Chapter 14, Problem 31RE

To determine

To find: The slope of the tangent line of the graph. Then graph the tangent line.

Answer to Problem 31RE

  $m_{\tan }=0$

Explanation of Solution

Given: $f(x)=x^2+2x-3at(-1,-4)$

  $f(x)=x^2+2x-3$

Finding $f(-1)$

  $\begin{array}{l}f(-1)={(}^{-}+2(-1)-3\\ f(-1)=1-2-3\\ f(-1)=-4\end{array}$

Also

Finding $f\text{'}(-1)$

We get

  $\begin{array}{l}f\text{'}(-1)=\underset{x\to -1}{\lim }\frac{f(x)-f(-1)}{x-(-1)}\\ =\underset{x\to -1}{\lim }\frac{x^2+2x-3-(-4)}{x+1}\\ =\underset{x\to -1}{\lim }\frac{{(x+1)}^2}{x+1}\\ =\underset{x\to -1}{\lim }x+1\\ =-1+1\\ =0\\ i.e.\\ m_{\tan }=0\end{array}$

Thus

Equation of the line is

  $\begin{array}{l}y-(-4)=0(x-(-1)\\ y+4=0\\ y=-4\end{array}$

The graph of the function with tangent line is shown below