You are given the parametric equations x = 2t^3 + 3t^2 - 12t, y = 2t^3 + 3t^2 +

Question

You are given the parametric equations x = 2t^3 + 3t^2 - 12t, y = 2t^3 + 3t^2 + 1. List all of the points (x, y) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of x. If two or more points share the same value of x, list those points starting with the smallest value of y. If any blanks are unused, type an upper-case "N" in them. List all of the points (x, y) where the tangent line is vertical. In entering your answer, list the points starting with the smallest value of x. If two or more points share the same value of x, list those points starting with the smallest value of y. If any blanks are unused, type an upper-case "N" in them.

Explanation / Answer

x = 2t3 + 3t2 - 12t , y = 2t3 + 3t2 + 1

dx/dt = 2(3)t3-1 + 3(2)t2-1 - 12(1)      since d/dx xn = n xn-1

==> dx/dt = 6t2 + 6t - 12

dy/dt = 2(3)t3-1 + 3(2)t2-1 + 0

==> dy/dt = 6t2 + 6t

Tangent is horizontal ==> dy/dt = 0 , dx/dt 0

==> 6t2 + 6t = 0

==> 6t(t + 1) = 0

==> t = 0 , t = -1

t = 0 ==> x = 2(0)3 + 3(0)2 - 12(0) = 0 , y = 2(0)3 + 3(0)2 + 1 = 1

t = -1 ==> x = 2(-1)3 + 3(-1)2 - 12(-1) = 13 , y = 2(-1)3 + 3(-1)2 + 1 = 2

Hence Tangent is horizontal at (0 , 0) and (13 , 2)

Tangent is vertical ==> dx/dt = 0 and dy/dt 0

==> 6t2 + 6t - 12 = 0

==> t2 + t - 2 = 0

==> t2 + 2t - t -2 = 0

==> t(t + 2) - 1(t + 2) = 0

==> (t + 2)(t - 1) = 0

==> t = -2 , 1

t = 1 ==> x = 2(1)3 + 3(1)2 - 12(1) = -7 , y = 2(1)3 + 3(1)2 + 1 = 6

t = -2 ==> x = 2(-2)3 + 3(-2)2 - 12(-2) = 20 , y = 2(-2)3 + 3(-2)2 + 1 = -3

Hence tangent is vertical at (-7 , 6) and (20 , -3)

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