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Section 1 - Quadratic Functions

Page history last edited by PBworks 15 years, 11 months ago

 

Section 2.1 Quadratic Functions

 

-This chapter is about quadratic functions

 

Polynomial Function is f(x)= ax^n+bx^(n-1)+cx^(n-2)...+z

 

Quadratic Fn: f(x)=ax²+bx+c

 

Standard Form of a parabola: f(x)=a(x-h)²+k

Vertex: (h,k)

Axis of Symmetry: x=h

x-int: set the equation to 0 and solve

 

Example 1) Find the vertex, Axis of Symmetry, and x-intercepts of f(x) = 2x²+8x+7

                 f(x) = 2(x²+4x)+7 *factor out "a" from only the 1st two terms

                       = 2(x²+4x+4)+7-8 *find half of the underlined term  and double it to find the

                                                   bold term; add or subtract the product of the italicized numbers

                                                   to the end of the problem

                       = 2(x+2)² - 1

 

vertex: (-2,-1) *the opposite sign of the bold number is your x and the underlined number is your y

 

 

AoS: x=-2 *the x in your vertex is the AoS

 

2(x+2)²-1=0 *set equal to 0 to find x-int

eq=sqrt{(x+2)^2 }=eq=sqrt{frac{1}{2}} *square both sides to get rid of the exponent

x+2=+eq=sqrt{frac{1}{2}}

X=-2+eq=sqrt{frac{1}{2}} *subtract 2 from both sides to find x-int

 

 

 

How to find the standard form of a palabora:

 

Ex: Vertex (3,2) and it goes through a point (1,-6)

Reminder: Standard Form: f(x)=a(x-h)²+k

          -6=a(1-3)²+2         *add in all the terms given to you in the problem

      -6=a(2)+2

      -8=2a          *subtract 2 from both sides

      -4=a           *divide both sides by 4 to find a

 

 

Final answer: f(x)=-4(x-3)²+2

 

 

To find maximum or minimum for vertex: x= -b/2a

 

Comments (1)

Anonymous said

at 8:23 pm on Jan 8, 2008

Can you find the word "palabra"?
Also, it's hard not to get bogged down in your example problems... changing the spacing/formatting would be nice. 18/20

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