Math-Circles

I am familiar with finding the vertex of a polynomial but I don't know how to solve this equation.
Give the vertex of x^2+9
can you please explain in steps?
Thank you.

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  1. An equation requires an equal sign.

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    posted by PsyDAG
  2. If you mean

    y = x^2+9

    you can see that since x^2 can never be less than zero, the vertex is at x=0. Since y=9 there, the vertex is at (0,9)

    Or, recalling that the vertex of

    y = (x-h)^2 + k

    is at (h,k), note that we have h=0 and k=9.

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    posted by Steve

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