How do you graph y=(4-x)/(5x^2-4x-1) using asymptotes, intercepts, end behavior?

Nov 10, 2016

The vertical asymptotes are $x = - \frac{1}{5}$ and $x = 1$
The horizontal asymptote is $y = 0$
There is no slant asymptote:
The intercepts are $\left(4 , 0\right) \mathmr{and} \left(0 , - 4\right)$

Explanation:

Let's factorise the denominator
$5 {x}^{2} - 4 x - 1 = \left(5 x + 1\right) \left(x - 1\right)$
As you cannot divide by $0$, the vertical asymptotes are
$x = - \frac{1}{5}$ and $x = 1$
As the degree of the numerator is $<$ the degree of the denominator, there is no slant asymptote.

${\lim}_{x \to - \infty} y = {\lim}_{x \to - \infty} - \frac{1}{5 x} = {0}^{+}$

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} - \frac{1}{5 x} = {0}^{-}$

So $y = 0$ is a horizontal asymptote
The intercepts are
x-axis when $y = 0$$\implies$$x = 4$

and y-axis when $x = 0$$\implies$$y = - 4$
graph{(4-x)/(5x^2-4x-1) [-20.27, 20.27, -10.15, 10.15]}