91Ó°ÊÓ

We have to find the slope of the tangent line $f\left(x\right)=x^3+x^2$at localid="1647249482483" $(c,f(c\left)\right)=(2,12)$.

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

Now let us substitute the values in equation.

That is,

$$\begin{gathered} \begin{array}{rl}{m}_{\tan }& =\underset{x→2}{lim}\frac{f\left(x\right)−f\left(2\right)}{x−\left(2\right)}\\ & =\underset{x→2}{lim}\frac{x^3+x^2−12}{x−2}\\ & =\underset{x→2}{lim}\frac{(x^2+3x+6)(x−2)}{x−2}\\ & =\underset{x→2}{lim}(x^2+3x+6)\end{array} \\ \begin{array}{rl}& =(2{)}^2+3(2)+6\\ & =4+6+6\\ & =16\end{array} \end{gathered}$$

The slope of the tangent line to $f\left(x\right)=x^3+x^2$at $(2,12)$is $m_{\tan }=16$.

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

Here,

$$\begin{gathered} m_{\tan }=16 \\ (x_1,y_1)=(2,12) \end{gathered}$$

Therefore,

$\begin{array}{rl}y−y_1& =m_{\tan }(x−x_1)\\ y−12& =16(x−2)\\ y−12& =16x−32\\ y& =16x−20\end{array}$

Hence the equation of the tangent line is$y=16x-20$

The graph of the original function $f\left(x\right)=x^3+x^2$and the tangent line$y=16x-20$at $(2,12)$.