# How to find domain, asymptotes, holes, intercepts for f(x) = x / (3x(x-1))?

Sep 26, 2015

Domain: $x \ne 0$ AND $x \ne 1$
Asymptote: $x = 1$
Hole: $x = 0$
X-intercept: None
Y-intercept: None

#### Explanation:

Obvious temptation is to reduce the fraction by $x$ that is a multiplier in the numerator and denominator.
But, prior to do that, we have to make sure that the function does not change its domain by this transformation. That means, we have to specify that the new function has exactly the same domain as the original one.

The original function has two points, $x = 0$ and $x = 1$ where its denominator equals to zero. Therefore, we have to specify the domain as
$x \ne 0$ AND $x \ne 1$
Another representation of the same domain would be
$x \ne 0 \cap x \ne 1$
or
$- \infty < x < 0 \cup 0 < x < 1 \cup 1 < x < + \infty$
or
$\left(- \infty , 0\right) \cup \left(0 , 1\right) \cup \left(1 , + \infty\right)$

Now, when we properly restricted the domain, we can reduce the fraction by $x$ getting
$\frac{1}{3 \left(x - 1\right)}$

This is a familiar hyperbola shifted to the right by $1$ and stretched vertically by a factor of $\frac{1}{3}$ (so, it actually shrinks by a factor of $3$). The only nuance is that it's undefined at $x = 0$, that is the value $x = 0$ is a hole.

Our hyperbola has an asymptote at $x = 1$ since it tends to infinity around this point.

It does not intersect the X-axis, so there is no X-intercept.

To find Y-intercept, we have to find the value of this function at $x = 0$. But this point is outside the function domain, so we can say that there is no Y-intercept.