Find the standard equation of the parabola which satisfies the given condition

Question

 vertex (0,7), vertical axis of symmetry, through the point P(4,5)

My answer: (y-k)^2 = -4c(x-h) (y-7)^2 = -16x

Please kindly correct me if I have any mistake.

Answer 1

So I think the most simple thing here is to use point slope.

Given the vertex $(h,k)$ of a parabola, we can write

$y=a(x-h)^2+k$.

Given vertex $(0,7)$, we have $\displaystyle y=ax^2+7$.

Now, we plug in the point $(4,5)=(x,y)$, to get $5=a(4^2)+7$.

We have $5=16a+7$, so $-2=16a$, and $\displaystyle a=-\frac{1}{8}$

We have $\boxed{\displaystyle y=-\frac{1}{8}x^2+7}$.

I am confused as to why you still have the variable $h$ in your answer.