Q:

Finding where the function is defined

$f(x) $= $\sqrt{ax^2+9}$ 
 a is a parameter that is greater than 0, 

I need to find the place the domain of this function,

so to do that I know that the numbers under the sqrt aren't negative at all times,
so

$ax^2$+9 $=$0
and I'm stuck right here, what is the next step I should do next?

A:

I teach my students to solve these kinds of questions like so:

Consider $x \in \mathbb{R}$ and $a \in (0,\infty)$.
Then the following are equivalent:
$\sqrt{ax^2+9}  \in \mathbb{R}$
$ax^2+9  \in [0,\infty)$
$ax^2  \in [-9,\infty)$
$x^2  \in [\frac{-9}{a},\infty)$
$\mathrm{true}$
So for all $a \in (0,\infty)$, we have that for all $x \in \mathbb{R},$ the expression $\sqrt{ax^2+9}$ makes sense and denotes a real number iff $\mathrm{true}$. In other words, for all $a \in (0,\infty)$, the domain of $f(x) = \sqrt{ax^2+9}$ is $\mathbb{R}$.

I should add that I don't like this talk of "finding" the domain of a function. The domain is always given; if you're trying to find something, what you're trying to find is the support of the entity of interest, and the entity of interest is a partial function, and not really a function at all.
For example, define a partial function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ by the formula $f(x) = \sqrt{1-x^2}$. Then the support of $f$ is $[-1,1]$. The domain of $f$ is just $\mathbb{R}$. See what I mean? The domain is always given; you don't "find" it. What you're really finding is the support.

