## Precalculus (6th Edition) Blitzer

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 $y=4x+2$.
(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 $y=4x+2$.