Question

f(x)=x+1x2−1
and i got va… x=1 and ha…. y=0….. is this correct?

Answer 1

if it is \[f(x)=\frac{x+1}{x^2-1}\] then no
your answers are correct, but not complete

Answer 2

a common mistake is to set \(x^2-1=0\) then say \(x^2=1\) and so \(x=1\) but the solution to \(x^2-1=0\) is \(x=1\) OR \(x=-1\) so there are two vertical asympotes

Answer 3

ok thanks