use the given parameters to answer the following questions. if you have a graphing device, graph the curve…

Explanation:

Step1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Differentiate $x = 2t^{3}+3t^{2}-12t$ with respect to $t$: $\frac{dx}{dt}=6t^{2}+6t - 12$. Differentiate $y = 2t^{3}+3t^{2}+4$ with respect to $t$: $\frac{dy}{dt}=6t^{2}+6t$.

Step2: Find horizontal - tangent points

A horizontal tangent occurs when $\frac{dy}{dt}=0$ and $\frac{dx}{dt}\neq0$. Set $\frac{dy}{dt}=6t^{2}+6t = 6t(t + 1)=0$. Solving $6t(t + 1)=0$ gives $t = 0$ or $t=-1$. When $t=-1$, $x=2(-1)^{3}+3(-1)^{2}-12(-1)=-2 + 3+12 = 13$, $y=2(-1)^{3}+3(-1)^{2}+4=-2 + 3+4 = 5$. When $t = 0$, $x=2(0)^{3}+3(0)^{2}-12(0)=0$, $y=2(0)^{3}+3(0)^{2}+4 = 4$. Check $\frac{dx}{dt}$: when $t=-1$, $\frac{dx}{dt}=6(-1)^{2}+6(-1)-12=6 - 6-12=-12\neq0$; when $t = 0$, $\frac{dx}{dt}=6(0)^{2}+6(0)-12=-12\neq0$.

Step3: Find vertical - tangent points

A vertical tangent occurs when $\frac{dx}{dt}=0$ and $\frac{dy}{dt}\neq0$. Set $\frac{dx}{dt}=6t^{2}+6t - 12=6(t^{2}+t - 2)=6(t + 2)(t - 1)=0$. Solving $6(t + 2)(t - 1)=0$ gives $t=-2$ or $t = 1$. When $t=-2$, $x=2(-2)^{3}+3(-2)^{2}-12(-2)=-16 + 12+24 = 20$, $y=2(-2)^{3}+3(-2)^{2}+4=-16 + 12+4 = 0$. When $t = 1$, $x=2(1)^{3}+3(1)^{2}-12(1)=2 + 3-12=-7$, $y=2(1)^{3}+3(1)^{2}+4=2 + 3+4 = 9$. Check $\frac{dy}{dt}$: when $t=-2$, $\frac{dy}{dt}=6(-2)^{2}+6(-2)=24-12 = 12\neq0$; when $t = 1$, $\frac{dy}{dt}=6(1)^{2}+6(1)=12\neq0$.

Answer:

(a) $(-1,5)$ (smaller $t$), $(0,4)$ (larger $t$) (b) $(-2,0)$ (smaller $t$), $(1,9)$ (larger $t$)