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

Question

1124 views around the world

You can reuse this answer
Creative Commons License

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]}