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Quadratic Problem


Date: 12 Jul 1995 11:05:54 -0400
From: Margie Salaz
Subject: Question

Here is a question: "Find quadratic functions satisfying the given 
conditions:  The vertex is (3,-1) and one x-intercept is x=1."

This is what I did:  Since the vertex is (3,-1)  x=3 and y=-1.  To come up 
with x=3 you must have (x - 3)^2  -1.  The only quadratic equation I could 
come up with to satisfy these conditions was x^2 - 6x + 8, which will give 
you the required point of (3,-1), however it  does not satisfy the condition 
that one x-intercept =1.  What can I do to find the proper quadratic 
equation that will satisfy both conditions??

Thanks for your help.

Sincerely,
Margie Salaz
Cuba, NM


Date: 12 Jul 1995 22:17:33 -0400
From: Dr. Ken
Subject: Re: Question

Hello there!

Here's a hint: what about the equation 2(x - 3)^2 - 1 = y?  It's not the
answer, but maybe it will help you get there.

-K
    
Associated Topics:
High School Basic Algebra

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