Sections 4.3 and 4.4
Quadratic Functions
and Their Properties
Quadratic Function
A quadratic function of the variable x is a function
that can be written in the form
f ( x)  ax  bx  c
2
 a  0
where a, b, and c are fixed real numbers. Also, the
domain of f is all real numbers, namely, (, ).
Example:
f ( x)  12x  3x 1
2
Quadratic Function
The graph of a quadratic function is a parabola.
f ( x)  ax  bx  c
2
a>0
 a  0
a<0
Vertex, Intercepts, Symmetry
Vertex coordinates are:
b
x ,
2a
 b 
y  f  
 2a 
y – intercept is:
x0
yc
x – intercepts are solutions of
symmetry
ax  bx  c  0
b
x
2a
2
Graph of a Quadratic Function
Example 1: Sketch the graph of
f ( x)  x  2x  8
Vertex:
b
2
x
   1 y  f (1)  9
2a
2
y – intercept
x0
y  8
x – intercepts
x  2x  8  0
2
x  4, 2
2
Graph of a Quadratic Function
Example 1: Sketch the graph of
f ( x)  x  2x  8
Absolute Minimum of f is:
b
y  f (1)  9 at x  
 1
2a
Absolute Maximum of f is:
Does not exist
Range of f is: [ 9, )
2
Graph of a Quadratic Function
Example 2: Sketch the graph of f ( x)  4x 12x  9
2
Vertex:
b
12 3
x


y  f (3 / 2)  0
2a
2 4 2
y – intercept
x0
y 9
x – intercepts
4 x 12 x  9  0
2
x  3/ 2
Graph of a Quadratic Function
Example 2: Sketch the graph of f ( x)  4x 12x  9
2
Absolute Minimum of f is:
b
y  f (3 / 2)  0 at x  
 3/ 2
2a
Absolute Maximum of f is:
Does not exist
Range of f is: [0, )
Graph of a Quadratic Function
1 2
Example 3: Sketch the graph of g ( x)   x  4 x  12
2
Vertex:
b
x
 4 y  f (4)  4
2a
y – intercept
x0
y  12
x – intercepts
1 2
 x  4 x  12  0
2
no solutions
Graph of a Quadratic Function
1 2
Example 3: Sketch the graph of g ( x)   x  4 x  12
2
Absolute Minimum of f is:
Does not exist
Absolute Maximum of f is:
b
y  f (4)  4 at x  
4
2a
Range of f is: (, 4]
Applications
Example: For the demand equation below, express the
total revenue R as a function of the price p per item
and determine the price that maximizes total revenue.
q( p)  3 p  600
R( p)  pq  p  3 p  600
 3 p  600 p
2
Maximum is at the vertex, p = $100
Applications
Example: As the operator of Workout Fever Health Club,
you calculate your demand equation to be q 0.06p + 84
where q is the number of members in the club and p is the
annual membership fee you charge.
1. Your annual operating costs are a fixed cost of $20,000 per
year plus a variable cost of $20 per member. Find the annual
revenue and profit as functions of the membership price p.
2. At what price should you set the membership fee to obtain the
maximum revenue? What is the maximum possible revenue?
3. At what price should you set the membership fee to obtain the
maximum profit? What is the maximum possible profit? What is
the corresponding revenue?
Solution
The annual revenue is given by
R( p)  pq  p  0.06 p  84
2
 0.06 p  84 p
The annual cost as function of q is given by
C (q)  20000  20q
The annual cost as function of p is given by
C( p)  20000  20q  20000  20  0.06 p  84 
 1.2 p  21680
Solution
Thus the annual profit function is given by
P( p )  R  C  ( 0.06 p  84 p )   1.2 p  21680 
2
 0.06 p 2  85.2 p  21680
The graph of the revenue function R  0.06 p2  84 p is
b
84
Maximum is at the vertex p  

 $700
2a
2( 0.06)
The graph of the revenue function R  0.06 p2  84 p is
Maximum revenue is R(700)  $29, 400
The profit function is P( p)  0.06 p2  85.2 p  21680
b
85.2
Maximum is at the vertex p  

