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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