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

1 Answer
Mar 2, 2016

Answer:

There is a horizontal asymptote: #y = 3#

There is a vertical asymptote: #x = -1#

Explanation:

You can rewrite the expression.

#frac{3x-2}{x + 1} = frac{3x + 3 - 5}{x + 1}#

#= frac{3x + 3}{x + 1} + frac{-5}{x + 1}#

#= 3 - frac{5}{x + 1}#

From this, you can see that

#lim_{x -> oo} frac{3x-2}{x + 1} = lim_{x -> oo} (3 - frac{5}{x + 1}) = 3#

Similarly,

#lim_{x -> -oo} frac{3x-2}{x + 1} = lim_{x -> -oo} (3 - frac{5}{x + 1}) = 3#

There is a horizontal asymptote: #y = 3#

You can also see that #x = -1# results in division by zero. When #x# approaches #-1# from the left, the denominator will become infinisimally less than zero, while the numerator is negative. So,

#lim_{x -> -1^-} frac{3x-2}{x + 1} = oo#

Similarly,

#lim_{x -> -1^+} frac{3x-2}{x + 1} = -oo#

There is a vertical asymptote: #x = -1#

Below is a graph for your reference.
graph{(3x-2)/(x + 1) [-20, 20, -10, 10]}