Chapter 8.5, Problem 57SR

To determine

To find any vertical asymptotes and holes in the graph of each rational function

Answer to Problem 57SR

The vertical asymptotes are $x=-3\text{ and }x=-2$ and there is no holes in the function.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{1}{x^2+5x+6}$

The denominator of $f\left(x\right)$ cannot be zero as this would make $f\left(x\right)$ undefined. Equate the denominator to zero and solve the equation then this will give the value of $x$ .If the numerator is non-zero for these values then they are vertical asymptotes.

Now, solve the denominator

  $\begin{array}{l}{x}^2+5x+6=0\\ \Rightarrow x^2+\left(3+2\right)x+6=0\\ \Rightarrow x^2+3x+2x+6=0\\ \Rightarrow \left(x+3\right)\left(x+2\right)=0\\ \text{Either }\left(x+3\right)=0\text{ or }\left(x+2\right)=0\\ x=-3\text{ or }x=-2\end{array}$

The two values of $x$ are $-3$ and $-2$

Therefore, the asymptotes are $x=-3\text{ and }x=-2$

Holes occur when a common factor is removed from the numerator and denominator. This is not in the case here.

Hence there are no holes.