SOLUTION: Find a quadratic function y=a(x-h)^2 for which the graph includes the given points. What are the steps for solving this kind of problem? Example (-2,1),(2,1)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Find a quadratic function y=a(x-h)^2 for which the graph includes the given points. What are the steps for solving this kind of problem? Example (-2,1),(2,1)      Log On


   

Question 1080303: Find a quadratic function y=a(x-h)^2 for which the graph includes the given points.
What are the steps for solving this kind of problem?
Example
(-2,1),(2,1)
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
a%28x-h%29%5E2 is like a%28x-h%29%5E2%2Bk if k=0.

Two values of x having the same value of y, in (-2,1) and (2,1).
Symmetry axis and vertex are in the middle of x=-2 and x=2, which is x=0.
This means, h=0.

y=a%28x-0%29%5E2

y=ax%5E2

Pick either point to determine coefficient a.
ax%5E2=y
a=y%2Fx%5E2
a=1%2F2%5E2
a=1%2F4

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highlight%28y=%281%2F4%29x%5E2%29

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y=a%28x-h%29%5E2
given points:(-2,1),(2,1)
use given points to find a and h
y=a%28x-h%29%5E2 .....if (-2,1)=>x=-2 and y=1
1=a%28-2-h%29%5E2.....solve for a
a=1%2F%28-2-h%29%5E2...........eq.1

y=a%28x-h%29%5E2 .....if (2,1)=>x=2 and y=1
1=a%282-h%29%5E2....solve for a
a=1%2F%282-h%29%5E2...........eq.2
from eq.1 and eq2 we have
1%2F%28-2-h%29%5E2=1%2F%282-h%29%5E2...cross multiply
%282-h%29%5E2=%28-2-h%29%5E2.............solve for h
%282-h%29%5E2-%28-2-h%29%5E2=0
-8h+=+0
h=0
find a
a=1%2F%28-2-h%29%5E2...........eq.1
a=1%2F%28-2-0%29%5E2
a=1%2F%28-2%29%5E2

a=1%2F4
so your equation is:y=%281%2F4%29x%5E2