How do you find vertical, horizontal and oblique asymptotes for (3x^2+2x-1 )/(x^2-4)?

1 Answer
Apr 22, 2018

$\text{vertical asymptotes at } x = \pm 2$
$\text{horizontal asymptote at } y = 3$

Explanation:

$\text{the denominator of "f(x)" cannot be zero as this would}$
$\text{make "f(x)" undefined. Equating the denominator to zero}$
$\text{and solving gives the values that x cannot be and if the}$
$\text{numerator is non-zero for these values then they are}$
$\text{vertical asymptotes}$

$\text{solve "x^2-4=0rArrx=+-2" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

$\text{divide the terms on the numerator/denominator by}$
$\text{the highest power of x that is } {x}^{2}$

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

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

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

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence no oblique asymptote.
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}