# How do you find the asymptotes for #(3x-2) / (x+1)#?

vertical asymptote at x = - 1

horizontal asymptote at y = 3

The denominator of the function cannot equal zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

Horizontal asymptotes occur as

divide terms on numerator/denominator by x

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