# How do you evaluate h(x)= -5/3absx + 6  for x=-2?

Jul 29, 2017

$h \left(x\right) = \frac{13}{3}$ at x = -2
$h \left(x\right)$ is just a way of writing "a function (call it h) in x", meaning whatever is on the other side of the equation. Given a particular value (-2) you just put it into the equation and solve it.
h(x)=−(5/3)|x|+6
h(x)=−(5/3)|-2|+6 ; h(x)=−(5/3)*2+6
h(x)=−(10/6)+6 ; h(x)=−(10/6)+36/6
$h \left(x\right) = \frac{26}{6}$ ; $h \left(x\right) = \frac{13}{3}$ or 4.33