Answer
$$\frac{3}{2 \sqrt{2}}$$$$y=\frac{3}{2 \sqrt{2}}t +\frac{7}{2 \sqrt{2}}$$
Work Step by Step
Let $f(x)=\sqrt{3t+5}$. Then
\begin{aligned}
f^{\prime}(-1)&=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{3 h+2}-\sqrt{2}}{h} \\
&=\lim _{h \rightarrow 0} \frac{\sqrt{3 h+2}-\sqrt{2}}{h} \cdot \frac{\sqrt{3 h+2}+\sqrt{2}}{\sqrt{3 h+2}+\sqrt{2}}\\
&=\lim _{h \rightarrow 0} \frac{3 h}{h(\sqrt{3 h+2}+\sqrt{2})}\\
&=\lim _{h \rightarrow 0} \frac{3}{\sqrt{3 h+2}+\sqrt{2}}=\frac{3}{2 \sqrt{2}}
\end{aligned}
The tangent line at $a=-1$ is
\begin{aligned}
y&=f^{\prime}(-1)(t+1)+f(1)\\
&=\frac{3}{2 \sqrt{2}}(t+1)+\sqrt{2}\\
&=\frac{3}{2 \sqrt{2}}t +\frac{7}{2 \sqrt{2}}
\end{aligned}
