How do you identify all asymptotes or holes for f(x)=(x-4)/(x^2-3x)?

Aug 30, 2016

Vertical Asymptotes at $x = 0 \text{ and } x = 3$
Horizontal Asymptotes at $y = 0$

Explanation:

To determine asymptotes you have understand the behaviour of the equation. Some equations are more challenging than others.

$\textcolor{b l u e}{\text{Denominator approaching 0}}$

It is 'bad news' to have a denominator become zero.

None technical term for this is 'not allowed'. Correct name is: undefined.

So solve for ${x}^{2} - 3 x = 0$

$x \left(x - 3\right) = 0 \implies x = 0 \text{ and } x = 3$

Consider the case $x = 0$

$\textcolor{b r o w n}{\text{Suppose "x" tended to 0 from the right (positive)}}$
Then writing the equation as:

$f \left(x\right) = \frac{x - 4}{x \left(x - 3\right)} \to \frac{\frac{x}{x} - \frac{4}{x}}{x - 3}$

Then this is almost but not quite the same as:

${\lim}_{x \to 0} f \left(x\right) = L \to \frac{1 - \infty}{-} 3 \to + \infty$

$\textcolor{b r o w n}{\text{Suppose "x" tended to 0 from the left (negative value"->-x")}}$

$f \left(x\right) = \frac{x - 4}{x \left(x - 3\right)} \to \frac{\frac{- x}{- x} - \frac{4}{- x}}{\left(- x\right) - 3}$

Then this is almost but not quite the same as:

${\lim}_{x \to 0} f \left(x\right) = L \to \frac{1 + \infty}{-} 3 \to - \infty$

$\textcolor{g r e e n}{\text{You have the same situation for "x" tending to 3}}$
$\textcolor{g r e e n}{\text{See if you can work that one out using the same method as above}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{x \text{ tending to } \pm \infty}$

$f \left(x\right) = \frac{x - 4}{x \left(x - 3\right)} \to \frac{\frac{x}{x} - \frac{4}{x}}{x - 3}$

As $x$ becomes larger and larger the 3 in $x - 3$ has less and less effect. In the end it becomes so insignificant you can forget about it

In the same way $\frac{4}{x}$ becomes smaller and smaller the larger $x$ becomes. In the end it tends to 0

So ${\lim}_{x \to + \infty} f \left(x\right) \to \frac{1}{\infty} \to 0$

So ${\lim}_{x \to - \infty} f \left(x\right) \to \frac{1}{- \infty} \to 0$ 