How do you find the asymptotes for #(4x)/(x-3)#?

1 Answer
Mar 1, 2016

Answer:

There is a horizontal asymptote: #y = 4#

There is a vertical asymptote: #x = 3#

Explanation:

You can rewrite the expression.

#frac{4x}{x - 3} = 4 * frac{x}{x - 3}#

#= 4 * frac{(x - 3) + 3}{x - 3}#

#= 4 * (frac{x - 3}{x - 3} + frac{3}{x - 3})#

#= 4 * (1 + frac{3}{x - 3})#

#= 4 + frac{12}{x - 3}#

From this, you can see that

#lim_{x -> oo} frac{4x}{x - 3} = lim_{x -> oo} (4 + frac{12}{x - 3}) = 4#

Similarly,

#lim_{x -> -oo} frac{4x}{x - 3} = lim_{x -> -oo} (4 + frac{12}{x - 3}) = 4#

There is a horizontal asymptote: #y = 4#

You can also see that #x = 3# results in division by zero. When #x# approaches #3# from the left, the denominator will become infinisimally less than zero. So,

#lim_{x -> 3^-} frac{4x}{x - 3} = -oo#

Similarly,

#lim_{x -> 3^+} frac{4x}{x - 3} = oo#

There is a vertical asymptote: #x = 3#

Below is a graph for your reference.
graph{(4x)/(x - 3) [-40, 40, -20, 20]}