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

1 Answer
Aug 16, 2016

#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]}