How do you identify all vertical asymptotes for #f(x)=(x^2-5x+4)/(x^2-4)#?

2 Answers
Aug 9, 2017

Answer:

Vertical asymptotes are at #x = 2 and x = -2#

Explanation:

# f(x) = (x^2 -5x + 4)/(x^2-4) = ((x-4)(x-1))/((x+2)(x-2))#

For vertical asymptotes to form denominator is zero.

#x+2=0 :. x = -2 and x-2 = 0 or x =2 #

So vertical asymptotes are at #x = 2 and x = -2#

graph{(x^2-5x+4)/(x^2-4) [-40, 40, -20, 20]} [Ans]

Aug 9, 2017

Answer:

#"vertical asymptotes at "x=+-2#

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^2-4=0rArr(x-2)(x+2)=0#

#rArrx=+-2" are the asymptotes"#
graph{(x^2-5x+4)/(x^2-4) [-10, 10, -5, 5]}