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

Explanation:

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

Differentiate $x = 2t^{3}+3t^{2}-72t$ with respect to $t$: $\frac{dx}{dt}=6t^{2}+6t - 72$. Differentiate $y = 2t^{3}+3t^{2}+2$ 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 $t = 0$: $x=2(0)^{3}+3(0)^{2}-72(0)=0$, $y=2(0)^{3}+3(0)^{2}+2 = 2$. When $t=-1$: $x=2(-1)^{3}+3(-1)^{2}-72(-1)=-2 + 3+72=73$, $y=2(-1)^{3}+3(-1)^{2}+2=-2 + 3+2=3$. Check $\frac{dx}{dt}$: When $t = 0$, $\frac{dx}{dt}=6(0)^{2}+6(0)-72=-72\neq0$. When $t=-1$, $\frac{dx}{dt}=6(-1)^{2}+6(-1)-72=6 - 6-72=-72\neq0$.

Step3: Find vertical - tangent points

A vertical tangent occurs when $\frac{dx}{dt}=0$ and $\frac{dy}{dt}\neq0$. Set $\frac{dx}{dt}=6t^{2}+6t - 72=6(t^{2}+t - 12)=6(t + 4)(t - 3)=0$. Solving $6(t + 4)(t - 3)=0$ gives $t=-4$ or $t = 3$. When $t=-4$: $x=2(-4)^{3}+3(-4)^{2}-72(-4)=-128 + 48+288=208$, $y=2(-4)^{3}+3(-4)^{2}+2=-128+48 + 2=-78$. When $t = 3$: $x=2(3)^{3}+3(3)^{2}-72(3)=54 + 27-216=-135$, $y=2(3)^{3}+3(3)^{2}+2=54+27 + 2=83$. Check $\frac{dy}{dt}$: When $t=-4$, $\frac{dy}{dt}=6(-4)^{2}+6(-4)=96-24 = 72\neq0$. When $t = 3$, $\frac{dy}{dt}=6(3)^{2}+6(3)=54 + 18=72\neq0$.

Answer:

(a) $(-1,3)$ (smaller $t$), $(0,2)$ (larger $t$) (b) $(-4, - 78)$ (smaller $t$), $(3,83)$ (larger $t$)