find x intercepts of parabola with vertex(-4,12) and y intercept(0,36)

Question

i need help with this math problem. it says " find the x intercepts of the parabola with a vertex (-4,12) and y intercept (0,36)

round to the nearest hundredth if nessisary.

Answer 1

Given : vertex of parabola  (h, k ) = (- 4, 12) and y - intercept is (0, 36).

The standard form of the parabola with vertex (h, k) and axis of symmetry x = h is y = a(x - h)² + k.

Substitute the values of (h, k ) = (- 4, 12) and (x, y ) = (0, 36) in y = a(x - h )^2 + k.

36 = a(0 - (- 4))^2 + 12

36 = a( 4)^2 + 12

36 = 16a + 12

16a = 36 - 12 = 24

a = 24/16 = 3/2.

Therefore, the parabola is y = (3/2)(x + 4)^2 + 12.

To find x - intercept substitute y = 0 in y = (3/2)(x + 4)^2 + 12.

0 = (3/2)(x + 4)^2 + 12

(3/2)(x + 4)^2 = - 12

(x + 4)^2 = - 8

(x + 4)^2 = (± √- 8)^2

x + 4 = ± √- 8

Substitute - 1 = i ^2.

x + 4 = ± √(8i ^2)

x + 4 = ± 2√2 i

x = - 4 ± 2√2 i.

Therefore, x - intercepts of the parabola are - 4 + 2√2 i and - 4 - 2√2 i.