Answer
a) The slope of the tangent to the graph of $f\left( x \right)=2x+3$ at $\left( 1,5 \right)$ is $2$.
b) The slope intercept equation of the tangent line is $y=2x+3$.
Work Step by Step
(a)
Consider the function $f\left( x \right)=2x+3$ and the point $\left( 1,5 \right)$
Here, $\left( a,f\left( a \right) \right)=\left( 1,5 \right)$
Now, compute the slope of the tangent line for the function $f\left( x \right)=2x+3$ as follows:
Put $a=4$ ,
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
To compute $f\left( 1+h \right)$, substitute $x=1+h$ in the function $f\left( x \right)=2x+3$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 2\left( 1+h \right)+3 \right]-\left[ 2\left( 1 \right)+3 \right]}{h}$
Now, simplify $2\left( 1+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 2+2h+3 \right]-5}{h}$
Combine the like terms in the numerator.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{2h}{h}$
Cancel out h and apply the limits.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,2 \\
& =2
\end{align}$
A window showing the graph of the function and the tangent will appear as shown below.
The graph of the function $f\left( x \right)=2x+3$ is a line and there is no tangent drawn in the graph because a tangent line to a straight line is the line itself. Thus, the slope of the tangent to the graph of $f\left( x \right)=2x+3$ at $\left( 1,5 \right)$ is $2$.
(b)
Consider the function $f\left( x \right)=2x+3$ and the point $\left( 1,5 \right)$
From part (a), the slope of the tangent to the graph of $f\left( x \right)=2x+3$ at $\left( 1,5 \right)$ is $2$.
Now compute the slope intercept equation of the tangent line as follows:
Here, the tangent line passes through the point $\left( 1,5 \right)$ which implies ${{x}_{1}}=1$ and ${{y}_{1}}=5$
Substitute $m=2,{{x}_{1}}=1,{{y}_{1}}=5$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-5=2\left( x-1 \right) \\
& y-5=2x-2 \\
& y=2x-2+5 \\
& y=2x+3
\end{align}$
The tangent line of the function $f\left( x \right)=2x+3$ is same as the function itself which is shown in the graph. Thus, the slope intercept equation of the tangent line is $y=2x+3$.
