Question

What is the equation of the parabola that has a vertex at `(-4, 4)` and passes through point `(6,104)`?

Answer 1

`y=(x+4)^2 + 4` or
`y=x^2 + 8*x + 20`

Explanation:

Start with the vertex form of the quadratic equation.

`y=a*(x-x_{vertex})^2 + y_{vertex}`.

We have `(-4,4)` as our vertex, so right off the bat we have

`y=a*(x-(-4))^2 + 4` or

`y=a*(x+4)^2 + 4`, less formally.

Now we just need to find "`a`."
To do this we sub in the values for the second point `(6,104)` into the equation and solve for `a`.
Subbing in we find
`(104)=a*((6)+4)^2 + 4`
or
`104=a*(10)^2+4`.

Squaring `10` and subtracting `4` from both sides leaves us with
`100=a*100` or `a=1`.

Thus the formula is `y=(x+4)^2 + 4`.

If we want this in standard form ( `y=a*x^2 + b*x +c` ) we expand the square term to get

`y=(x^2 + 8*x + 16) + 4` or

`y=x^2 + 8*x + 20`.