Suppose you have a tennis ball and you toss it in the air. Then, you catch it with the other hand at the same height at which you started…What path does the ball take as it travels from one hand to the other?
What shape is formed if the above scenario is graphed?

Parabolas have a “U” shape when you graph them on paper
(2-dimensional).
However a “real-life” parabola is more like a bowl shape.
A satellite is an example of a parabola.

These graphs are called Quadratic Functions.
A quadratic function can be described by an equation of the following form:
2
There are 2 ways to write Quadratic Functions
Standard Form and Vertex Form
The above equation is in Standard Form. We will look at this form first.

f(x)= Ax 2 +Bx+C
Ax 2
Quadratic Term
Bx
Linear Term
C
Constant Term

For a more straightforward explanation on the affect of A, B, and C. Look at the following Gizmo.
Gizmo: How A, B, and C affect the quadratic function.
To explore the affect the variables A, B, and C have on the graph, play the
Zap It game found at the following link.
Game

2
-
2
What do you see? What is A in the above two equations and how does it affect the graph?

In this equation, A = -1 In this equation, A = 1
Gizmo: Standard
Form, Vertex, and
Intercepts
Gizmo: Standard
Form, Vertex, and
Intercepts (B)

f(x) = Ax 2 + Bx + C
If A = 0, the graph is not quadratic, it is linear.
If A is not 0, B and/or C can be 0 and the function is still a quadratic function.
If A < 0,
The parabola opens down, like an umbrella.
If A > 0,
The parabola opens up, like an bowl.
In this equation, A = -1 In this equation, A = 1

When "a" is positive, the graph of y = ax 2 + bx + c opens upward and the vertex is the lowest point on the curve.
As the value of the coefficient "a" gets larger, |a| > 1, the parabola narrows.
As the value of the coefficient
“a” gets smaller, | a | is between
0 and 1, the parabola widens.

For the graph of y = Ax 2 + Bx + C
If C is positive, the graph shifts up C units
If C is negative, the graph shifts down C units y

9 .
6 x
2 
5 Since C is negative, -5, the graph shifts down 5 units y
 
2 x
2  x

12 Since C is positive, 12, the graph shifts up 12 units.
Note: This graph would open upside down since A is -2 y
 x
2

2 x

8
3
Since C is negative, -8, the graph shifts down 8 units

•
The graph of a quadratic function is a U-shaped graph called a parabola.
•
The vertex is the highest or lowest point on the graph. The vertex is the lowest point on the parabola if the parabola opens upward and is the highest point on the parabola if the parabola opens downward.
•
The vertex represents the maximum or minimum value of a function.
•
In the graphs, the parabolas are symmetric about the axis of symmetry. In general, the axis of symmetry is a vertical line through the vertex that divides the parabola into two symmetrical parts.
•
Each point on the graph has a corresponding point on its mirror image. These points are equidistant from the axis of symmetry.

y = 2x 2 -8x + 8
Vertex (0,2)
Axis of Symmetry x= 2
This is also the minimum
(lowest point on this graph)

Standard Form
• y = Ax 2 + Bx + C
•
Axis of symmetry
• x = B/2A
•
Vertex: (x,y)
–
To find:
•
The number from the axis of symmetry is your x coordinate
•
Plug x into the equation to solve for you y coordinate
•
Y-intercept: C

y=Ax 2 + Bx + C
To find the axis of symmetry and vertex of a Quadratic Function follow these steps.
1. Name A, B, C
2. The axis of symmetry is x
 
B
2 A
3. To find the vertex , take the x value you received from the axis of symmetry and plug it in to the given quadratic equation and solve to find the value of y.
For example, y=2x 2 -3x+5
1. A= 2, B= -3, C= 5
2. Axis of Symmetry : x
 
B
2 A x
 
(

3 )
2 ( 2 )

3
4
The y-intercept is (0, 5). You can find the y-intercept by letting x equal 0 just like you did in linear equations.
3. Vertex ( ¾ , ??) y=2( ¾ ) 2 -3( ¾) + 5 y =
31
8


3
4
,
31



Name the axis of symmetry and vertex of y=2x 2 -4x+1
1. A = 2
B= -4
C= 1
2. Axis of Symmetry

4 x =
2 ( 2 )
 x =4/4  x = 1
The y-intercept is (0, 1).
3. Vertex
( 1 , ???) y = 2(1) 2 – 4(1) + 1 y = -1
Vertex: (1, -1) ; Minimum