How to find the asymptotes of $f (x) = \frac{x^{2} + 4}{6 x - 5 x^{2}}$ $f(x)= (x^2+4)/(6x-5x^2)$ ?

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Answer 1 · 2016-01-09T15:16:17

Vertical asymptotes at $x = 0, \frac{6}{5}$ $x=0,6/5$ , horizontal asymptote at $y = - \frac{1}{5}$ $y=-1/5$

Vertical Asymptotes

These will occur when the denominator equals $0$ $0$ .

$6 x - 5 x^{2} = 0$ $6x-5x^2=0$

$x (6 - 5 x) = 0$ $x(6-5x)=0$

Split this into two equations.

$x = 0$ $x=0$

and

$6 - 5 x = 0$ $6-5x=0$

$x = \frac{6}{5}$ $x=6/5$

The vertical asymptotes occur at $x = 0, \frac{6}{5}$ $x=0,6/5$ .

Horizontal Asymptotes

When the numerator and denominator have the same degree, the horizontal asymptote will be the terms with the largest degree divided.

$\frac{x^{2}}{- 5 x^{2}} = - \frac{1}{5}$ $x^2/(-5x^2)=-1/5$

There is a horizontal asymptote at $y = - \frac{1}{5}$ $y=-1/5$ .

The function graphed:

graph{(x^2+4)/(6x-5x^2) [-10.35, 12.15, -4.37, 6.88]}

How to find the asymptotes of f(x)= (x^2+4)/(6x-5x^2)f(x)=x2+46x−5x2f(x)= (x^2+4)/(6x-5x^2)?