Question

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

Answer 1

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]

Answer 2

`"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]}