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. $$f(x)=\frac{x^{2}-1}{2 x+4}$$

Short answer

The function \(f(x) = \frac{{x^{2} - 1}}{{2x + 4}}\) has one vertical asymptote at \(x = -2\). As \(x\) approaches -2 from the left (x < -2), the function goes to negative infinity, and as \(x\) approaches -2 from the right (x > -2), the function goes to positive infinity.

Step-by-step solution

Identify when the denominator equals zero

A vertical asymptote occurs when the denominator of the rational function equals zero. Hence solve for \(x\) from the equation \(2x + 4 = 0\). Subtract 4 from both sides and then divide by 2 to isolate \(x\).

Calculate the values of x

After simplifying the equation from step 1, \(x = -2\) is obtained. Thus, \(x = -2\) is the vertical asymptote for the function \(f(x) = \frac{{x^{2} - 1}}{{2x + 4}}\).

Behavior of the function on the left and right of the asymptote

To describe the behavior of the function around the vertical asymptote, substiture values that are slightly above and slightly below the asymptote into the function. When \(x\) is slightly less than -2, \(x= -2.01\) for instance, \(f(x)\) becomes a very large negative number. This means that as \(x\) approaches -2 from the left, \(f(x)\) goes to negative infinity. Conversely, when \(x\) is slightly more than -2, \(x = -1.99\) for instance, \(f(x)\) becomes a very large positive number. This translates that as \(x\) approaches -2 from the right, \(f(x)\) goes to positive infinity.