Given: $p(x) = -5absx + abs(-x + 1$ , $h(x) = -x^2 - 3x$ , and $g(x) = 5 - sqrt(x+30)$ , how do you find h(-4) ?

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Answer 1 · 2015-07-19T02:56:13

Substitute $-4$ in place of $x$ in $h(x) = -x^2-3x$ and simplify.

$h(x) = -(x)^2-3(x)$

Whatever goes in for $(x)$ on the left goes in for each and every $(x)$ on the right.

$h(5) = -(5)^2-3(5) = -25-15 = -40$

and

$h(Delta) = -(Delta)^2-3(Delta)$

So

$h(-4) = -(-4)^2-3(-4)$

$= -(16)+12$

$= -16+12$

$= -4$

Yes, it's a special case: $h(-4) = -4$

$p(x)$ and $g(x)$ are not involved at all in finding $h(-4)$ (nor in finding $h( "any number")$ )

Given: p(x) = -5absx + abs(-x + 1p(x) = -5absx + abs(-x + 1, h(x) = -x^2 - 3xh(x) = -x^2 - 3x, and g(x) = 5 - sqrt(x+30)g(x) = 5 - sqrt(x+30), how do you find h(-4) ?