Question

rational function relay #4
$f(x) = \frac{x}{x^2 + 1}$
step 1: find all vertical asymptotes and/or holes and describe the function behavior around them.
step 2: evaluate $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$. write the equations of any horizontal or slant asymptotes, if applicable.
step 3: find all x- and y-intercepts.
step 4: use the work of your group members to graph $y = f(x)$. choose an appropriate scale.
(graph with x and y axes, grid lines)

Explanation:

Step 1: Vertical Asymptotes/Holes

To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 1 = 0\). Solving \(x^2 = -1\), there are no real solutions, so no vertical asymptotes. For holes, check if numerator and denominator have common factors. The numerator is \(x\), denominator is \(x^2 + 1\) (no common factors), so no holes.

Step 2: Limits and Horizontal Asymptotes

For \(\lim_{x \to \infty} \frac{x}{x^2 + 1}\), divide numerator and denominator by \(x^2\): \(\lim_{x \to \infty} \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = \frac{0}{1 + 0} = 0\). Similarly, \(\lim_{x \to -\infty} \frac{x}{x^2 + 1}\), divide by \(x^2\) (note \(x = -\sqrt{x^2}\) for \(x < 0\)): \(\lim_{x \to -\infty} \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = \frac{0}{1 + 0} = 0\). So horizontal asymptote is \(y = 0\).

Step 3: Intercepts

• x - intercept: Set \(f(x) = 0\), so \(\frac{x}{x^2 + 1} = 0\) implies \(x = 0\).
• y - intercept: Set \(x = 0\), \(f(0) = \frac{0}{0 + 1} = 0\).

Step 4: Graphing

The function is odd (since \(f(-x) = -f(x)\)), symmetric about the origin. It passes through (0,0), has horizontal asymptote \(y = 0\), and no vertical asymptotes. As \(x\) increases from 0 to \(\infty\), \(f(x)\) increases to 0; as \(x\) decreases from 0 to \(-\infty\), \(f(x)\) decreases to 0.

Answer:

• Step 1: No vertical asymptotes, no holes.
• Step 2: \(\lim_{x \to \infty} f(x) = 0\), \(\lim_{x \to -\infty} f(x) = 0\); horizontal asymptote \(y = 0\).
• Step 3: x - intercept (0,0), y - intercept (0,0).
• Step 4: Graph is symmetric about origin, passes through (0,0), approaches \(y = 0\) as \(x \to \pm\infty\).