Questions: Determine the following information given that f(x) = 5x / (16-x^2) Separate multiple answers by commas. If an asymptote or intercept does not exist, write none. Note that you may click a graph to enlarge it. (a) Horizontal Asymptote(s): y = (b) Hole(s): (Type an ordered pair.) (c) Vertical Asymptote(s): x = (d) x-intercept(s): (Type an ordered pair.) (e) y-intercept: (Type an ordered pair.)

Solution Steps

To solve the given problem, we need to analyze the function \( f(x) = \frac{5x}{16 - x^2} \) for its horizontal asymptotes, holes, vertical asymptotes, and intercepts.

1. Horizontal Asymptote: Determine the behavior of the function as \( x \) approaches infinity.
2. Holes: Identify any values of \( x \) that make both the numerator and denominator zero.
3. Vertical Asymptote: Find the values of \( x \) that make the denominator zero but not the numerator.
4. x-intercept: Find the value of \( x \) that makes the numerator zero.
5. y-intercept: Evaluate the function at \( x = 0 \).

To find the horizontal asymptote of the function \( f(x) = \frac{5x}{16 - x^2} \), we evaluate the limit as \( x \) approaches infinity: \[ \lim_{x \to \infty} f(x) = 0 \] Thus, the horizontal asymptote is \( y = 0 \).

A hole occurs in the function where both the numerator and denominator are zero. In this case, the numerator \( 5x \) is zero when \( x = 0 \), but the denominator \( 16 - x^2 \) is not zero at this point. Therefore, there are no holes in the function: \[ \text{Holes: none} \]

Vertical asymptotes occur where the denominator is zero and the numerator is not zero. The denominator \( 16 - x^2 = 0 \) gives: \[ x^2 = 16 \implies x = \pm 4 \] Thus, the vertical asymptotes are at \( x = 4 \) and \( x = -4 \).

The x-intercept is found by setting the numerator equal to zero: \[ 5x = 0 \implies x = 0 \] Thus, the x-intercept is \( (0, 0) \).

The y-intercept is found by evaluating the function at \( x = 0 \): \[ f(0) = \frac{5 \cdot 0}{16 - 0^2} = 0 \] Thus, the y-intercept is \( (0, 0) \).

Final Answer