parabolic equation from vertex and y-intercept

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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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

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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Matthew Conroy has already given a good answer. Here is an alternative that is not as elegant.

A quadratic function has the form $f(x)=ax^2+bx+c$.

The $y$-intercept tells you $c$.

From the vertex formula, you have $-\frac{b}{2a}=8$. Since $(8,-1)$ is on the graph, you also have $-1=a\cdot 8^2 + b\cdot 8 +c$. Since you already know $c$ from the $y$-intercept, this last equation has only $a$ and $b$ as unknowns. Now you have a system of $2$ equations and $2$ unknowns to solve for $a$ and $b$.

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solve -2x^2+8x+7 find vertex and y intercept

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This does not answer the question asked. In fact, $-2x^2+8x+7=15-2(x-2)^2$, so the vertex is $(2,15)$ and has a $y$-intercept of $7$. –  robjohn Jan 17 '13 at 6:41