Question: Graph the following function. Determine the asymptotes, if any. List the domain and the…

Graph the following function. Determine the asymptotes, if any. List the domain and the x-intercepts and the y-intercepts. Check your work using a graphing calculator.

\[ f(x) = \frac{5}{x^2 + 3} \]

B. The function has two vertical asymptotes. The leftmost asymptote is \(\boxed{\text{ }}\), and the rightmost asymptote is \(\boxed{\text{ }}\). (Type equations.)

C. The function has no vertical asymptotes.

Find any horizontal asymptotes. Select the correct choice below, and if necessary, fill in the answer box(es) to complete your choice.

A. The function has one horizontal asymptote, \(\boxed{\text{ }}\). (Type an equation.)

B. The function has two horizontal asymptotes. The top asymptote is \(\boxed{\text{ }}\), and the bottom asymptote is \(\boxed{\text{ }}\). (Type equations.)

C. The function has no horizontal asymptotes.

Solution

Step 1: Identify the function. The given function is: \[ f(x) = \frac{5}{x^2 + 3} \] Step 2: Determine the domain. The function is defined for all real numbers except where the denominator is zero. \[ x^2 + 3 \neq 0 \] Since \(x^2 + 3\) is always positive, there are no restrictions. \[ \text{Domain: } (-\infty, \infty) \] Step 3: Determine the vertical asymptotes. Vertical asymptotes occur where the denominator is zero. \[ x^2 + 3 = 0 \] \[ x^2 = -3 \] There are no real solutions, so there are no vertical asymptotes. \[ \text{Vertical Asymptotes: None} \] Step 4: Determine the horizontal asymptote. For rational functions, the horizontal asymptote is determined by the degrees of the numerator and the denominator. \[ \text{Degree of numerator} = 0 \\ \text{Degree of denominator} = 2 \] Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\). \[ \text{Horizontal Asymptote: } y = 0 \] Step 5: Find the y-intercept. To find the y-intercept, set \(x = 0\). \[ f(0) = \frac{5}{0^2 + 3} = \frac{5}{3} \] \[ \text{y-intercept: } \left(0, \frac{5}{3}\right) \] Step 6: Find the x-intercepts. To find the x-intercepts, set \(f(x) = 0\). \[ \frac{5}{x^2 + 3} = 0 \] Since the numerator is never zero, there are no x-intercepts. \[ \text{x-intercepts: None} \] Step 7: Graph the function. The graph of \(f(x) = \frac{5}{x^2 + 3}\) is a horizontal curve approaching the horizontal asymptote \(y = 0\) with a y-intercept at \(\left(0, \frac{5}{3}\right)\). There are no vertical or x-intercepts. You can verify the graph using a graphing calculator.