Question

directions: determine if the following rational functions have a horizontal asymptote, slant asymptote, or neither. 1. $f(x)=\frac{2x^{2}-3x + 1}{x^{3}+1}$ 2. $g(x)=\frac{x^{2}+x - 5}{3x^{2}-2x + 1}$ 3. $h(x)=\frac{x^{3}+4x^{2}+5x - 6}{3x^{2}-7}$ 4. $k(x)=\frac{-2x^{4}+3x^{3}+8}{x^{2}-6x - 11}$ 5. $r(x)=\frac{(2x + 1)(3x - 4)}{(x + 3)(x - 1)}$ 6. $y=\frac{(x - 1)^{3}(2x + 3)^{2}}{(x - 2)^{2}(x + 4)^{3}}$ 7. $y=\frac{(x - 1)^{4}}{-3x(x - 8)^{2}}$ 8. $y=\frac{(x^{2}+1)^{2}}{5x^{2}(x + 2)}$ 9. $y=\frac{6}{x + 3}$

Explanation:

Step1: Recall asymptote rules

For a rational function $y = \frac{f(x)}{g(x)}$ where $f(x)=a_nx^n+\cdots+a_0$ and $g(x)=b_mx^m+\cdots + b_0$, if $n < m$, the horizontal asymptote is $y = 0$; if $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n=m + 1$, there is a slant asymptote; if $n>m+1$, there is neither a horizontal nor a slant asymptote.

Step2: Analyze $f(x)=\frac{2x^2 - 3x + 1}{x^3+1}$

Here $n = 2$ and $m = 3$. Since $n

Step3: Analyze $g(x)=\frac{x^2+x - 5}{3x^2-2x + 1}$

Here $n = 2$ and $m = 2$. So $y=\frac{1}{3}$ is the horizontal asymptote.

Step4: Analyze $h(x)=\frac{x^3+4x^2+5x - 6}{3x^2-7}$

Here $n = 3$ and $m = 2$, $n=m + 1$, so there is a slant asymptote.

Step5: Analyze $k(x)=\frac{-2x^4+3x^3 + 8}{x^2-6x-11}$

Here $n = 4$ and $m = 2$, $n>m + 1$, so there is neither a horizontal nor a slant asymptote.

Step6: Analyze $r(x)=\frac{(2x + 1)(3x - 4)}{(x + 3)(x - 1)}=\frac{6x^2-5x-4}{x^2+2x-3}$

Here $n = 2$ and $m = 2$, so $y = 6$ is the horizontal asymptote.

Step7: Analyze $y=\frac{(x - 1)^3(2x+3)^2}{(x - 2)^2(x + 4)^3}$

Expand the numerator and denominator to find the degrees. The degree of the numerator $n=3 + 2=5$ and the degree of the denominator $m=2+3 = 5$. After expanding and finding the leading - coefficients, we can find the horizontal asymptote.

Step8: Analyze $y=\frac{(x - 1)^4}{-3x(x - 8)^2}$

Expand to find $n = 4$ and $m=1 + 2=3$, $n>m+1$, so there is neither a horizontal nor a slant asymptote.

Step9: Analyze $y=\frac{(x^2+1)^2}{5x^2(x + 2)}$

Expand, $n = 4$ and $m=2 + 1=3$, $n>m+1$, so there is neither a horizontal nor a slant asymptote.

Step10: Analyze $y=\frac{6}{x + 3}$

Here $n = 0$ and $m = 1$, so $y = 0$ is the horizontal asymptote.

Answer:

1. Horizontal asymptote $y = 0$
2. Horizontal asymptote $y=\frac{1}{3}$
3. Slant asymptote
4. Neither
5. Horizontal asymptote $y = 6$
6. Horizontal asymptote (after expansion and coefficient - comparison)
7. Neither
8. Neither
9. Horizontal asymptote $y = 0$