urgent calculus question please help

Question

Find the value of t for which  = 0 for the parametric equations x = t³ + 6t² + 9t + 1 and y = 2t³ + 9t² - 24t + 6.

	t = 1, t = 3 and t = 4
	t = -1
	t = -3
	t = 1

I can't decide between A and D

Answer 1

dy/dt = 6t² + 18t - 24

dx/dt = 3t² + 12t + 9

dy/dx = (dy/dt)/(dx/dt) = (6t² + 18t - 24)/(3t² + 12t + 9)

(6t² + 18t - 24)/(3t² + 12t + 9) = 0 only when the numerator equals zero

6t² + 18t - 24 = 0

6(t² + 3t - 4) = 0

6(t + 4)(t - 1) = 0

t = -4, t = 1 we can throw ot t = -4 so the answer is t = 1