How do you identify all vertical asymptotes for #f(x)=(3x^2)/(x^2-1)#?

1 Answer
Jan 14, 2017

Answer:

#"vertical asymptotes at " x=+-1#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : #x^2-1=0rArrx^2=1rArrx=+-1#

#rArrx=-1" and " x=1" are the asymptotes"#
graph{(3x^2)/(x^2-1) [-10, 10, -5, 5]}