In Problems $30-32$, find the slope of the tangent line to the graph of $f$at the given point. Graph $f$and the tangent line.

$\text{31.}f\left(x\right)=x^2+2x-3$at $(-1,-4)$

We have to find the slope of the tangent line $f\left(x\right)=x^2+2x-3$at $(-1,-4)$.

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

Now let us substitute the values in equation.

That is,

role="math" localid="1647248058997" $$\begin{gathered} \begin{array}{rl}{m}_{\tan }& =\underset{x\to -1}{lim}\frac{f\left(x\right)-f(-1)}{x-(-1)}\\ & =\underset{x\to -1}{lim}\frac{x^2+2x-3-(-4)}{x+1}\\ & =\underset{x\to -1}{lim}\frac{x^2+2x+1}{x+1}\\ & =\underset{x\to -1}{lim}\frac{(x+1)(x+1)}{x+1}\end{array} \\ \begin{array}{rl}& =\underset{x\to -1}{lim}(x+1)\\ & =-1+1\\ & =0\end{array} \end{gathered}$$

The slope of the tangent line to $f\left(x\right)=x^2+2x-3$at $(-1,-4)$is$m_{\tan }=0$.

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,

$$\begin{gathered} m_{\tan }=0 \\ (x_1,y_1)=(-1,-4) \end{gathered}$$

Therefore,

$\begin{array}{rl}y-y_1& =m_{\tan }(x-x_1)\\ y-(-4)& =0(x-(-1\left)\right)\\ y+4& =0\\ y& =-4\end{array}$

Hence the equation of the tangent line is$y=-4$.

The graph of the original function $f\left(x\right)=x^2+2x-3$and the tangent line $y=-4$at $(-1,-4)$.