Question

Find the equation of a parabola that has a vertex of (-2,-3) and contains the point (4,1)?

Answer 1

See below.

Explanation:

The equation in `color(blue)("vertex form")` of a parabola is:

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

Where:

`bba` is the coefficient of `bb(x^2)`

`bbh` is the axis of symmetry.

`bbk` is the maximum/minimum value of the function.

We are given the coordinates of the vertex `(-2,-3)`. The `bbx` coordinate of the vertex is also the axis of symmetry `h` and the `bby` coordinate is the maximum/minimum value `bbk`

Plugging in all known information, we have:

`1=a((4)-(-2))^2-3`

We need to solve for `bba`:

`1=a(36)-3`

`a=4/36=1/9`

So our equation is now:

`y=1/9(x+2)^2-3`

We can either leave it in this form, or expand and simplify it:

`y=1/9(x^2+4x+4)-3`

`y=1/9x^2+4/9x+4/9-3`

`y=1/9x^2+4/9x-23/9`

The graph confirms these points.