Answer
b) The slope intercept equation of the tangent line is $y=3x+4$.
Work Step by Step
(b)
Consider the function, $f\left( x \right)={{x}^{3}}+2$
The $x$ coordinate for the tangent is $-1$.
Substitute $x=-1$ in the function $f\left( x \right)={{x}^{3}}+2$.
$\begin{align}
& f\left( -1 \right)={{\left( -1 \right)}^{3}}+2 \\
& =-1+2 \\
& =1
\end{align}$
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 $f\left( x \right)={{x}^{3}}+2$ and replace $f\left( -1 \right)=1$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( -1+h \right)}^{3}}+2 \right]-1}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( h-1 \right)}^{3}}+2-1}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( h-1 \right)}^{3}}+1}{h}
\end{align}$
Use the property
${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{h}^{3}}-1-3{{h}^{2}}+3h \right]+1}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{3}}-3{{h}^{2}}+3h}{h}
\end{align}$
Make the factor of the numerator and simplify.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( {{h}^{2}}-3h+3 \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\left( {{h}^{2}}-3h+3 \right)
\end{align}$
Apply the limits.
$\begin{align}
& {{m}_{\tan }}={{0}^{2}}-3\left( 0 \right)+3 \\
& =3
\end{align}$
Thus, the slope of the tangent to the graph of $f\left( x \right)={{x}^{3}}+2$ at $\left( -1,1 \right)$ is $3$.
Substitute ${{x}_{1}}=-1$, ${{y}_{1}}=1$ and $m=3$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-1=3\left[ x-\left( -1 \right) \right] \\
& y-1=3\left( x+1 \right) \\
& y=3x+3+1 \\
& y=3x+4
\end{align}$
Therefore, the slope intercept equation of the tangent line is $y=3x+4$.
