Answer
a) The slope of the tangent to the graph of $f\left( x \right)=4x+2$ at $\left( 1,6 \right)$ is $4$.
b) The slope intercept equation of the tangent line is $y=4x+2$.
Work Step by Step
(a)
Consider the function $f\left( x \right)=4x+2$ and the point $\left( 1,6 \right)$
Here, $\left( a,f\left( a \right) \right)=\left( 1,6 \right)$
Now, compute the slope of the tangent line for the function $f\left( x \right)=4x+2$ as follows:
Put $a=1$ ,
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
To compute $f\left( 1+h \right)$, substitute $x=1+h$ in the function $f\left( x \right)=4x+2$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 4\left( 1+h \right)+2 \right]-\left[ 4\left( 1 \right)+2 \right]}{h}$
Now, simplify $2\left( 1+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 4+4h+2 \right]-6}{h}$
Combine the like terms in the numerator.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{4h}{h}$
Cancel out h and apply the limits.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,4 \\
& =4
\end{align}$
Thus, the slope of the tangent to the graph of $f\left( x \right)=4x+2$ at $\left( 1,6 \right)$ is $4$.
(b)
Consider the function $f\left( x \right)=4x+2$ and the point $\left( 1,6 \right)$
From part (a), the slope of the tangent to the graph of $f\left( x \right)=4x+2$ at $\left( 1,6 \right)$ is $4$.
Now compute the slope intercept equation of the tangent line as follows:
Here, the tangent line passes through the point $\left( 1,6 \right)$ which implies ${{x}_{1}}=1$ and ${{y}_{1}}=6$
Substitute $m=4,{{x}_{1}}=1,{{y}_{1}}=6$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-6=4\left( x-1 \right) \\
& y-6=4x-4 \\
& y=4x-4+6 \\
& y=4x+2
\end{align}$
Thus, the slope intercept equation of the tangent line is $y=4x+2$.
