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

1 Answer
SCooke
Jul 29, 2017

#h(x)= 13/3# at x = -2

Explanation:

#h(x)# 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(x)= 26/6# ; #h(x)= 13/3# or 4.33

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