How do you find vertical, horizontal and oblique asymptotes for #f(x) = (x^2-4x+1) /( 2x-3)#?

1 Answer
May 28, 2017

Answer:

The vertical asymptote is #x=3/2#
The oblique asymptote is #y=1/2x-5/4#
No horizontal asymptote

Explanation:

As we cannot divide by #0#, #x!=3/2#

The vertical asymptote is #x=3/2#

We perform a long division

#color(white)(aaaa)##2x-3##color(white)(aaaa)##|##x^2-4x+1##color(white)(aaaa)##|##1/2x-5/4#

#color(white)(aaaaaaaaaaaaaaa)##x^2-3/2x#

#color(white)(aaaaaaaaaaaaaaaa)##0-5/2x+1#

#color(white)(aaaaaaaaaaaaaaaaaa)##-5/2x-15/4#

#color(white)(aaaaaaaaaaaaaaaaaaa)##-0-11/4#

Therefore,

#f(x)=(x^2-4x+1)/(2x-3)=(1/2x-5/4)-(11/4)/(2x-3)#

#lim_(x->-oo)f(x)-(1/2x-5/4)=lim_(x->-oo)-(11/4)/(2x-3)=0^+#

#lim_(x->+oo)f(x)-(1/2x-5/4)=lim_(x->+oo)-(11/4)/(2x-3)=0^-#

The oblique asymptote is #y=1/2x-5/4#
graph{(y-(x^2-4x+1)/(2x-3))(y-1/2x+5/4)(y-1000(x-3/2))=0 [-14.24, 14.24, -7.12, 7.12]}