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

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Answer 1 · 2017-05-28T09:41:52

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

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]}

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