Question 1113240
y=a(x-h)^2+k is the standard form, so h=-1 and k=-16
y=a(x+1)^2-16
putting (-1, -16) into vertex makes y+16=a(x+1)^2
x intercepts are where y=0.  Where y=-15, x=0, so 1=a(1)^2. Therefore, a=1

a(x+1)^2=16
x+1=+/-4, since a=1
x=3, -5
(3, 0) and (-5, 0) ANSWER
The parabola is x^2+2x-15
That gives a vertex of (-1, -16) and the y-intercept of -15
{{{graph(300,300,-10,10,-20,20,x^2+2x-15)}}}