Answer
b) The slope intercept equation of the tangent line is $y=x+1$.
Work Step by Step
(b)
Consider the function, $f\left( x \right)=-\frac{1}{x+3}$
The $x$ coordinate for the tangent is $-2$.
Substitute $x=-2$ in the function $f\left( x \right)=-\frac{1}{x+3}$.
$\begin{align}
& f\left( -2 \right)=-\frac{1}{\left( -2 \right)+3} \\
& =-\frac{1}{1} \\
& =-1
\end{align}$
Thus, the tangent line passes through $\left( -2,-1 \right)$.
The value of $a$ from the point $\left( a,f\left( a \right) \right)$ or $\left( -2,-1 \right)$ is $-2$.
Substitute $a=-2$ in ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( -2+h \right)-f\left( -2 \right)}{h}$
Substitute $x=-2+h$ in the function $f\left( x \right)=-\frac{1}{x+3}$ and replace $f\left( -2 \right)=-1$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ -\frac{1}{\left( -2+h \right)+3} \right]-\left( -1 \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ -\frac{1}{h+1} \right]+1}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{-1+h+1}{h\left( h+1 \right)} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\left( h+1 \right)}
\end{align}$
Apply the limits and simplify.
$\begin{align}
& {{m}_{\tan }}=\frac{1}{\left( 0+1 \right)} \\
& =1
\end{align}$
Thus, the slope of the tangent to the graph of $f\left( x \right)=-\frac{1}{x+3}$ at $\left( -2,-1 \right)$ is $1$.
Substitute ${{x}_{1}}=-2$, ${{y}_{1}}=-1$ and $m=1$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-\left( -1 \right)=1\left[ x-\left( -2 \right) \right] \\
& y+1=\left( x+2 \right) \\
& y=x+2-1 \\
& y=x+1
\end{align}$
Therefore, the slope intercept equation of the tangent line is $y=x+1$.
