Question

Find the domain of each function. Find any horizontal, vertical, or oblique asymptotes.

$r\left(x\right)=\frac{x^2+2x-3}{x+1}$

Short answer

The domain of the function is $\left\{x|x\ne 1\right\}$, vertical asymptote is $x=-1$ and oblique asymptote is at$y=x-1$

Step-by-step solution

Step 1. Given Information

The given function is$r\left(x\right)=\frac{x^2+2x-3}{x+1}$

Step 2. Calculation 

Substitute the denominator expression equal to 0 to find the domain.

$$\begin{gathered} x+1=0 \\ x=-1 \end{gathered}$$

Thus, the domain of the function is$\left\{x|x\ne -1\right\}$

Step 3. Calculation 

The zeroes of denominator $x+1$is $-1$

Thus, line $x=-1$is the vertical asymptote.

The rational function is improper since degree of numerator is greater than the degree of denominator.

We will be using the long division method to express the function as $P\left(x\right)=f\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)}$

Thus, $P\left(x\right)=x-1+\frac{-2}{x+1}$

As $x\to \infty orx\to -\infty$,

$\frac{-2}{x+1}\approx \frac{-2}{x}\to 0$

Thus,$P\left(x\right)\to x-1$. The oblique asymptote is at$y=x-1$