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Check it out:

f ( x ) = ( x^2 - x - 6 ) / ( x^2 - 1 )

denominator = 0 gives x^2 - 1 = 0 which gives ( x - 1 ) ( x + 1 ) = 0 which gives x - 1 = 0 or x + 1 = 0 which gives x = 1 and x = -1

So, the vertical asymptotes are the lines

x = -1 and x = 1

They look like

a graph with vertical asymptotes at x = -1 and x = 1

We draw them with dashes since they are really invisible.


TRY IT:

Find (and draw) the vertical asymptotes of

f ( x ) = ( x - 3 ) / ( x^2 - 3x - 10 )