Answer
Tangent line: $\quad y=x-4$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-1}=\frac{1}{2}$
Work Step by Step
The slope of the tangent line at $t=-1$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=-1}$
$\displaystyle \left\{\begin{array}{l}
\frac{dx}{dt}=4t,\\
\\
\frac{dy}{dt}=4t^{3}
\end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t^{3}}{4t}=t^{2}$
$\left. \displaystyle \frac{dy}{dx} \right|_{t=-1}=1$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-1}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[t^{2}]=2t$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{2t}{4t}=\frac{1}{2}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-1}=\frac{1}{2}$
The point $P$ on the curve corresponding to $t=-1$ is
$\left\{\begin{array}{l}
x=2(1)+3=5\\
\\
y=1
\end{array}\right.\quad, P(5,1)$
Use the point-slope equation for the tangent line.
$y-1=1\cdot(x-5)$
$y=x-4$
