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If I am given $V(8, -1)$ and $y_{intercept} = -17$, how would I go about finding the equation for that?

Thank you very much!

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closed as off-topic by Weapon of Choice, Jonas Meyer, Claude Leibovici, PVAL, hardmath Aug 18 at 4:47

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The answer is not determined from the information you've given us, but presumably you want $y$ expressed as a quadratic function of $x$. There is a unique solution once the axis is determined, in this case presumably $x=8$. E.g., $x=-\frac{1}{32}(y+1)^2+8$ is also the equation of a parabola with the stated conditions, but there the axis is $y=-1$. –  Jonas Meyer May 10 '11 at 2:01

1 Answer 1

The vertex form of a quadratic function is $$ y=a(x-h)^2+k $$ where $(h,k)$ is the vertex. Use your given vertex, and solve for $a$ using the fact that you know the y-intercept.

Note that the vertex and any other point on the curve would be sufficient.

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