1. Warm-up: Sketch y = 3|x – 1| – 2
HW: page 214 (1-12, , , 47-50)

2. Graphing Quadratic Functions
Objective:
Graph a quadratic function
Write a quadratic function in standard form
Find the equation of a parabola

3. Definition of a Polynomial Function
Let n be a nonnegative integer and let
be real numbers with
The function given by
is called a polynomial function of x with degree n.
f(x) = degree of constant function
f(x) = 9x degree of 1 linear function
f(x) = 9x2 + 5x degree of 2 quadratic function
f(x) = 9x3 + 5x2 + x + 1 degree of 3 cubic function

4. The graph of a quadratic function is a parabola.
Every parabola is symmetrical about a line called the
axis of symmetry. x y
f (x) = ax2 + bx + c
vertex
Axis of symmetry
Vertex the intersection point of the parabola
and the axis of symmetry

5. The leading coefficient of ax2 + bx + c is a.y
a > 0 opens upward
When the leading coefficient is positive (+)
f(x) = ax2 + bx + c
vertex minimum x y
When the leading
coefficient is negative (-)
vertex maximum
f(x) = ax2 + bx + c
a < 0 opens downward

6. The simplest quadratic functions are of the form f (x) = ax2 (a  0)
a < 1 wider a > 1 narrower
These are most easily graphed by comparing them with the parent function y = x2.
Example: Compare the graphs of
, and 5 y x -5
g(x) = 2x2

7. Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis.
f (x) = (x – 3)  shift upwards two units.
g (x) = (x – 3)  shift right three units.
- 4 x y 4
f (x) = (x – 3)2 + 2
y = x 2
vertex
(3, 2)

8. a > 0  parabola opens upward like y = 2x2. Shift left 1 and down 3
The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a  0)
The graph is a parabola opening upward if a  0 and opening downward if a  0. The axis of symmetry is x = h, and the vertex is (h, k).
Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find the axis of symmetry and vertex. x y
f (x) = 2x2 + 4x – 1
x = –1
f (x) = 2x2 + 4x – original equation
f (x) = 2( x2 + 2x) – factor out 2
f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square
f (x) = 2( x + 1)2 – standard form
a > 0  parabola opens upward like y = 2x2. Shift left 1 and down 3
(–1, –3)
axis x = –1, vertex (–1, –3).

9. a < 0  parabola opens downward. axis x = 3, vertex (3, 16).
Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7. x y 4
f (x) = – x2 + 6x original equation
(3, 16)
x = 3
f (x) = – ( x2 – 6x) factor out –1
f (x) = – ( x2 – 6x + 9) complete the square
f (x) = – ( x – 3) standard form
a < 0  parabola opens downward.
axis x = 3, vertex (3, 16).
Find the x-intercepts by solving –x2 + 6x + 7 = 0
(7, 0)
(–1, 0)
x2 – 6x – 7 = 0 divide by -1
(x – 7)( x + 1) = factor
x = 7, x = –1 x-intercepts (7, 0), (–1, 0)
f(x) = –x2 + 6x + 7

10. Finding the Equation of a Parabola
Find the equation for a parabola whose vertex is (1, 2) and that passes through the point (3, -2).
Because the parabola has a vertex at (h, k)  (1, 2), the standard form equation is
Now because the parabola passes through the point (3, -2), it follows that f(3) = -2

11. Now we can finish the standard form equation using a = -1 and
Changing to quadratic form
Quadratic form

12. Vertex of a Parabola
The vertex of the graph of f (x) = ax2 + bx + c (a  0)
Example: Find the vertex of the graph of f (x) = x2 – 10x + 22.
f (x) = x2 – 10x original equation
a = 1, b = –10, c = 22
At the vertex,
So, the vertex is (5, -3).

13. Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is:
The path is a parabola opening downward. The maximum height occurs at the vertex.
At the vertex,
So, the vertex is (9, 15).
The maximum height of the ball is 15 feet.

14. Graphing Quadratic Functions
Summary:
Graph a quadratic function
Write a quadratic function in standard form
Find the equation of a parabola

15. Sneedlegrit:
Find an equation of the parabola in standard form given the vertex at (5, 4) and that it goes through the point (6, 7).
HW: page 214 (1-12, , , 47-50)