# How do you find vertical, horizontal and oblique asymptotes for f(x)= (5x^2-17x-12)/(2x^2-7x-4)?

Oct 23, 2016

The horizontal asymptote is $y = \frac{5}{2}$, the vertical asymptote is $x = - \frac{1}{2}$ and there is no oblique asymptote.

#### Explanation:

$f \left(x\right) = \frac{5 {x}^{2} - 17 x - 12}{2 {x}^{2} - 7 x - 4}$

This problem is a bit tricky because there is a "hole" at $x = 4$.

Start by factoring both numerator and denominator.

$f \left(x\right) = \frac{\left(5 x + 3\right) \left(x - 4\right)}{\left(2 x + 1\right) \left(x - 4\right)}$

Note that factor $\left(x - 4\right)$ cancels. This means there is a "hole" or undefined point in the graph at $x = 4$.

To find the asymptotes, use the function that results after the factor cancels.

$f \left(x\right) = \frac{5 x + 3}{2 x + 1}$

To find the horizontal asymptote (HA), compare the degree of the numerator to the degree of the denominator.

If the degrees are the same, the HA is $y =$ leading coefficient of the numerator divided by the leading coefficient of the denominator.

If the degree of the denominator is greater, the HA is $y = 0$.

If the degree of the numerator is greater, there is an oblique asymptote.

$f \left(x\right) = \frac{\textcolor{b l u e}{5} {x}^{\textcolor{red}{1}} + 3}{\textcolor{b l u e}{2} {x}^{\textcolor{red}{1}} + 1}$

In this example, both the numerator and denominator have a degree of $\textcolor{red}{1}$. (The degree is the highest exponent in standard form).

The HA is then $y = \frac{\textcolor{b l u e}{5}}{\textcolor{b l u e}{2}}$

There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.

To find the vertical asymptote (VA), find the value of $x$ where the denominator equals zero. This value of $x$ will result in dividing by zero, which is undefined.

$2 x + 1 = 0 \textcolor{w h i t e}{a a a}$Solve for x

$2 x = - 1$

$x = - \frac{1}{2} \textcolor{w h i t e}{a a a}$This is the equation of the VA.