# How do you find the Vertical, Horizontal, and Oblique Asymptote given g(x)=(3x^2+2x-1 )/( x^2-4)?

Jul 10, 2016

vertical asymptotes x = ± 2
horizontal asymptote y = 3

#### Explanation:

The denominator of g(x) cannot be zero. This would give division by zero which is undefined. Setting the denominator equal to zero and solving for x gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: x^2-4=0rArr(x-2)(x+2)=0rArrx=±2

$\Rightarrow x = - 2 , x = 2 \text{ are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , g \left(x\right) \to c \text{ (a constant)}$

Divide terms on numerator/denominator by the highest exponent of x , that is ${x}^{2}$

$\frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{3 + \frac{2}{x} - \frac{1}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

as $x \to \pm \infty , g \left(x\right) \to \frac{3 + 0 - 0}{1 - 0}$

$\Rightarrow y = 3 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2) Hence there are no oblique asymptotes.
graph{(3x^2+2x-1)/(x^2-4) [-20, 20, -10, 10]}