3.3 Quadratic Functions
Objectives:
1.
Define three forms for quadratic functions.
2.
Find the vertex and intercepts of a quadratic function and
sketch its graph.
3.
Convert one form of a quadratic function to another.
Parabolas
vertex
y
9
The graph of a quadratic function is
called a parabola and it always:
8
7
6
5
1. Opens upward or downward
2. Has a vertex which is a maximum
or minimum
3. Has exactly 1 y-intercept
4. Has either 0, 1, or 2 x-intercepts
5. Is symmetric through its vertex
x-intercepts
4
3
2
1
–9
–8
–7
–6
–5
y-intercept
–4
–3
–2
–1
–1
1
2
3
4
–2
–3
–4
–5
–6
–7
–8
–9
Line of
Symmetry
5
6
7
8
9
x
The Three Forms of a Quadratic Function

Transformation Form (Vertex Form):
f(x) = a(x – h)2 + k

Polynomial Form (Standard Form):
f(x) = ax2 + bx + c

x-Intercept Form (Factored Form):
f(x) = a(x – s)(x – t)
Graphing from the Three Quadratic Forms
Name
Form
Vertex
Transformation
f(x)  a(x  h) 2  k f(x)  ax 2  bx  c
(h,k)
x-Intercepts
h
y-Intercept
Polynomial
k
a
ah 2  k
 b  b 
  , f   
 2a  2a  
b 
b 2  4ac
2a
c
x-Intercept
f(x)  a(x  s)(x  t)
 s  t  s  t 
,f


 2 
 2
s and t
ast
Example #1
Transformation Form f(x) = a(x – h)2 + k

Find the vertex and the x- and y-intercepts for the following
function. Then graph the function.
f ( x)  4x  2  5
2
Vertex:
x-Intercept:
y-Intercept:
(h, k)
Let f(x) = 0
Let x = 0
y
Vertex:
(2, 5)
10
8
y-intercept:
x-intercept:
2
0  4 x  22  5 f (0)  40  2   5
 44   5
 5  4 x  2 2
5
  x  2 2
4
5

x2
2
5
x2
2
 11
The graph opens
down because a
is negative.
6
4
2
–10 –8
–6
–4
–2
–2
–4
–6
–8
–10
2
4
6
8
10
x
Example #2
Polynomial Form

f(x) = ax2 + bx + c
Find the vertex and the x- and y-intercepts for the following
function. Then graph the function.
f(x)  x  4x  5
2

The vertex can be found from polynomial form using:
b
h
2a
h
 b 
k  f  
 2a 
4
 2
21
k  f  2   2  4 2  5  1
2
Vertex :
 2,1
The graph opens up
because a is positive.
Example #2
Polynomial Form

f(x) = ax2 + bx + c
Find the vertex and the x- and y-intercepts for the following
function. Then graph the function.
f(x)  x  4x  5
2

x-Intercept:
Let f(x) = 0
y
10
0  x2  4x  5
x
4

y-Intercept:
8
6
42  415
21
4 4

 None
2
c
4
2
–10 –8
–6
–4
–2
–2
–4
–6
–8
–10
5
(2,1)
Vertex :
2
4
6
8
10
x
Example #3
x-Intercept Form

f(x) = a(x – s)(x – t)
Find the vertex and the x- and y-intercepts for the following
function. Then graph the function.
f(x)  2(x  3)(x  1)
Vertex:
st
h
2
s  3, t  1
3  1
h
1
2
k  f 1  21  31  1  8
Vertex : (1,8)
st
k  f

 2 
Graph opens down
because a is negative.
Example #3
x-Intercept Form

f(x) = a(x – s)(x – t)
Find the vertex and the x- and y-intercepts for the following
function. Then graph the function.
f(x)  2(x  3)(x  1)
Vertex :
(1,8)
y
10
x-Intercept:
s&t
8
6
x  3, x  1
4
2
y-Intercept:
ast
  2 3 1
6
–10 –8
–6
–4
–2
–2
–4
–6
–8
–10
2
4
6
8
10
x
Example #4
Changing to Polynomial and x-Intercept Form

