Question

select the correct choices that completes the sentence below. if the graph of a function f approaches b as x increases or decreases without bound, then the line y = b is of the graph of f. the equation of such a line for the graph of $y = \frac{x - 10}{3x^2 + x + 1}$ is the equation of such a line for the graph of $y = \frac{x^2 - 10}{3x^2 + x + 1}$ is

Explanation:

Step1: Recall Horizontal Asymptote Definition

A horizontal asymptote of a function \( y = f(x) \) is a horizontal line \( y = b \) such that \( \lim_{x \to \infty} f(x) = b \) or \( \lim_{x \to -\infty} f(x) = b \). So the first blank is "a horizontal asymptote".

Step2: Find Horizontal Asymptote for \( y=\frac{x - 10}{3x^2+x + 1} \)

For rational functions \( \frac{N(x)}{D(x)} \), if the degree of \( N(x) \) (denoted \( \deg N \)) is less than the degree of \( D(x) \) (denoted \( \deg D \)), then \( \lim_{x \to \pm\infty} \frac{N(x)}{D(x)} = 0 \). Here, \( \deg N = 1 \), \( \deg D = 2 \), so \( \lim_{x \to \pm\infty} \frac{x - 10}{3x^2+x + 1}=0 \). Thus, the horizontal asymptote is \( y = 0 \).

Step3: Find Horizontal Asymptote for \( y=\frac{x^2 - 10}{3x^2+x + 1} \)

If \( \deg N=\deg D \), then \( \lim_{x \to \pm\infty} \frac{N(x)}{D(x)}=\frac{\text{leading coefficient of } N}{\text{leading coefficient of } D} \). Here, leading coefficient of \( N \) is \( 1 \), leading coefficient of \( D \) is \( 3 \), so \( \lim_{x \to \pm\infty} \frac{x^2 - 10}{3x^2+x + 1}=\frac{1}{3} \). Thus, the horizontal asymptote is \( y=\frac{1}{3} \).

Answer:

First blank: a horizontal asymptote; Second blank: \( y = 0 \); Third blank: \( y=\frac{1}{3} \)