Chapter 9.5, Problem 58STP

To determine

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

Answer to Problem 58STP

The vertical asymptotes are $x=-4\text{ and }x=1$ and there is no holes in the function.

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{x+2}{x^2+3x-4}$

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

  $\begin{array}{l}{x}^2+3x-4=0\\ \Rightarrow x^2+\left(4-1\right)x-4=0\\ \Rightarrow x^2+4x-x-4=0\\ \Rightarrow x\left(x+4\right)-1\left(x+4\right)=0\\ \Rightarrow \left(x+4\right)\left(x-1\right)=0\\ \text{Either }\left(x+4\right)=0\text{ or }\left(x-1\right)=0\\ x=-4\text{ or }x=1\end{array}$

The two values of $x$ are $-4$ and 1

Therefore, the asymptotes are $x=-4\text{ and }x=1$

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

Hence there are not holes.