Question

let $f(x) = \frac{p(x)}{q(x)}$ be a rational function, where p and q are polynomials in standard form with no common factors other than 1. if the degree of p is less than the degree of q, how can you determine the horizontal asymptote of f, if any?

\bigcirc function of this type do not have a horizontal asymptote.

\bigcirc divide the degree of p by the degree of q to find the y - value of the asymptote.

\bigcirc divide the leading coefficient of p by the leading coefficient of q to find the y - value of the asymptote.

\bigcirc for this case, the horizontal asymptote is always $y = 0$

question 12 1 pts

what is the horizontal asymptote, if any, of each rational function? if there is no horizontal asymptote, write
one\. remember to write your answer as an equation. if needed, use \a/b\ as fraction notation.

$y = \frac{3x^2 - 8x - 1}{6x + 4}$

$y = \frac{3x^3 + 2x^2 + 4}{6x^3 - 9}$

$y = \frac{2x^2 + 4x}{x^3 + 2x^2 + 5x + 1}$

$y = \frac{2x - 10}{4x^2 + 8}$

$y = \frac{8x^2 - 7x + 9}{4x^2 - x + 1}$

$y = \frac{2x^3}{x^2 - 5x + 2}$

Explanation:

Step1: Recall Horizontal Asymptote Rules

For a rational function \( f(x)=\frac{p(x)}{q(x)} \) (where \( p(x) \) and \( q(x) \) are polynomials with no common factors):

• If \( \text{deg}(p) < \text{deg}(q) \), horizontal asymptote is \( y = 0 \).
• If \( \text{deg}(p)=\text{deg}(q) \), horizontal asymptote is \( y=\frac{\text{leading coefficient of } p}{\text{leading coefficient of } q} \).
• If \( \text{deg}(p) > \text{deg}(q) \), no horizontal asymptote (oblique/slant asymptote may exist).

Step2: Analyze Each Function

1. \( y=\frac{3x^{2}-8x - 1}{6x + 4} \)

• \( \text{deg}(p)=2 \), \( \text{deg}(q)=1 \). Since \( \text{deg}(p)>\text{deg}(q) \), no horizontal asymptote. So answer: none.

2. \( y=\frac{3x^{3}+2x^{2}+4}{6x^{3}-9} \)

• \( \text{deg}(p)=3 \), \( \text{deg}(q)=3 \) (equal degrees). Leading coefficient of \( p \) is \( 3 \), leading coefficient of \( q \) is \( 6 \). So \( y=\frac{3}{6}=\frac{1}{2} \).

3. \( y=\frac{2x^{2}+4x}{x^{3}+2x^{2}+5x + 1} \)

• \( \text{deg}(p)=2 \), \( \text{deg}(q)=3 \). Since \( \text{deg}(p)<\text{deg}(q) \), horizontal asymptote is \( y = 0 \).

4. \( y=\frac{2x - 10}{4x^{2}+8} \)

• \( \text{deg}(p)=1 \), \( \text{deg}(q)=2 \). Since \( \text{deg}(p)<\text{deg}(q) \), horizontal asymptote is \( y = 0 \).

5. \( y=\frac{8x^{2}-7x + 9}{4x^{2}-x + 1} \)

• \( \text{deg}(p)=2 \), \( \text{deg}(q)=2 \) (equal degrees). Leading coefficient of \( p \) is \( 8 \), leading coefficient of \( q \) is \( 4 \). So \( y=\frac{8}{4}=2 \).

6. \( y=\frac{2x^{3}}{x^{2}-5x + 2} \)

• \( \text{deg}(p)=3 \), \( \text{deg}(q)=2 \). Since \( \text{deg}(p)>\text{deg}(q) \), no horizontal asymptote. So answer: none.

Answer:

1. \( y = \text{none} \)
2. \( y=\frac{1}{2} \)
3. \( y = 0 \)
4. \( y = 0 \)
5. \( y = 2 \)
6. \( y = \text{none} \)