Question

(1 point) You are given the parametric equations x=2t3+3t2−12t,y=2t3+3t2+1. (a) 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.

Answer 1

Answer:

(0,1) and (13,2)

Step-by-step explanation:

We are given that parametric equations

$x=2t^3+3t^2-12t$

$y=2t^3+3t^2+1$

a.We have to find the points where tangent line is horizontal.

Differentiate x and y w.r.t.time

$\frac{dx}{dt}=6t^2+6t-12$

$\frac{dy}{dt}=6t^2+6t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Substitute the values

$\frac{dy}{dx}=\frac{6t^2+6t}{6t^2+6t-12}$

We know that when tangent line is horizontal then

$\frac{dy}{dx}=0$

$\frac{6t^2+6t}{6t^2+6t-12}=0$

$6t^2+6t=0$

$t^2+t=0$

$t(t+1)=0$

$t=0, t+1=0$

$t+1=0\implies t=-1$

Substitute the t=0 then we get

$x=2(0)^3+3(0)^2-12(0)=0$

$y=2(0)^3+3(0)^2+1=1$

Substitute t=-1

$x=2(-1)^3+3(-1)^2-12(-1)=-2+3+12=13$

$y=2(-1)^3+3(-1)^2+1=-2+3+1=2$

The points  (0,1) and (13,2) where tangent line is horizontal .