Question

In Exercises \(27-34,\) (a) find the vertical asymptotes of the graph of \(f(x) .(\) b) Describe the behavior of \(f(x)\) to the left and right of each vertical asymptote. $$f(x)=\frac{x^{2}-2 x}{x+1}$$

Short answer

The vertical asymptote is at \(x = -1\). As \(x\) approaches -1 from the left, the function decreases without bound reaching towards \(-\infty\), and as \(x\) approaches -1 from the right, it increases without bound moving towards \(+\infty\).

Step-by-step solution

Identifying the Vertical Asymptote(s)

To find the vertical asymptotes of a rational function, one needs to set the denominator equal to zero and solve for \(x\). Here, this involves solving \(x+1=0\). This gives the solution \(x=-1\). So there is one vertical asymptote at \(x=-1\).

Describing the behavior of the function to the left and right of the vertical asymptote

To determine the behavior of the function as it approaches the vertical asymptote from the left and from the right, one needs to consider the limit of the function as \(x\) approaches the asymptote value from the left \(x \to -1^-\) and from the right \(x \to -1^+\). If we substitute values slightly less than -1 (e.g., -1.1, -1.01, -1.001), it can be observed that when \(x\) approaches -1 from the left, the function approaches \(-\infty\) meaning the function will decrease without bound. On the other hand, if we substitute values slightly more than -1 (e.g., -0.9, -0.99, -0.999), it is noticeable that when \(x\) approaches -1 from the right, the function approaches \(+\infty\) meaning the function will increase without bound.