# How do you find the domain of p(x)=x^2+4?

There are no restrictions in the argument of the function, so the domain is $x \in \left(- \infty \to + \infty\right)$
The range is another story. Since ${x}^{2} \ge 0$:
$p \left(x\right) \in \left[4 \to + \infty\right)$