SOLUTION: Maybe I'm thinking too hard. This seems easy enough but the more I think about it, the more I think I'm wrong. The question is: 'Find the equation of a quadratic function with a

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Question 261422: Maybe I'm thinking too hard. This seems easy enough but the more I think about it, the more I think I'm wrong. The question is:
'Find the equation of a quadratic function with a vertex at the point (-1,-5) that also contains the point (1,3).'
My answer is (x+1)^2-5
Am I right or at least close? I don't recall doing a problem like this is class and I haven't found one like it in my textbooks. Thanks for the help guys!
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Unfortunately, that is not correct. Why? Since the quadratic goes through the point (1,3), this means that when x=1 then y=3. Plugging in x=1 gets us y=%281%2B1%29%5E2-5=2%5E2-5=4-5=-1 which is not what we're looking for.


What you need to do is set up the equation y=a%28x%2B1%29%5E2-5, plug in the values x=1 and y=3 and solve for 'a'.


Your answer should be a=2 which means that the equation of the quadratic is y=2%28x%2B1%29%5E2-5. Let me know if you need help getting to that answer.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Not quite. Here's how I know:

Substitute -1 for and evaluate:






So the point (-1,-5) is on the curve. So far, so good. Now substitute 1 for for and evaluate:





and that means that the point (1,3) is NOT on the graph. Back to the drawing board.

You are looking for a quadratic function of the form:




such that if , then and if , then . Furthermore, since (-1,-5) is specified as the vertex, the -coordinate of the vertex, namely -1, must be equal to .

That gives rise to a system of three linear equations:







Doing a little tidying up, we get:







You should be able to take it from here.

John