Answer
a) The slope of the tangent to the graph of $f\left( x \right)=\sqrt{x}$ at $\left( 16,4 \right)$ is $\frac{1}{8}$.
b) The slope intercept equation of the tangent line is $y=\frac{1}{8}x+2$.
Work Step by Step
(a)
Consider the function $f\left( x \right)=\sqrt{x}$ and the point $\left( 16,4 \right)$
Here, $\left( a,f\left( a \right) \right)=\left( 16,4 \right)$
Now, compute the slope of the tangent line for the function $f\left( x \right)=\sqrt{x}$ as follows:
Put $a=16$ ,
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 16+h \right)-f\left( 16 \right)}{h}$
To compute $f\left( 16+h \right)$, substitute $x=16+h$ in the function $f\left( x \right)=\sqrt{x}$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{16+h}-\sqrt{16}}{h}$
Now, multiply and divide with the conjugate of $\sqrt{16+h}-\sqrt{16}$ that is $\sqrt{16+h}+\sqrt{16}$ and simplify using the property $\left( A-B \right)\left( A+B \right)={{A}^{2}}-{{B}^{2}}$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{16+h}-\sqrt{16}}{h}\cdot \frac{\sqrt{16+h}+\sqrt{16}}{\sqrt{16+h}+\sqrt{16}} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( \sqrt{16+h} \right)}^{2}}-16}{h\left( \sqrt{16+h}+4 \right)} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{16+h-16}{h\left( \sqrt{16+h}+4 \right)}
\end{align}$
Combine the like terms in the numerator and cancel out h.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\left( \sqrt{16+h}+4 \right)}$
Apply the limits and simplify.
$\begin{align}
& {{m}_{\tan }}=\frac{1}{\left( \sqrt{16+0}+4 \right)} \\
& =\frac{1}{4+4} \\
& =\frac{1}{8}
\end{align}$
Thus, the slope of the tangent to the graph of $f\left( x \right)=\sqrt{x}$ at $\left( 16,4 \right)$ is $\frac{1}{8}$.
(b)
Consider the function $f\left( x \right)=\sqrt{x}$ and the point $\left( 16,4 \right)$
From part (a), the slope of the tangent to the graph of $f\left( x \right)=\sqrt{x}$ at $\left( 16,4 \right)$ is $\frac{1}{8}$.
Now compute the slope intercept equation of the tangent line as follows:
Here, the tangent line passes through the point $\left( 16,4 \right)$ which implies ${{x}_{1}}=16$ and ${{y}_{1}}=4$
Substitute $m=\frac{1}{8},{{x}_{1}}=16,{{y}_{1}}=4$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-4=\frac{1}{8}\left( x-16 \right) \\
& y-4=\frac{1}{8}x-2 \\
& y=\frac{1}{8}x-2+4 \\
& y=\frac{1}{8}x+2
\end{align}$
