SOLUTION: Find the quadratic equation with a maximum value of 10 and x-intercepts of 1 and 3

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Question 378471: Find the quadratic equation with a maximum value of 10 and x-intercepts of 1 and 3
Found 2 solutions by solver91311, robertb:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


I can't quite do exactly what you ask. A quadratic equation is something of the form:



which has two roots, but does NOT have a maximum value or intercepts of any kind.

On the other hand, a quadratic function is a thing of the form:



And DOES have a maximum/minimum at the vertex, x intercepts (sometimes, provided ), a y-intercept, and an axis of symmetry. I'm going to proceed on the assumption that is what you meant in the first place.

You have specified 3 points through which the graph of the function passes. First, we have the two intercepts, and . By symmetry, the vertex, (the maximum point in this case) must lie midway between the intercepts. Hence the -coordinate of the vertex must be 2. And since the value of the function at the maximum is 10, the vertex must be at

Using this data we can create three linear equations:







which is to say:







Solve the system for , , and to get the coefficients of your desired quadratic function.


John

My calculator said it, I believe it, that settles it
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Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The y-coordinate of the vertex is 10. To get the x-coordinate, just get the average of the x-intercepts, which is (1+3)/2 = 2. Hence the vertex is (2, 10).
From the vertex form of the equation of the parabola, we get y+=+a%28x+-+2%29%5E2+%2B+10. We have to solve for a. Then using either one of the x-intercepts (I will use (1,0)) we get
0+=+a%281-2%29%5E2+%2B+10;
0+=+a+%2B10, or
a = -10.
Therefore the equation is y+=+-10%28x+-+2%29%5E2+%2B+10.
=D