 $710
2a
2(0.06)
The profit function is P( p)  0.06 p2  85.2 p  21680
Maximum profit is
P(710)  $8,566
Corresponding Revenue is R(710)  $29,394
Vertex Form of a Parabola
Vertex Form of a Parabola
To get the vertex form of the parabola we complete the
square in x as indicated in the next steps:
Vertex Form of a Parabola
Standard form
Vertex form
Vertex Form of a Parabola
Example: Find the vertex form of f ( x)  2 x  8x  5
2
f ( x)  2  x 2  4 x  __   5  2  __
f ( x)  2  x  4 x  4   5  2  4 
2
f ( x)  2  x  2   3
2

Vertex Form of a Parabola
Use the vertex form of f ( x)  2 x2  8x  5 to graph
the parabola
f ( x)  2  x  2   3
2
Vertex Form of a Parabola
Use the vertex form of f ( x)  2 x2  8x  5 to graph
the parabola
f ( x)  2  x  2   3
2
Find a Quadratic Function
Given Its Vertex and One
Other Point
Vertex Form of a Parabola
Determine the quadratic function whose vertex is (2, 3)
and whose y-intercept is 1.
f  x   a  x  h   k  a  x  2  3
2
2
y
Using the fact that the y-intercept

is 1: 1  a  0  2   3

1
Thus 1  4a  3 and a  
2

2
1
2
f  x     x  2  3
2

x













More Examples
Maximizing Revenue
The marketing department at Widgets Inc. found that, when certain
widgets are sold at a price of p dollars per unit, the number x of
widgets sold is given by the demand equation
x = 1500  30p
1. Find a model that expresses the revenue R as a function of the
price p.
2. What is the domain of R?
3. What unit price should be used to maximize revenue?
4. If this price is charged, what is the maximum revenue?
Maximizing Revenue
The marketing department at Widgets Inc. found that, when certain
widgets are sold at a price of p dollars per unit, the number x of
widgets sold is given by the demand equation
x = 1500  30p
1. Find a model that expresses the revenue R as a function of the
price p.
2. What is the domain of R?
1. Revenue R  xp  1500  30 p  p  30 p2 +1500p
2. x  0 so 1500  30 p  0 30 p  1500 p  50
The domain of R is  p 0  p  50  [0,50]
Maximizing Revenue
The marketing department at Widgets Inc. found that, when certain
widgets are sold at a price of p dollars per unit, the number x of
widgets sold is given by the demand equation
x = 1500  30p
3. What unit price should be used to maximize revenue?
4. If this price is charged, what is the maximum revenue?
b
1500
3. p  

 $25
2a
2(30)
4. R (25)  30  25   1500  25   $18, 750
2
Maximizing Revenue
The marketing department at Widgets Inc. found that, when certain
widgets are sold at a price of p dollars per unit, the number x of
widgets sold is given by the demand equation
x = 1500  30p
5. How many units are sold at this price?
6. Graph R.
7. What price should Widgets Inc. charge to collect at least $12,000
in revenue?
Maximizing Revenue
The marketing department at Widgets Inc. found that, when certain
widgets are sold at a price of p dollars per unit, the number x of
widgets sold is given by the demand equation
x = 1500  30p
5. x  1500  30  25 = 750
7. 12000  30 p2  1500 p
30 p2 1500 p  12000  0
30  p 2  50 p  40   0
30( p  10)( p  40)  0
p  10 or p  40
So the company should charge between $10 and $40 to earn at least
$12,000 in revenue.
Maximizing Area
A farmer has 1600 yards of fence to enclose a rectangular field. What
are the dimensions of the rectangle that encloses the most area?
A  xw
2 x  2w  1600
w  800  x
A  x(800  x)   x  800x
2
b
800
x

 400
2a
2(1)
w  800  400  400
The farmer should make the rectangle 400 yards by 400 yards to
enclose the most area.
Projectile Motion
1. Find the maximum height of the projectile.
b
1
1
5000
x



 2500
2a
2
 32 
 1 
2
2


2 
 400 
 5000 
1
2
h  2500  
 2500   2500  500  1750 ft
5000
Projectile Motion
2. How far from the base of the cliff will the projectile strike the water?
h  x 
1 2
x  x  500  0
5000
 1 
1  1  4 
  500 
 5000 
x
1 

2 

5000


2
x  458 or 5458
Solution cannot be negative so the projectile will hit the water about
5458 feet from the base of the cliff.
The Golden Gate Bridge
The Golden Gate Bridge
The Golden Gate Bridge