How do you find the vertical, horizontal or slant asymptotes for $f(x)=( 2x - 3)/( x^2 - 1)$ ?

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Answer 1 · 2017-02-05T11:01:44

The vertical asymptotes are $x=1$ and $x=-1$
The horizontal asymptote is $y=0$
No slant asymptote

Let's factorise the denominator

$x^2-1=(x-1)(x+1)$

Therefore,

$f(x)=(2x-3)/(x^2-1)=(2x-3)/((x-1)(x+1))$

As you cannot divide by $0$ , $x!=1$ and $x!=-1$

The vertical asymptotes are $x=1$ and $x=-1$

As the degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote

$lim_(x->-oo)f(x)=lim_(x->-oo)(2x)/x^2=im_(x->-oo)2/x=0^-$

$lim_(x->+oo)f(x)=lim_(x->+oo)(2x)/x^2=im_(x->+oo)2/x=0^+$

The horizontal asymptote is $y=0$

graph{(2x-3)/(x^2-1) [-16.02, 16.01, -8.01, 8.01]}

How do you find the vertical, horizontal or slant asymptotes for f(x)=( 2x - 3)/( x^2 - 1)f(x)=( 2x - 3)/( x^2 - 1)?