Question

Given the range for a quadratic function is (-∞,5] and two of the points on the function are (-2,3) and (1,-3) determine the equation for the function in the form of y=a(x-h)^2+k

Answer 1

This is the vertex form.
k=5 (the range)
substitute (-2,3) and (1,-3)
3=a(-2-h)^2+5 (1)
-2=a(h+2)^2

-3=a(1-h)^2+5 (2)
-8=a(h-1)^2

(2)/(1)
4=[(h-1)^2]/[(h+2)^2]
4h^2+16h+16=h^2-2h+1
3h^2+18h+15=0
h=-1 or-5

y=a(x-h)^2+5

-2=a(-2+1)^2
a=-2

y=-2(x+2)^2+5

-2=a(-5+2)^2
a=-2/9

y=-2/9(x+5)^2+5

I don't know how to reject one of the h