Question

What is the equation of the parabola that has a vertex at `(77, 7)` and passes through point `(82,32)`?

Answer 1

`y=(x-77)^2+7`

Explanation:

The vertex form of a parabola is `y=a(x-h)^2+k`, where the vertex is `(h,k)`.

Since the vertex is at `(77,7)`, `h=77` and `k=7`. We can rewrite the equation as:

`y=a(x-77)^2+7`

However, we still need to find `a`. To do this, substitute the given point `(82, 32)` in for the `x`- and `y`-values.

`32=a(82-77)^2+7`

Now, solve for `a`.

`32=a(82-77)^2+7`
`32=a(5)^2+7`
`32=25a+7`
`25=25a`
`a=1`

The final equation is `y=1(x-77)^2+7`, or `y=(x-77)^2+7`.