How do you find the Vertical, Horizontal, and Oblique Asymptote given (x^2+4)/(6x-5x^2)?

Apr 27, 2017

$\text{vertical asymptotes at " x=0" and } x = \frac{6}{5}$

$\text{horizontal asymptote at } y = - \frac{1}{5}$

Explanation:

$\text{let } f \left(x\right) = \frac{{x}^{2} + 4}{6 x - 5 {x}^{2}}$

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

$\text{solve } 6 x - 5 {x}^{2} = 0 \Rightarrow x \left(6 - 5 x\right) = 0$

$\Rightarrow x = 0 \text{ and " x=6/5" are the asymptotes}$

Horizontal asymptotes occur as.

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{4}{x} ^ 2}{\frac{6 x}{x} ^ 2 - \frac{5 {x}^{2}}{x} ^ 2} = \frac{1 + \frac{4}{x} ^ 2}{\frac{6}{x} - 5}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{0 - 5}$

$\Rightarrow y = - \frac{1}{5} \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 2 ) Hence there are no oblique asymptotes.
graph{(x^2+4)/(6x-5x^2) [-10, 10, -5, 5]}