# What are the asymptotes for f(x) = (x+3)/(x-3)?

Jul 3, 2018

The vertical asymptote is $x = 3$ and the horizontal asymptote is $y = 1$

#### Explanation:

The denominator must be $\ne 0$

$x - 3 \ne 0$

$x \ne 3$

Therefore, the vertical asymptote is $x = 3$

As, the degree of the numerator $=$ to the degree of the denominator, There is no slant asymptote.

Also,

${\lim}_{x \to + \infty} \frac{x + 3}{x - 3} = {\lim}_{x \to + \infty} \left(\frac{x}{x}\right) = 1$

${\lim}_{x \to - \infty} \frac{x + 3}{x - 3} = {\lim}_{x \to - \infty} \left(\frac{x}{x}\right) = 1$

The horizontal asymptote is $y = 1$

graph{(y-(x+3)/(x-3))(y-1)=0 [-18.02, 18.03, -9.01, 9.01]}

Jul 3, 2018

$\text{vertical asymptote at "x=3}$
$\text{horizontal asymptote at } y = 1$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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

$\text{solve "x-3=0rArrx=3" is the asymptote}$

$\text{Horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

$\text{divide terms on numerator/denominator by } x$

$f \left(x\right) = \frac{\frac{x}{x} + \frac{3}{x}}{\frac{x}{x} - \frac{3}{x}} = \frac{1 + \frac{3}{x}}{1 - \frac{3}{x}}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{1 - 0}$

$y = 1 \text{ is the asymptote}$
graph{(x+3)/(x-3) [-20, 20, -10, 10]}