Find the value of t for which = 0 for the parametric equations

Question

Find the value of t for which  = 0 for the parametric equations x = t³ + 6t² + 9t + 1 and y = 2t³ + 9t² - 24t + 6. (5 points)

answer choices

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

Answer 1

y / x = ( 2 x ^ ^{3 +} 9 t ^ ² – 24 t + 6 ) ⁄ ( t ^ ³ + 6 t ^ ² + 9 t + 1 )

d y / d x = (d y / d t ) / ( d x / d t )

= ( 6 t ^ ² + 18 t – 24 ) ⁄ ( 3 t ^ ² + 12 t + 9 )

= 6 (t ^ ² + 3 t – 4 ) ⁄ 3 ( t ^ ² + 4 t + 3 )

= [2 ( t + 4 ) ( t – 1 ) ] ⁄ [ ( t + 3 ) ( t + 1 ) ]

= t = - 4 ; t = 1