# How do you find the vertical, horizontal and slant asymptotes of: f(x)= (4x^2+ 4x-24)/(x^4- 2x^3 - 9x^2+ 18x)?

Jun 18, 2016

$f \left(x\right)$ has vertical asymptotes $x = 0$ and $x = 3$, a horizontal asymptote $y = 0$ and holes (removable singularities) at $x = 2$ and $x = - 3$.

#### Explanation:

The numerator factors like this:

$4 {x}^{2} + 4 x - 24 = 4 \left({x}^{2} + x - 6\right) = 4 \left(x + 3\right) \left(x - 2\right)$

The denominator factors like this:

${x}^{4} - 2 {x}^{3} - 9 {x}^{2} + 18 x$

$= x \left(\left({x}^{3} - 2 {x}^{2}\right) - \left(9 x - 18\right)\right)$

$= x \left({x}^{2} \left(x - 2\right) - 9 \left(x - 2\right)\right)$

$= x \left({x}^{2} - 9\right) \left(x - 2\right)$

$= x \left(x - 3\right) \left(x + 3\right) \left(x - 2\right)$

Note the common factors $\left(x - 2\right)$ and $\left(x + 3\right)$ in the numerator and denominator, so $f \left(x\right)$ can be simplified, noting that $x = 2$ and $x = - 3$ are excluded from the domain:

$f \left(x\right) = \frac{4 {x}^{2} + 4 x - 24}{{x}^{4} - 2 {x}^{3} - 9 {x}^{2} + 18 x}$

$= \frac{4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 3\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}}{x \left(x - 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x + 3\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}}$

$= \frac{4}{x \left(x - 3\right)}$

So when $x = 0$ or $x = 3$, the denominator of $f \left(x\right)$ is $0$ and the numerator is non-zero, resulting in a vertical asymptote.

As $x \to \pm \infty$ the denominator $\to + \infty$ resulting in a horizontal asymptote $y = 0$.

$f \left(x\right)$ has removable singularities (holes) at $x = 2$ and $x = - 3$

graph{(4x^2+4x-24)/(x^4-2x^3-9x^2+18x) [-10, 10, -5, 5]}