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

1 Answer
Narad T.
Nov 10, 2016

The vertical asymptotes are x=-1/5 and x=1
The horizontal asymptote is y=0
There is no slant asymptote:
The intercepts are (4,0) and (0,-4)

Explanation:

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

lim_(x->-oo)y=lim_(x->-oo)-1/(5x)=0^+

lim_(x->+oo)y=lim_(x->+oo)-1/(5x)=0^-

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

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

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