Question

How do you find a function whose graph is a parabola with vertex (-2,2) and that passes through the point (1,-4)?

Answer 1

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

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 ,2) so we can write a 'partial' equation.

`rArry=a(x+2)^2+2`

To find a, substitute x = 1 , y = -4 from (1 ,-4) into the partial equation.

`a(1+2)^2+2=-4rArr9a=-6rArra=-2/3`

`rArry=-2/3(x+2)^2+2" is the equation"`
graph{-2/3(x+2)^2+2 [-10, 10, -5, 5]}