Quadratic ProblemDate: 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 |
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