# How do you find the domain of p(x) =-(x-1)^2-1?

$p \left(x\right) = - {\left(x - 1\right)}^{2} - 1$
$p \left(x\right) = - \left({x}^{2} - 2 x + 1\right) - 1$
$p \left(x\right) = - {x}^{2} + 2 x - 1 - 1$
$p \left(x\right) = - {x}^{2} + 2 x - 2$
This is a quadratic equation, so the domain (x-value range) is all real numbers, or $- \infty \to \infty$.