Identifying Horizontal and Slant Asymptotes

For these solutions, we will use $f(x)=\frac{p(x)}{q(x)},q(x)\ne 0.$

1. ⓐ $g(x)=\frac{6x^3-10x}{2x^3+5x^2}:$ The degree of $p=\text{degree of}q=3,$ so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at $y=\frac{6}{2}$ or $y=3.$
2. ⓑ $h(x)=\frac{x^2-4x+1}{x+2}:$ The degree of $p=2$ and degree of $q=1.$ Since $p>q$ by 1, there is a slant asymptote found at $\frac{x^2-4x+1}{x+2}.$
   $$\begin{array}{l}\hfill \\ \begin{array}{c}-2\\ \end{array}\overline{)\begin{array}{rrr}\hfill 1& \hfill -4& \hfill 1\\ \hfill & \hfill -2& \hfill 12\end{array}}\hfill \\ \begin{array}{lll}\text{ 1}\hfill & -6\hfill & 13\hfill \end{array}\hfill \end{array}$$

   The quotient is $x–6$ and the remainder is 13. There is a slant asymptote at

3. ⓒ $k(x)=\frac{x^2+4x}{x^3-8}:$ The degree of $p=2<$ degree of $q=3,$ so there is a horizontal asymptote $y=0.$