Question

worksheet a: (topic 1.7) rational functions and end behavior
name:
directions: for each of the following, determine if the given rational function has a horizontal asymptote. if it does, write the equation of the horizontal asymptote.

1. (f(x)=\frac{3x - 1}{2x^{2}+5x + 7}) horizontal asymptote: y or n if yes, equation:
2. (g(x)=\frac{x^{2}+x + 4}{5x^{2}+7x + 8}) horizontal asymptote: y or n if yes, equation:
3. (h(x)=\frac{5x^{2}-2x - 1}{x - 6}) horizontal asymptote: y or n if yes, equation:
4. (k(x)=\frac{6x^{2}+2x + 3}{2x^{2}-11x + 4}) horizontal asymptote: y or n if yes, equation:
5. (r(x)=\frac{(2x - 1)(5x + 6)}{(x + 3)(x - 6)}) horizontal asymptote: y or n if yes, equation:
6. (q(x)=\frac{x^{3}-3}{3x^{3}+4x^{2}+7}) horizontal asymptote: y or n if yes, equation:
7. (p(x)=\frac{(3x - 1)^{2}}{2x^{2}+5x + 5}) horizontal asymptote: y or n if yes, equation:
8. (y=\frac{(x - 2)(4 - x)}{(x + 3)^{2}}) horizontal asymptote: y or n if yes, equation:
9. (s(x)=\frac{(2x^{2}+3)^{2}(x - 4)}{(x^{2}+5)(x - 2)}) horizontal asymptote: y or n if yes, equation:
10. (y=\frac{2x - 5x + 6}{6x^{4}+10x^{2}-4x - 12}) directions: write limit statements for the end behavior of the following rational functions.
11. (y=\frac{(4x + 3)^{2}}{(3x - 1)(2x + 5)}) left: right:
12. (y=\frac{x^{3}}{(x - 1)^{3}}) left: right:

created by bryan passwater
rational functions and end behavior

Explanation:

Step1: Recall rules for horizontal asymptotes

For a rational function $f(x)=\frac{a_nx^n + \cdots+a_0}{b_mx^m+\cdots + b_0}$, if $n < m$, the horizontal asymptote is $y = 0$; if $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n>m$, there is no horizontal asymptote.

Step2: Analyze function 1: $f(x)=\frac{3x - 1}{2x^2+5x + 7}$

Here $n = 1$ (degree of numerator) and $m=2$ (degree of denominator). Since $n

Step3: Analyze function 2: $g(x)=\frac{x^2+2x + 4}{5x^2+7x + 8}$

Here $n = 2$ and $m = 2$. The leading - coefficient of the numerator $a_n=1$ and the leading - coefficient of the denominator $b_m = 5$. So the horizontal asymptote is $y=\frac{1}{5}$.

Step4: Analyze function 3: $h(x)=\frac{5x^3-2x^2 - 1}{x^2-6}$

Here $n = 3$ and $m = 2$. Since $n>m$, there is no horizontal asymptote.

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

Here $n = 2$ and $m = 2$. The leading - coefficient of the numerator $a_n=6$ and the leading - coefficient of the denominator $b_m = 2$. So the horizontal asymptote is $y = 3$.

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

Expand the numerator: $(2x - 1)(5x+6)=10x^2+12x-5x - 6=10x^2 + 7x-6$. Expand the denominator: $(x + 3)(x - 6)=x^2-6x+3x - 18=x^2-3x - 18$. Here $n = 2$ and $m = 2$, and $\frac{a_n}{b_m}=\frac{10}{1}=10$. So the horizontal asymptote is $y = 10$.

Step7: Analyze function 6: $q(x)=\frac{x^3-3}{3x^3+4x^2+7}$

Here $n = 3$ and $m = 3$. The leading - coefficient of the numerator $a_n=1$ and the leading - coefficient of the denominator $b_m = 3$. So the horizontal asymptote is $y=\frac{1}{3}$.

Step8: Analyze function 7: $p(x)=\frac{(3x - 1)^2}{2x^2+3x + 5}$

Expand the numerator: $(3x - 1)^2=9x^2-6x + 1$. Here $n = 2$ and $m = 2$, and $\frac{a_n}{b_m}=\frac{9}{2}$. So the horizontal asymptote is $y=\frac{9}{2}$.

Step9: Analyze function 8: $y=\frac{(x - 2)(4 - x)}{(x + 3)^2}$

Expand the numerator: $(x - 2)(4 - x)=4x-x^2-8 + 2x=-x^2+6x - 8$. Expand the denominator: $(x + 3)^2=x^2+6x + 9$. Here $n = 2$ and $m = 2$, and $\frac{a_n}{b_m}=-1$. So the horizontal asymptote is $y=-1$.

Step10: Analyze function 9: $s(x)=\frac{(2x^2+3)^2(x - 4)}{(x^2+5)(x - 2)}$

The degree of the numerator $n=2\times2 + 1=5$ and the degree of the denominator $m=2 + 1=3$. Since $n>m$, there is no horizontal asymptote.

Step11: Analyze function 10: $y=\frac{2x^3-5x + 6}{6x^4+10x^2-4x - 12}$

Here $n = 3$ and $m = 4$. Since $n

Step12: Analyze function 11: $y=\frac{(4x + 3)^2}{(3x - 1)(2x+5)}$

Expand the numerator: $(4x + 3)^2=16x^2+24x + 9$. Expand the denominator: $(3x - 1)(2x+5)=6x^2+15x-2x - 5=6x^2+13x - 5$. Here $n = 2$ and $m = 2$, and $\frac{a_n}{b_m}=\frac{16}{6}=\frac{8}{3}$. So the horizontal asymptote is $y=\frac{8}{3}$.

Step13: Analyze function 12: $y=\frac{x^3}{(x - 1)^3}$

Here $n = 3$ and $m = 3$. The leading - coefficient of the numerator $a_n=1$ and the leading - coefficient of the denominator $b_m = 1$. So the horizontal asymptote is $y = 1$.

Answer:

1. Horizontal Asymptote: Y, Equation: $y = 0$
2. Horizontal Asymptote: Y, Equation: $y=\frac{1}{5}$
3. Horizontal Asymptote: N
4. Horizontal Asymptote: Y, Equation: $y = 3$
5. Horizontal Asymptote: Y, Equation: $y = 10$
6. Horizontal Asymptote: Y, Equation: $y=\frac{1}{3}$
7. Horizontal Asymptote: Y, Equation: $y=\frac{9}{2}$
8. Horizontal Asymptote: Y, Equation: $y=-1$
9. Horizontal Asymptote: N
10. Horizontal Asymptote: Y, Equation: $y = 0$
11. Horizontal Asymptote: Y, Equation: $y=\frac{8}{3}$
12. Horizontal Asymptote: Y, Equation: $y = 1$