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