Question

problems 13 - 16, describe the end behavior for the graph of the rational functions. analyze using limit notation to explain your answer. 13. $f(x)=\frac{x^{4}+6x^{2}}{x - 1}$ 14. $g(x)=\frac{8x - x^{6}}{3 + x^{2}}$ 15. $h(x)=\frac{4x^{5}+3x}{7 - x^{2}}$ 16. $k(x)=\frac{2x^{3}+x^{4}}{3x^{2}-5}$

Explanation:

Step1: Recall end - behavior using limits

To find the end - behavior of a rational function $y = \frac{f(x)}{g(x)}$, we consider $\lim_{x
ightarrow\pm\infty}\frac{f(x)}{g(x)}$. We divide both the numerator and denominator by the highest - power of $x$ in the denominator.

Step2: Analyze $f(x)=\frac{x^{4}+6x^{2}}{x - 1}$

The highest power of $x$ in the denominator is $x^1$. Divide numerator and denominator by $x$: $f(x)=\frac{x^{3}+6x}{1-\frac{1}{x}}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{x^{3}+6x}{1 - \frac{1}{x}}=\infty$ since the numerator $x^{3}+6x
ightarrow\infty$ and the denominator $1-\frac{1}{x}
ightarrow1$.
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{x^{3}+6x}{1-\frac{1}{x}}=-\infty$ since the numerator $x^{3}+6x
ightarrow-\infty$ and the denominator $1-\frac{1}{x}
ightarrow1$.

Step3: Analyze $g(x)=\frac{8x - x^{6}}{3 + x^{2}}$

The highest power of $x$ in the denominator is $x^{2}$. Divide numerator and denominator by $x^{2}$: $g(x)=\frac{\frac{8}{x}-x^{4}}{\frac{3}{x^{2}}+1}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{\frac{8}{x}-x^{4}}{\frac{3}{x^{2}}+1}=-\infty$ since the numerator $\frac{8}{x}-x^{4}
ightarrow-\infty$ and the denominator $\frac{3}{x^{2}}+1
ightarrow1$.
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{\frac{8}{x}-x^{4}}{\frac{3}{x^{2}}+1}=-\infty$ since the numerator $\frac{8}{x}-x^{4}
ightarrow-\infty$ and the denominator $\frac{3}{x^{2}}+1
ightarrow1$.

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

The highest power of $x$ in the denominator is $x^{2}$. Divide numerator and denominator by $x^{2}$: $h(x)=\frac{4x^{3}+\frac{3}{x}}{\frac{7}{x^{2}}-1}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{4x^{3}+\frac{3}{x}}{\frac{7}{x^{2}}-1}=\infty$ since the numerator $4x^{3}+\frac{3}{x}
ightarrow\infty$ and the denominator $\frac{7}{x^{2}}-1
ightarrow - 1$.
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{4x^{3}+\frac{3}{x}}{\frac{7}{x^{2}}-1}=-\infty$ since the numerator $4x^{3}+\frac{3}{x}
ightarrow-\infty$ and the denominator $\frac{7}{x^{2}}-1
ightarrow - 1$.

Step5: Analyze $k(x)=\frac{2x^{3}+x^{4}}{3x^{2}-5}$

The highest power of $x$ in the denominator is $x^{2}$. Divide numerator and denominator by $x^{2}$: $k(x)=\frac{2x + x^{2}}{\frac{3x^{2}}{x^{2}}-\frac{5}{x^{2}}}=\frac{2x + x^{2}}{3-\frac{5}{x^{2}}}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{2x + x^{2}}{3-\frac{5}{x^{2}}}=\infty$ since the numerator $2x + x^{2}
ightarrow\infty$ and the denominator $3-\frac{5}{x^{2}}
ightarrow3$.
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{2x + x^{2}}{3-\frac{5}{x^{2}}}=\infty$ since the numerator $2x + x^{2}
ightarrow\infty$ and the denominator $3-\frac{5}{x^{2}}
ightarrow3$.

Answer:

1. $\lim_{x

ightarrow\infty}f(x)=\infty$, $\lim_{x
ightarrow-\infty}f(x)=-\infty$

1. $\lim_{x

ightarrow\infty}g(x)=-\infty$, $\lim_{x
ightarrow-\infty}g(x)=-\infty$

1. $\lim_{x

ightarrow\infty}h(x)=\infty$, $\lim_{x
ightarrow-\infty}h(x)=-\infty$

1. $\lim_{x

ightarrow\infty}k(x)=\infty$, $\lim_{x
ightarrow-\infty}k(x)=\infty$