Write the following functions in polynomial and x-intercept form, if possible.
A.
Polynomial:
f(x)  0.3(x  2) 2  5
f ( x)  0.3 x  2 x  2  5
 0.3( x 2  4 x  4)  5
 0.3x 2  1.2 x  1.2  5
x-intercept:
 0.3x 2  1.2 x  6.2
b 2  4ac
  1.2 2  40.36.2 
 1.44  7.44
 6
Since the discriminant is
negative there are no xintercepts so the function
cannon be written into xintercept form.
Example #4
Changing to Polynomial and x-Intercept Form

Write the following functions in polynomial and x-intercept form, if possible.
B.
g(x)  2x  5.8x  27
Polynomial:
x-intercept:
2
Already in polynomial form.

g ( x)  2 x 2  2.9 x  13.5
x
 2 .9 

2.92  41 13.5
21
 2 .9  7 . 9

 2.5 and  5.4
2
g ( x)  2 x  2.5 x  5.4 
Example #4
Changing to Polynomial and x-Intercept Form

Write the following functions in polynomial and x-intercept form, if possible.
C.
h(x)  3(x  6)(x  1)
Polynomial:
h( x)  3x
2
 5x  6
 3 x  15 x  18
2
x-intercept:
Already in x-intercept form.

Example #5
Changing to Transformation Form

Write the following functions in transformation form:
f(x)  5x 2  6x  1
A.
To write into transformation form we must use completing the square.
First factor out the a term.
Then complete the square
and add this value to the
inside of the parentheses.
2
b  6

      2
2  5

2
9
 3
   
25
 5
2
 2 6 
f ( x)  5 x  x   1
5 

Because of the -5 out front we are
subtracting 9/5 so we must add 9/5
to the outside to keep it balanced.
9 
9
 2 6
 5 x  x    1 
5
25 
5

2
3 4

 5 x   
5 5

 9 
 5 
 25 
9

5
Example #5
Changing to Transformation Form

B.
Write the following functions in transformation form:
g(x)  0.2(x  3)(x  2)
2
b  1
    
2  2
1
  0.25
4
0.20.25 
 0.05
2

g ( x)  0.2 x  x  6

 0.2x
2

 0.2 x 2  x  1.2
2


 x  0.25  1.2  0.05
 0.2 x  0.5  1.25
2
Example #6
Maximum Area for a Fixed Perimeter

Find the dimensions of a rectangular parking lot
that can be enclosed with 2400 feet of fence
and that has the largest possible area.
A( x)   x 2  1200 x
b
1200
x 
 600
2a
2 1
y  600  1200
 600
The largest area would occur if
the parking lot were a square
with 600 ft on each side and an
area of 360,000 square feet.
Example #6
Maximum Area for a Fixed Perimeter

Find the dimensions of a rectangular parking lot
that can be enclosed with 2400 feet of fence
and that has the largest possible area.
Let x = length & y = width
Perimeter = x + x + y + y = 2x + 2y
Area = xy
2 x  2 y  2400
2 y  2 x  2400
y   x  1200
A( x)  xy
A( x)  x( x  1200 )
  x 2  1200 x
This is a parabola opening downward so the maximum area
should occur at its vertex.
Example #7
Maximizing Profit

The owner of a concession stand sells 150
lunches per day at a price of $3 each. The cost
to the owner is $2.25 per lunch. Each $0.25
price increase decreases sales by 30 lunches per
day. What price should be charged to maximize
profit?
x = number of $0.25 price increases
$3 – $2.25 = $0.75 per lunch
The profit on each lunch is then 0.25x +0.75
The number of lunches sold per day is 150 – 30x
Total Profit:
P(x) = (0.25x + 0.75)(150 – 30x)
Example #7
Maximizing Profit

The owner of a concession stand sells 150
lunches per day at a price of $3 each. The cost
to the owner is $2.25 per lunch. Each $0.25
price increase decreases sales by 30 lunches per
day. What price should be charged to maximize
profit?
P( x)  0.25 x  0.75 150  30 x 
 37.5 x  7.5 x 2  112 .5  22.5 x
 7.5 x 2  15 x  112 .5
b
15
x 
1
2a
2 7.5
The maximum profit
occurs with just one
price increase with each
lunch priced at $3.25.
Example #8
Number Problems

Find two numbers whose difference is 7 and
whose product is a minimum.
x y7
xy  MIN
 y  x  7
y x7
MIN  x x  7   x 2  7 x
Even though is problem is looking
for a minimum, the function is still a
parabola so the vertex will be the
lowest point.
b
7
x 
 3.5
2a 21
y  3.5  7  3.5
The two numbers are 3.5 and −3.5.