Answer
a) horizontal tangent line: at $\frac{dy}{dx}=0$
$$\frac{dy}{dt}=0$$$$2t+1=0$$$$t=-\frac{1}{2}$$
b) vertical tangent line: at $\frac{dx}{dt}=0$
$$\frac{dx}{dt}=6t^2-30t+24=0$$$$t=1,4$$
Work Step by Step
1) differentiate with x with respect to t: $\frac{dx}{dt}=6t^2-30t+24$
2) differentiate with y with respect to t: $\frac{dy}{dt}=2t+1$
3) find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ & $\frac{dx}{dt}$ :$\frac{dy}{dx}=\frac{2t+1}{6(t-1)(t-4)}$
4) horizontal tangent line: at $\frac{dy}{dx}=0$
$$\frac{dy}{dt}=0$$$$2t+1=0$$$$t=-\frac{1}{2}$$
5) vertical tangent line: at $\frac{dx}{dt}=0$
$$\frac{dx}{dt}=6t^2-30t+24=0$$$$6(t^2-5t+4)=6(t-1)(1-4)=0$$$$t=1,4$$
