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by **pistolpete** on Thu May 28, 2009 7:01 pm

I am having a really hard time with this section. My teacher will be absent tomorrow and I have a test on Monday. I don't understand this. So, for practice could anyone help me out and show me step by step how to do some problems?

In my book, it says:
"use the Vertex Form: y - k = a (x - h)^2
Write an equation of the parabola having V as the vertex and containing P."

Ex.1
V(3, 5) ; P(4,7)

Ex.2
V(-1 ,1) ; P(1, 3)

I don't understand what to do. Do I just plug these in to the above formula and that's the answer or do I solve to get a = something?

**Sponsor**

by **stapel_eliz** on Thu May 28, 2009 7:16 pm

You are given the formula $y\, =\, a(x\, -\,h)^2\, +\, k,$ and are given values for x, y, h, and k. Plug these values in, and solve for "a".

Then rewrite the formula, plugging in the values for h, k, and a. :wink:

by **pistolpete** on Thu May 28, 2009 8:58 pm

Thanks stapel for the simple explanation.

Here's the work I did for the 1st example.

y = a(x - h)^2 + k
7 = a(4 - 3)^2 + 5
7 = a(1)^2 +5
7= a + 5
2 = a

And the final answer is: y = 2(x-3)^2 + 5

For the second one I got 1/2 = a
and the final answer/equation is: y = 1/2(x - (-1))^2 + 1

Are these correct ?

by **stapel_eliz** on Thu May 28, 2009 9:57 pm

The only adjustment I would recommend is simplifying inside the parentheses in the second one, to get:

$y\, =\, \frac{1}{2}\left(x\, +\, 1)^2\, + \,1$

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Graphing Quadratic Functions (vertex formula)

Graphing Quadratic Functions (vertex formula)

Re: Graphing Quadratic Functions (vertex formula)