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)?

1 Answer
Jul 28, 2016

#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]}