```Sullivan PreCalculus
Section 3.1
Objectives
• Graph a Quadratic Function Using Transformations
• Identify the Vertex and Axis of Symmetry of a Quadratic
Function
• Graph a Quadratic Function Using the Vertex, Axis of
Symmetry, and Intercepts
• Use a Graphing Utility to Find the Quadratic Function of
Best Fit
A quadratic function is a function of the
form:
f ( x)  ax  bx  c
2
where a, b, and c are real numbers and
a  0. The domain of a quadratic function
consists of all real numbers.
The graph of a quadratic function is called
a parabola.
Graphs of a quadratic function f(x) = ax2 + bx + c, a  0
Axis of symmetry
Vertex is lowest point
a>0
Opens up
Vertex is highest point
Axis of symmetry
a<0
Opens down
2
f
(
x
)
2
x
 -  8x - 1
Graph the function
Find the vertex and axis of symmetry.
First, rewrite the function by completing the square
f ( x )  - 2 x  8 x - 1  - 2 x - 4 x  - 1
2
2
 -2 x - 4 x  _  - 1
2



2
 - 4  4
 
 2 
 - 2 x - 4 x  4  - 1  8
2
 - 2 x - 2   7
2
Now, graph the function using the
transformations discussed in Chapter 3.
15
15
(2,4)
(0,0)
10
(0,0) 0
10
10
0
10
(2, -8)
15
15
yx
2
y  -2 x 2
15
15
(2, 7)
(2, 0)
10
0
10
10
0
(4, -1)10
(4, -8)
15
y  - 2 x - 2 
15
2
y  -2 x - 2  7
2
Vertex : (2,7)
2
f(x)  ax  bx  c
 - b  - b 
Vertex  
, f
 
 2a  2a  
-b
Axis of Symmetry : The line x 
2a
Parabola opens up if a > 0. Parabola opens
down if a < 0.
Given the function f ( x)  2x2 12x  5, determine
whether the graph opens upward or downward.
Find the vertex, axis of symmetry, the xintercepts, and the y-intercept.
a = 2, b = 12, c = 5
a = 2 > 0, the parabola opens up
x-coord. of vertex:
- b - 12

 -3
2a
22 
-b 

y-coord. of vertex: f    f -3  2(-3)2 12(-3)  5  -13
 2a 
Vertex: (-3, -13)
b
x

 -3
Axis of symmetry:
2a
2
f ( x)  2x  12x  5
y-intercepts: f(0) = 5; so the y-intercept is (0,5)
2
x-intercepts: Solve the equation f ( x)  2x  12x  5 = 0
- b  b 2 - 4ac
x
2a
- 12  104

2(2)
26
 -3 
2
The x-intercepts are approximately (-5.6,0) and (-.45,0)
Summary: Parabola opens up
Vertex (-3, -13)
y-intercept: (0,5)
x-intercepts: (-0.45, 0) and (-5.55,0)
Now, graph the function using the information
found in the previous steps.
10
(0, 5)
(-5.55, 0)
(-0.45, 0)
5
0
10
Vertex: (-3, -13)
If a mathematical model of a real world situation leads to a
quadratic function, the properties of the function can be
applied to the model.
Example: The John Deere Company has found that the
revenue from sales of heavy duty tractors is a function of
the unit price p that it charges. If the revenue R is
-1 2
R( p) 
p  1900 p
2
what unit price p should be charged to maximize
revenue? What is the maximum revenue?
Note that the revenue function is a quadratic function that
opens downward, since a < 0. So, the maximum revenue
will be achieved at the vertex of the function.
x-coord. of vertex:
-b
- 1900

 1900
2a 2 - 1
2


y-coord. of vertex:
-1
 -b 
2
f    f 1900  1900  1900(1900)  1805000
2
 2a 
So, the maximum revenue occurs when they charge \$1900.
The maximum revenue is \$1,805,000.
Suppose you throw a ball straight up and record the
height of the ball in 0.5 second intervals and obtain
the following data:
time, t
(in seconds)
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
height, s
(in feet)
51
89
120
141
155
162
160
149
133
106
73
29
Use a graphing utility to complete the following:
a.) Draw a scatter diagram of the data.
b.) Find the quadratic function of best fit.
c.) Draw the quadratic function of best fit on the
scatter diagram.
```