91Ó°ÊÓ

The given function is$\frac{x^2+2x-3}{x+1}$

The denominator of the rational function cannot be factored further. So, factor the numerator as follows,

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

The domain of the function is $\left\{x|x≠-1\right\}$

Since 0 is the domain, find the y-intercept.

$$\begin{gathered} r\left(0\right)=\frac{(0+3)(0-1)}{(0+1)} \\ =-3 \end{gathered}$$

Thus, the y-intercept is$(0,-3)$

The real zero of the numerator that satisfies the equation $x+3=0andx-1=0$. Hence, the x-intercepts are $-3and1$

Determine the behavior of the graph near the x-intercept.

First substitute $x=-3$in the denominator.

$$\begin{gathered} r\left(x\right)=\frac{(x+3)(x-1)}{(x+1)} \\ =\frac{(x+3)(x-1)}{-3+1} \\ =\frac{(x+3)(x-1)}{-2} \\ ≈-\frac{1}{2}(x+3)(x-1) \end{gathered}$$

Now, substitute $x=1$in the denominator.

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

The line$x=1$represents the vertical asymptote.

The degree of numerator is greater than the degree of denominator. So the function is improper. Use the long division to find the horizontal asymptote.

Thus, $r\left(x\right)=x+1-\frac{4}{x+1}$

Thus, line $y=x+1$is an oblique asymptote of the graph.

The graph does not intersect the horizontal asymptote at any point.

Construct the table with four intervals,

Plot the points and lines on the graph.