problems 13 - 16, describe the end behavior for the graph of the rational functions. analyze using limit…

Explanation:

Step1: Recall end - behavior using limits

To find the end - behavior of a rational function $y = \frac{f(x)}{g(x)}$, we consider $\lim_{x\rightarrow\pm\infty}\frac{f(x)}{g(x)}$. 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 $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{x^{3}+6x}{1 - \frac{1}{x}}=\infty$ since the numerator $x^{3}+6x\rightarrow\infty$ and the denominator $1-\frac{1}{x}\rightarrow1$. As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{x^{3}+6x}{1-\frac{1}{x}}=-\infty$ since the numerator $x^{3}+6x\rightarrow-\infty$ and the denominator $1-\frac{1}{x}\rightarrow1$.

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 $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{\frac{8}{x}-x^{4}}{\frac{3}{x^{2}}+1}=-\infty$ since the numerator $\frac{8}{x}-x^{4}\rightarrow-\infty$ and the denominator $\frac{3}{x^{2}}+1\rightarrow1$. As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{\frac{8}{x}-x^{4}}{\frac{3}{x^{2}}+1}=-\infty$ since the numerator $\frac{8}{x}-x^{4}\rightarrow-\infty$ and the denominator $\frac{3}{x^{2}}+1\rightarrow1$.

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 $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{4x^{3}+\frac{3}{x}}{\frac{7}{x^{2}}-1}=\infty$ since the numerator $4x^{3}+\frac{3}{x}\rightarrow\infty$ and the denominator $\frac{7}{x^{2}}-1\rightarrow - 1$. As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{4x^{3}+\frac{3}{x}}{\frac{7}{x^{2}}-1}=-\infty$ since the numerator $4x^{3}+\frac{3}{x}\rightarrow-\infty$ and the denominator $\frac{7}{x^{2}}-1\rightarrow - 1$.

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 $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{2x + x^{2}}{3-\frac{5}{x^{2}}}=\infty$ since the numerator $2x + x^{2}\rightarrow\infty$ and the denominator $3-\frac{5}{x^{2}}\rightarrow3$. As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{2x + x^{2}}{3-\frac{5}{x^{2}}}=\infty$ since the numerator $2x + x^{2}\rightarrow\infty$ and the denominator $3-\frac{5}{x^{2}}\rightarrow3$.

Answer:

13. $\lim_{x\rightarrow\infty}f(x)=\infty$, $\lim_{x\rightarrow-\infty}f(x)=-\infty$
14. $\lim_{x\rightarrow\infty}g(x)=-\infty$, $\lim_{x\rightarrow-\infty}g(x)=-\infty$
15. $\lim_{x\rightarrow\infty}h(x)=\infty$, $\lim_{x\rightarrow-\infty}h(x)=-\infty$
16. $\lim_{x\rightarrow\infty}k(x)=\infty$, $\lim_{x\rightarrow-\infty}k(x)=\infty$