Question

review a: (topics 1.7 - 1.11) rational functions name directions: for each of the following rational functions, write limit statements to describe the left and right end behaviors. graph of f(x) graph of g(x) left: right: left: right: h(x)=\frac{2x^{2}-2x + 1}{3x^{2}+5x + 7} 4. k(x)=\frac{2x(x - 3)}{(x + 2)^{2}(x - 1)} left: right: left: right: 5. r(x)=\frac{-2x^{4}+3x^{2}+x - 1}{5x^{2}+2x + 3} 6. m(x)=\frac{3(x - 1)^{2}(x + 5)}{(2x + 3)^{2}} left: right: left: right:

Explanation:

Step1: Recall end - behavior rules for rational functions

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 $n = m$, $\lim_{x
ightarrow\pm\infty}y=\frac{a_n}{b_m}$; if $n<m$, $\lim_{x
ightarrow\pm\infty}y = 0$; if $n>m$ and $n - m$ is odd, $\lim_{x
ightarrow\pm\infty}y=\pm\infty$; if $n>m$ and $n - m$ is even, $\lim_{x
ightarrow\pm\infty}y=\infty$ or $-\infty$ depending on the signs of $a_n$ and $b_m$.

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

Since the degree of the numerator $n = 2$ and the degree of the denominator $m = 2$, and $a_n = 2$, $b_m=3$. Then $\lim_{x
ightarrow-\infty}h(x)=\lim_{x
ightarrow\infty}h(x)=\frac{2}{3}$.
Left: $\lim_{x
ightarrow-\infty}h(x)=\frac{2}{3}$
Right: $\lim_{x
ightarrow\infty}h(x)=\frac{2}{3}$

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

The degree of the numerator $n=2$ and the degree of the denominator $m = 3$. So $\lim_{x
ightarrow-\infty}k(x)=\lim_{x
ightarrow\infty}k(x)=0$.
Left: $\lim_{x
ightarrow-\infty}k(x)=0$
Right: $\lim_{x
ightarrow\infty}k(x)=0$

Step4: Analyze $r(x)=\frac{-2x^4+3x^2+x - 1}{5x^2+2x + 3}$

The degree of the numerator $n = 4$ and the degree of the denominator $m = 2$. Since $n>m$ and $n - m=2$ (even), and $a_n=-2$, $b_m = 5$. Then $\lim_{x
ightarrow-\infty}r(x)=-\infty$ and $\lim_{x
ightarrow\infty}r(x)=-\infty$.
Left: $\lim_{x
ightarrow-\infty}r(x)=-\infty$
Right: $\lim_{x
ightarrow\infty}r(x)=-\infty$

Step5: Analyze $m(x)=\frac{3(x - 1)^2(x + 5)}{(2x+3)^2}$

The degree of the numerator $n = 3$ and the degree of the denominator $m = 2$. Since $n>m$ and $n - m = 1$ (odd), and the leading - coefficient of the numerator is positive and the leading - coefficient of the denominator is positive. Then $\lim_{x
ightarrow-\infty}m(x)=-\infty$ and $\lim_{x
ightarrow\infty}m(x)=\infty$.
Left: $\lim_{x
ightarrow-\infty}m(x)=-\infty$
Right: $\lim_{x
ightarrow\infty}m(x)=\infty$

Answer:

1. For $h(x)$:

• Left: $\lim_{x

ightarrow-\infty}h(x)=\frac{2}{3}$

• Right: $\lim_{x

ightarrow\infty}h(x)=\frac{2}{3}$

1. For $k(x)$:

• Left: $\lim_{x

ightarrow-\infty}k(x)=0$

• Right: $\lim_{x

ightarrow\infty}k(x)=0$

1. For $r(x)$:

• Left: $\lim_{x

ightarrow-\infty}r(x)=-\infty$

• Right: $\lim_{x

ightarrow\infty}r(x)=-\infty$

1. For $m(x)$:

• Left: $\lim_{x

ightarrow-\infty}m(x)=-\infty$

• Right: $\lim_{x

ightarrow\infty}m(x)=\infty$