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

Jul 12, 2018

$\text{vertical asymptote at } x = 4$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{simplify by cancelling}$

$y = \frac{\cancel{x + 3}}{\left(x - 4\right) \cancel{\left(x + 3\right)}} = \frac{1}{x - 4}$

$\text{the removal of "(x+3)" from numerator/denominator}$
$\text{indicates a hole at } x = - 3$

$\text{the graph of the simplification "1/(x-4)" is the same as}$

$\text{the graph of "y=(x+3)/((x-4)(x+3))" without the hole}$

The denominator of y cannot be zero as this would make y 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.

$\text{solve "x-4=0rArrx=4" is the asymptote}$

$\text{Horizontal asymptotes occur as}$

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

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

$y = \frac{\frac{1}{x}}{\frac{x}{x} - \frac{4}{x}} = \frac{\frac{1}{x}}{1 - \frac{4}{x}}$

$\text{as } x \to \pm \infty , y \to \frac{0}{1 - 0}$

$y = 0 \text{ is the asymptote}$
graph{1/(x-4) [-10, 10, -5, 5]}