# 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) ?

Jul 19, 2015

Substitute $- 4$ in place of $x$ in $h \left(x\right) = - {x}^{2} - 3 x$ and simplify.

#### Explanation:

$h \left(x\right) = - {\left(x\right)}^{2} - 3 \left(x\right)$

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

$h \left(5\right) = - {\left(5\right)}^{2} - 3 \left(5\right) = - 25 - 15 = - 40$

and

$h \left(\Delta\right) = - {\left(\Delta\right)}^{2} - 3 \left(\Delta\right)$

So

$h \left(- 4\right) = - {\left(- 4\right)}^{2} - 3 \left(- 4\right)$

$= - \left(16\right) + 12$

$= - 16 + 12$

$= - 4$

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

$p \left(x\right)$ and $g \left(x\right)$ are not involved at all in finding $h \left(- 4\right)$ (nor in finding $h \left(\text{any number}\right)$)