# How do you find the domain and range and intercepts for H(x) = [(x - 1) (x + 2) (x - 3)]/(x(x - 4)^2)?

Feb 22, 2018

The domain is all Real numbers with $x \ne 0 , 4$.
The intercepts are at $x = - 2 , 1 , 3$.
The range is all Real numbers.

#### Explanation:

First, the domain: This is where your function (y-values) can exist. The idea here is that you can never have a $0$ in the denominator of a fraction. (You can't divide anything into zero parts.) Set your denominator equal to $0$ to find the vertical asymptotes, which are the places your function can't exist:

$x {\left(x - 4\right)}^{2} = 0$ There are two terms here that can make this expression equal $0$. Either $x = 0$ or ${\left(x - 4\right)}^{2} = 0$.
So $x = 0$ is one of the "holes" in your domain.

Solve for the other expression by taking the square root of both sides: ${\left(x - 4\right)}^{2} = 0 \implies \sqrt{{\left(x - 4\right)}^{2}} = \sqrt{0}$
$x - 4 = 0$
Add $4$ to both sides, and the other place you need to exclude from the domain is at $x = 4$.

As far as the intercepts, think about what those are. We often call intercepts "roots" or "zeros" of a function. They are where $y = 0$. And you know that a fraction $= 0$ when the numerator $= 0$.
Set the numerator $= 0$ to find your intercepts:
$\left(x - 1\right) \left(x + 2\right) \left(x - 3\right) = 0$
In this case, if any of the three terms in parentheses $= 0$, the whole numerator$= 0$.
If $x - 1 = 0$, you can solve for $x$ by adding $1$ to both sides and get $x = 1$.
Doing the same for the other two terms gives you $x = - 2$ & $x = 3$.