The vertex of a parabola?

Question

A parabola is defined by (x-57)^2=-3y+288. How do I find the absolute value of the y-intercept of the parabola?

Answer 1

to find the "y-intercept", make note of the fact that this point is where x=0.

so just set x=0 and solve for "y"

in this case:

(x-57)^2=-3y+288

(0-57)^2 = 3y + 288
57^2 - 288 = 3y
3249 - 288 = 3y
2961 = 3y
987 = y

987 is also the "absolute value" of 987



I'm not sure where "vertex" came into play, but
there you have it

good luck

Answer 2

Find the y-intercept by plugging in x = 0.

(x - 57)² = -3y + 288
(0 - 57)² = -3y + 288
57² = -3y + 288
3y = 288 - 57² = 288 - 3249 = -2961
y = -2961/3 = -987

The absolute value of the y-intercept is |-987| = 987.