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