```2.1 Quadratic Functions
Completing the square
Definition of a Polynomial Function
F(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + .. + a 1 x 1 + a0
an , a n-1,… are numbers
In “x n” n is the exponents going down in degree.
This would be a polynomial function of x with degree n.
Linear functions are first degree
f(x) = ax2 + bx + c
Would be a graph of a Parabola
A parabola is
Axis of symmetry
symmetric; “x = h”
Where h is from the
vertex (h, k)
If a > 0, then the parabola opens up
If a < 0, then the parabola opens downward
What equation do you think of when
This equation?
Do you remember how to use it?
Standard form of a Quadratic equation
f(x) = a(x – h) 2 + k
How would you graph the equation
f(x) = 2(x + 3) 2 + 1
The vertex has moved off the origin 3 units to
the left and 1 unit up. Since a = 2,the parabola
opens up and gets skinny.
The vertex is at (-3, 1)
f(x) = 2(x + 3) 2 + 1
Finding the vertex when it is not in
Standard form
f(x) = x 2 + 12x – 9
We need to complete the square to find the
standard form
Take half of b squared and add and subtract to the
function.
f(x) = x 2 + 12x + (6)2+ ( -36) – 9
Factor the first three terms, then add the last terms
f(x) = (x + 6) 2 – 45
vertex ( - 6, -45)
Find the standard form of
f(x) = x 2 + 10x + 8
Write the equation with
the vertex ( - 4, 8) and the point (1, 4)
f(x) = a(x – h) 2 + k
from the vertex h = - 4, k = 8
from the point x = 1, f(x) = 4
4 = a(1 – (- 4)) 2 + 8
4 = a(5)2 + 8
4 = 25a + 8
-4 = 25a
-4/25 = a
Rewrite with a, h and k
f(x) = -4/25 (x +4)2 + 8
Find the vertex from f(x) = ax2 + bx + c
To find “h” we use “b” and “a”
To find “k” we place h back in the equation.
Lets find the vertex of the equation
f(x) = 4x 2 + 3x – 8
Now let h = -⅜
k = 4(- ⅜)2 + 3(⅜) – 8
Vertex is ( - ⅜, - 137/16)
k = -137/16
Homework
Page 116 – 117
#3, 11, 17, 19, 25, 31, 45, 59, 63
Homework
Page 116
#22, 33, 51, 64, 73, 77, 80, 83, 89, 91, 104
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