91Ó°ÊÓ

We have to find the slope of the tangent line $f\left(x\right)=2x^2+8x$at$(c,f(c\left)\right)=(1,10)$.

For that, we can use the formulalocalid="1647246256562" $m_{\tan }=\underset{x→c}{lim} \frac{f\left(x\right)−f\left(c\right)}{x−c}$

Now let us substitute the values in equation.

That is,

$$\begin{gathered} m_{\tan }=\underset{x→1}{lim} \frac{f\left(x\right)−f\left(1\right)}{x−1} \\ =\underset{x→1}{lim} \frac{2x^2+8x−10}{x−1} \\ =\underset{x→1}{lim} \frac{(2x+10)(x−1)}{x−1} \\ =\underset{x→1}{lim} (2x+10) \\ =2\left(1\right)+10 \\ =12 \end{gathered}$$

The slope of the tangent line to $f\left(x\right)=2x^2+8x$at$(1,10)$islocalid="1647246269096" $m_{\tan }=12$.

To find the equation of the tangent line, we can use point-slope formula $y−y_1=m_{\tan }\left(x−x_1\right)$ to find the equationof the tangent line.

Here,

Therefore,

$\begin{array}{rl}y−y_1& =m_{\tan }(x−x_1)\\ y−10& =12(x−1)\\ y−10& =12x−12\\ y& =12x−2\end{array}$

Hence the equation of the tangent line isrole="math" localid="1647246638551" $y=12x-2$.

The graph of the original function$f\left(x\right)=2x^2+8x$and the tangent line $y=12x-2$ at $(1,10)$.