# What is the domain for p(x) = x^2 - 2x + 9?

Jul 6, 2015

$p \left(x\right)$ is defined $\forall x \in \mathbb{R}$ (=for all real numbers)

#### Explanation:

The domain of definition (or simply the domain) of a function is the set of "input" or argument values for which the function is defined.

(Source : Domain of a function, Wikipedia)

Your function $p \left(x\right)$ is composed by 3 terms. The domain of $p \left(x\right)$ is the intersection of each domain of each term.

Therefore : Domain of $p \left(x\right)$ :
"Domain of ${x}^{2}$"$\cap$ "domain of $- 2 x$"$\cap$ "domain of $9$"

Domain of $9$ : No problem, it's well defined for every value of $x$
Domain of $9$ : $\mathbb{R}$

Domain of $- 2 x$ : No problem again, it's defined for every value of $x$
Domain of $- 2 x$ : $\mathbb{R}$

Domain of ${x}^{2}$ : Same thing, ${x}^{2}$ can be calculate for every value of $x$, no forbidden value here.
Domain of ${x}^{2}$ : $\mathbb{R}$

To conclude :

Domain of $p \left(x\right)$ : $\mathbb{R} \cap \mathbb{R} \cap \mathbb{R}$ = $\mathbb{R}$


Note : If you have to answer to this question in an exercise, you can conclude with one argument :

"Each term of the function is defined on $\mathbb{R}$ that's why the function is defined on $\mathbb{R}$ too."