How do you find the horizontal and vertical asymptote of this function (3x-2) / (x+1)?

May 7, 2015

The vertical asymptote is the line passing through the $x$ coordenate that makes the denominator equal to zero, i.e., the line of equation $x = - 1$.
The horizontal asymptote can be found by observing the behaviour of the functions for $x$ very large doing:
${\lim}_{x \to \infty} \frac{3 x - 2}{x + 1} = {\lim}_{x \to \infty} \frac{x \left(3 - \frac{2}{x}\right)}{x \left(1 + \frac{1}{x}\right)} = {\lim}_{x \to \infty} \frac{\cancel{x} \left(3 - \frac{2}{x}\right)}{\cancel{x} \left(1 + \frac{1}{x}\right)} =$
doing the limit:
${\lim}_{x \to \infty} \frac{3 - \frac{2}{x}}{1 + \frac{1}{x}} = \frac{3 - \cancel{\frac{2}{x}}}{1 + \cancel{\frac{1}{x}}} = 3$
The function approximates the value $3$ when $x$ becomes large.
So, the horizontal line of equation $y = 3$ will be a horizontal asymptote:

Graphically:
graph{(3x-2)/(x+1) [-43.4, 49.07, -17.12, 29.13]}