Asked by Stryder
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.
Give the vertex of x^2+9
can you please explain in steps?
Thank you.
Answers
Answered by
PsyDAG
An equation requires an equal sign.
Answered by
Steve
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.
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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