Question

Sketch the graph of the function in Problem 6. Label all intercepts, vertical asymptotes, horizontal asymptotes, and oblique asymptotes.

Short answer

The graph is

Step-by-step solution

Step 1. Given Information

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

Step 2. Calculation

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 $\left\{x|x\ne -1\right\}$

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$(0,-3)$

Step 3. Calculation

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.

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

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.

Step 4. Calculation

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.

Step5. Graph

Construct the table with four intervals,

Plot the points and lines on the graph.