. 3—Maximum Value
Example (c)
Since the coefficient of x2 is negative,
f has a maximum value.
• This is f(1/2) = 9/4.
Finding Maximum or Minimum Value
Expressing a quadratic function in standard
form helps us sketch its graph as well as find
its maximum or minimum value.
If we are interested only in finding
the maximum or minimum value, then
a formula is available for doing so.
Finding Maximum or Minimum Value
The formula is obtained by completing
the square for the general quadratic function:
f  x   ax 2  bx  c
 2 b 
 a x  x   c
a 

 2 b
 b2 
b2 
 a x  x  2   c  a 2 
a
4a 

 4a 
2
b 
b

 a x 
c 

2a 
4a

2
Finding Maximum or Minimum Value
This equation is in standard form with
h = –b/(2a) and k = c – b2/(4a)
• Since the maximum or minimum value occurs
at x = h, we have the following result.
Maximum or Minimum Value of a Quadratic Function
The maximum or minimum value of
a quadratic function f(x) = ax2 + bx + c
occurs at:
x = –b/2a
• If a > 0, then the minimum value is f(–b/2a).
• If a < 0, then the maximum value is f(–b/2a).
E.g. 4—Finding Maximum & Minimum Values
Find the maximum or minimum value
of each quadratic function.
(a) f(x) = x2 + 4x
(b) g(x) = –2x2 + 4x – 5
E.g. 4—Finding Max. & Min. Values Example (a)
f(x) = x2 + 4x is a quadratic function with
a = 1 and b = 4.
Thus, the maximum or minimum value
occurs at:
b
4
x

 2
2a
2 1
E.g. 4—Finding Max. & Min. Values Example (a)
Since a > 0, the function has the minimum
value
f(–2) = (–2)2 + 4(–2)
= –4
E.g. 4—Finding Max. & Min. Values Example (b)
g(x) = –2x2 + 4x – 5 is a quadratic function
with a = –2 and b = 4.
Thus, the maximum or minimum value
occurs at:
b
4
x

1
2a
2   2 
E.g. 4—Finding Max. & Min. Values Example (b)
Since a < 0, the function has the maximum
value
f(1) = –2(1)2 + 4(1) – 5
= –3
Modeling with Quadratic Functions
We study some examples of real-world
phenomena that are modeled by quadratic
functions.
• In the next example, we find the maximum
value of a quadratic function that models
the gas mileage for a car.
E.g. 5—Maximum Gas Mileage for a Car
Most cars get their best gas mileage when
traveling at a relatively modest speed.
The gas mileage M for a certain new car is
modeled by the function
1 2
M s   
s  3s  31
28
where:
• s is the speed in mi/h.
• M is measured in mi/gal.
(15  s  70)
E.g. 5—Maximum Gas Mileage for a Car
What is the car’s best gas mileage?
At what speed is it attained?
E.g. 5—Maximum Gas Mileage for a Car
The function M is a quadratic function
with a = –1/28 and b = 3.
• Thus, its maximum value occurs when
b
3
s


42
2a
2   281 
• The maximum is:
M  42   
1
28
 32
 42 
2
 3  42   31
E.g. 5—Maximum Gas Mileage for a Car
So, the car’s best gas mileage is
32 mi/gal, when it is traveling at 42 mi/h.
E.g. 6—Maximizing Revenue from Ticket Sales
A hockey team plays in an arena with
a seating capacity of 15,000 spectators.
• With the ticket price set at $14, average
attendance at recent games has been 9,500.
• A market survey indicates that, for each dollar
the ticket price is lowered, the average attendance
increases by 1,000.
E.g. 6—Maximizing Revenue from Ticket Sales
(a) Find a function that models the revenue
in terms of ticket price.
(b) Find the price that maximizes revenue
from ticket sales.
(c) What ticket price is so high that no one
attends and so no revenue is generated?
E.g. 6—Max. Rev. from Ticket Sales Example (a)
The model we want is:
• A function that gives the revenue
for any ticket price.
E.g. 6—Max. Rev. from Ticket Sales Example (a)
1. Express the model in words.
• We know that:
Revenue = Ticket price x Attendance
E.g. 6—Max. Rev. from Ticket Sales Example (a)
2. Choose the variable.
• There are two varying quantities—ticket price
and attendance.
• Since the function we want depends on price,
we let:
x = ticket price
E.g. 6—Max. Rev. from Ticket Sales Example (a)
• Next, we express attendance in terms of x.
In Words
In Algebra
Ticket price
x
Amount ticket price
is lowered
14 – x
Increase in attendance
1000(14 – x)
Attendance
9500 + 1000(14 – x)
= 23,500 – 10,000x
E.g. 6—Max. Rev. from Ticket Sales Example (a)
3. Set up the model.
• The model is the function R that gives
the revenue for a given ticket price x.
Revenue = Ticket price x Attendance
R(x) = x(23,500 – 1000x)
R(x) = 23,500x – 1000x2
E.g. 6—Max. Rev. from Ticket Sales
4. Use the model.
• We use the model to answer
the questions in parts (b) and (c).
E.g. 6—Max. Rev. from Ticket Sales Example (b)
Since R(x) = 23,500x – 1000x2 is a quadratic
function with a = –1000 and b = 23,500,
the maximum occurs at:
b
23,500
x

2a
2( 1000)
 11.75
E.g. 6—Max. Rev. from Ticket Sales Example (b)
So a ticket price of $11.75 gives the
maximum revenue.
• At this price, the revenue is:
R(11.75) = 23,500(11.75) – 1000(11.75)2
= $138,062.50
E.g. 6—Max. Rev. from Ticket Sales Example (c)
We want to find the ticket price x
for which:
R(x) = 23,500x – 1000x2
=0
• We can solve this quadratic equation
algebraically or graphically.
E.g. 6—Max. Rev. from Ticket Sales Example (c)
From this graph, we see that:
R(x) = 0 when x = 0 or x = 23.5
• So, according to our
model, the revenue
would drop to zero
if the ticket price is
$23.50 or higher.
• Of course, revenue is
also zero if the
ticket price is zero!