# Consider the rational functions: r(x)=(2x−1)/((x^2)−x−2) , s(x)=((x^3)+27)/((x^2)+4) , t(x)=((x^3)−9x)/(x+2)

Alyce Wilkinson 2021-09-11 Answered

Consider the following rational functions: $$\displaystyle{r}{\left({x}\right)}=\frac{{{2}{x}−{1}}}{{{\left({x}^{{2}}\right)}−{x}−{2}}}$$
$$\displaystyle{s}{\left({x}\right)}=\frac{{{\left({x}^{{3}}\right)}+{27}}}{{{\left({x}^{{2}}\right)}+{4}}}$$
$$\displaystyle{t}{\left({x}\right)}=\frac{{{\left({x}^{{3}}\right)}−{9}{x}}}{{{x}+{2}}}$$
$$\displaystyle{u}{\left({x}\right)}=\frac{{{\left({x}^{{2}}\right)}+{x}−{6}}}{{{\left({x}^{{2}}\right)}−{25}}}$$
$$\displaystyle{w}{\left({x}\right)}=\frac{{{\left({x}^{{3}}\right)}+{\left({6}{x}^{{2}}\right)}+{9}{x}}}{{{x}+{3}}}$$
Which of these rational functions has a horizontal asymptote?

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Ezra Herbert
$$\displaystyle{r}{\left({x}\right)}=\frac{{{2}{x}−{1}}}{{{\left({x}^{{2}}\right)}−{x}−{2}}}=\frac{{{2}{x}-{1}}}{{{\left({x}-{2}\right)}{\left({x}+{1}\right)}}}$$
$$\displaystyle{s}{\left({x}\right)}=\frac{{{\left({x}^{{3}}\right)}+{27}}}{{{\left({x}^{{2}}\right)}+{4}}}=\frac{{{\left({x}+{3}\right)}{\left({x}+{9}+{3}{x}\right)}}}{{{\left({x}^{{2}}\right)}+{4}}}$$
$$\displaystyle{t}{\left({x}\right)}=\frac{{{\left({x}^{{3}}\right)}−{9}{x}}}{{{x}+{2}}}=\frac{{{x}{\left({x}-{3}\right)}{\left({x}+{3}\right)}}}{{{x}+{2}}}$$
$$\displaystyle{u}{\left({x}\right)}=\frac{{{\left({x}^{{2}}\right)}+{x}−{6}}}{{{\left({x}^{{2}}\right)}−{25}}}=\frac{{{\left({x}-{2}\right)}{\left({x}+{3}\right)}}}{{{\left({x}-{5}\right)}{\left({x}+{5}\right)}}}$$
$$\displaystyle{w}{\left({x}\right)}=\frac{{{\left({x}^{{3}}\right)}+{\left({6}{x}^{{2}}\right)}+{9}{x}}}{{{x}+{3}}}={x}{\left({x}+{3}\right)}$$
r,t and u have horizontal asymptotes