what is the y intercept of a parabola with the vertex of (1,9)

Answer 1

y-9 = a(x-1)^2 if it's an upward opening parabola with a>0

then y=a(x^2-2x +1) +9 = ax^2 -2ax +a +9, with y intercept = a+9

but the parabola could open downward with a<0. Still the y intercept would be a+9

but the parabola could open to the right or left, then

x-1 = a(y-9)^2

x = ay^2 -18a +81a +1, then the x intercept is 81a +1. IF it opens to the right, there is no y intercept.

if it opens to the left, set x =0 and solve for y. There should be 2 y intercepts

-1/a = (y-9)^2

y-9 = + or - sqr(-1/a)

y = 9 + or - sqr(-1/a) since a<0 the square root is a real number