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