Question

How do you write the equation ins vertex form for the quadratic whose vertex is at (-2, 5) and has a point at (2,-2)?

Answer 1

`y=-7/16(x+2)^2+5`

Explanation:

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

`color(red)(|bar(ul(color(white)(a/a)color(black)(y=a(x-h)^2+k)color(white)(a/a)|)))`
where (h ,k) are the coordinates of the vertex and a, is a constant.

here (h ,k) = (-2 ,5)

`rArry=a(x+2)^2+5" is the partial equation"`

Since the parabola passes through (2 ,-2) then the coordinates of this point make the equation true.
Substitute x = 2 and y = -2 into the partial equation to find a.

`rArra(2+2)^2+5=-2rArr16a=-7rArra=-7/16`

`rArry=-7/16(x+2)^2+5" is the equation in vertex form"`
graph{-7/16(x+2)^2+5 [-10, 10, -5, 5]}