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1.7 rational functions and end behavior worksheet directions: for each …

Question

1.7 rational functions and end behavior worksheet
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. $f(x)=\frac{3x^{2}-1}{2x^{2}+5x + 7}$

horizontal asymptote: y or n
if yes, equation:

  1. $g(x)=\frac{x^{3}+2x^{2}+x + 4}{5x^{2}+7x + 8}$

horizontal asymptote: y or n
if yes, equation:

  1. $h(x)=\frac{5x^{3}-2x^{2}-1}{x^{4}-6}$

horizontal asymptote: y or n
if yes, equation:

  1. $k(x)=\frac{6x^{3}+2x + 3}{2x^{2}-11x + 4}$

horizontal asymptote: y or n
if yes, equation:

  1. $r(x)=\frac{(2x - 1)(5x + 6)}{(x + 3)(x - 6)}$

horizontal asymptote: y or n
if yes, equation:

  1. $q(x)=\frac{(x^{2}-3)^{2}}{3x^{3}+4x^{2}+7}$

horizontal asymptote: y or n
if yes, equation:

  1. $p(x)=\frac{(3x - 1)^{2}}{2x^{2}+3x + 5}$

horizontal asymptote: y or n
if yes, equation:

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

horizontal asymptote: y or n
if yes, equation:

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

horizontal asymptote: y or n
if yes, equation:
directions: write limit statements for the end - behavior of the following rational functions.

  1. $y=\frac{2x^{3}-5x + 6}{6x^{3}+10x^{2}-4x - 12}$

left:
right:

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

left:
right:

  1. $y=\frac{x^{2}}{(x - 1)^{3}}$

left:
right:

Explanation:

Step1: Recall horizontal - asymptote rules

For a rational function $f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, where $a_n
eq0$ and $b_m
eq0$:

  1. If $n < m$, the horizontal asymptote is $y = 0$.
  2. If $n=m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$.
  3. If $n>m$, there is no horizontal asymptote.
1. For $f(x)=\frac{3x^2 - 1}{2x^2+5x + 7}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$.

Step2: Find the horizontal - asymptote

Since $n=m$, the horizontal asymptote is $y=\frac{3}{2}$.

2. For $g(x)=\frac{x^3+2x^2+x + 4}{5x^2+7x + 8}$

Step1: Identify degrees of numerator and denominator

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

3. For $h(x)=\frac{5x^3-2x^2 - 1}{x^4-6}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 3$ and the degree of the denominator $m = 4$. Since $n

4. For $k(x)=\frac{6x^3+2x + 3}{2x^2-11x + 4}$

Step1: Identify degrees of numerator and denominator

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

5. For $r(x)=\frac{(2x - 1)(5x + 6)}{(x + 3)(x - 6)}=\frac{10x^2+7x-6}{x^2-3x - 18}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$.

Step2: Find the horizontal - asymptote

Since $n=m$, the horizontal asymptote is $y = 10$.

6. For $q(x)=\frac{(x^2-3)^2}{3x^3+4x^2+7}=\frac{x^4-6x^2 + 9}{3x^3+4x^2+7}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 4$ and the degree of the denominator $m = 3$. Since $n>m$, there is no horizontal asymptote.

7. For $p(x)=\frac{(3x - 1)^2}{2x^2+3x + 5}=\frac{9x^2-6x + 1}{2x^2+3x + 5}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$.

Step2: Find the horizontal - asymptote

Since $n=m$, the horizontal asymptote is $y=\frac{9}{2}$.

8. For $y=\frac{(x - 2)(4 - x)}{(x + 3)^2}=\frac{-x^2+6x - 8}{x^2+6x + 9}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$.

Step2: Find the horizontal - asymptote

Since $n=m$, the horizontal asymptote is $y=-1$.

9. For $s(x)=\frac{(2x^2+3)^2(x - 4)}{(x^2+5)(x - 2)}=\frac{(4x^4 + 12x^2+9)(x - 4)}{x^3-2x^2+5x - 10}=\frac{4x^5-16x^4+12x^3-48x^2+9x - 36}{x^3-2x^2+5x - 10}$

Step1: Identify degrees of numerator and denominator

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

10. For $y=\frac{2x^3-5x + 6}{6x^3+10x^2-4x - 12}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 3$ and the degree of the denominator $m = 3$.

Step2: Find the horizontal - asymptote

Since $n=m$, the horizontal asymptote is $y=\frac{2}{6}=\frac{1}{3}$.
Left - hand limit: $\lim_{x
ightarrow-\infty}\frac{2x^3-5x + 6}{6x^3+10x^2-4x - 12}=\frac{1}{3}$
Right - hand limit: $\lim_{x
ightarrow+\infty}\frac{2x^3-5x + 6}{6x^3+10x^2-4x - 12}=\frac{1}{3}$

11. For $y=\frac{(4x + 3)^2}{(3x - 1)(2x + 5)}=\frac{16x^2+24x+9}{6x^2+13x - 5}$

Step1: Identify degrees of numerator and denominator

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$.

Step2: Find the horizo…

Answer:

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

ightarrow-\infty}\frac{2x^3-5x + 6}{6x^3+10x^2-4x - 12}=\frac{1}{3}$, Right: $\lim_{x
ightarrow+\infty}\frac{2x^3-5x + 6}{6x^3+10x^2-4x - 12}=\frac{1}{3}$

  1. Left: $\lim_{x

ightarrow-\infty}\frac{(4x + 3)^2}{(3x - 1)(2x + 5)}=\frac{8}{3}$, Right: $\lim_{x
ightarrow+\infty}\frac{(4x + 3)^2}{(3x - 1)(2x + 5)}=\frac{8}{3}$

  1. Left: $\lim_{x

ightarrow-\infty}\frac{x^2}{(x - 1)^3}=0$, Right: $\lim_{x
ightarrow+\infty}\frac{x^2}{(x - 1)^3}=0$