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

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Answer 1 · 2015-05-21T23:23:46

$p(x) = -(x - 1)^2 - 1$

$p(x) = -(x^2 - 2x + 1) - 1$

$p(x) = -x^2 + 2x - 1 - 1$

$p(x) = -x^2 + 2x - 2$

This is a quadratic equation, so the domain (x-value range) is all real numbers, or $-oo -> oo$ .

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