Given $f(x)=\frac{x^3-7x-6}{x+1}$.   

1. What is the value of $f(–1)$?

   Hint (a):

   What happens when the denominator is $0$?

   $\left. \begin{array} { r } { x ^ { 2 } - x - 6 } \\ { x + 1 \longdiv { x ^ { 3 } + 0 x ^ { 2 } - 7 x - 6 } } \end{array} \right.$
   $\ \ \ \ \ \ \ \ \underline{- ( x ^ { 3 } + x ^ { 2 } )}$
   $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - x ^ { 2 } - 7 x$
   $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{ - ( - x ^ { 2 } - 1 x )}$
   $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 6 x - 6$
   $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline {- ( - 6 x - 6 )}$
   $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$

2. Evaluate $\lim\limits_{ x \rightarrow - 1 } f ( x )$. Hint: Use polynomial division.

   Hint (b):

   Substitute $x = −1$ into the quotient.

3. Use what your answer to part (b) to sketch $y=f(x)$ without using a calculator.

4. If $y = f(x)$ continuous at $x = -1$? Use the definition of continuity to justify your answer.

   Hint (d):

   You know it is a parabola with a hole at $x = −1$. Find the $x$-intercepts by factoring.