# How do you graph and find the discontinuities of y=(-4x-3)/(x-2)?

Aug 17, 2015

$f \left(x\right) = - \frac{4 x + 3}{x - 2}$
$= - \frac{4 x - 8 + 8 + 3}{x - 2}$
$= - 4 - \frac{11}{x - 2}$

A discontinuity at $x = a$ occurs when:
1. $f \left(a\right)$ does not exist
2. ${\lim}_{x \rightarrow {a}^{+}} f \left(x\right) \ne {\lim}_{x \rightarrow {a}^{-}} f \left(x\right)$
3. ${\lim}_{x \rightarrow a} f \left(x\right) \ne f \left(a\right)$

In this case, discontinuity at $x = 2$ since $f \left(2\right)$ does not exist.

To plot the graph:
$x$-axis intercept
$f \left(x\right) = 0$
$x = - \frac{3}{4}$

$y$-axis intercept
$f \left(0\right) = \frac{3}{2}$

Vertical aysmptote at $x = 2$
 lim_(x rarr 2^(""_+^-)) f(x) = +- oo

Horizontal asymptote at $y = - 4$
lim_(x rarr ""_+^(-)oo) f(x) = -4^(+-)

No stationary points since there are no real values of $x$ that satisfy $f ' \left(x\right) = 0$.

graph{(-4x-3)/(x-2) [-5, 5, -20, 10]}