If #h(x) = -x^2 - 3x#, how do you solve h(-4)?

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Answer 1 · 2018-04-14T08:50:05

h(-4) = -4

Since x is given (x = -4). Then, all you need to do is log in -4 for each value of x
h(-4) = #-(-4)^2# -3(-4)
h(-4) = -(16) + 12 = -4

Answer 2 · 2018-04-14T08:50:19

#h(-4)=-4#

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

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

#h(-4)=-16-(-12))#

#h(-4)=-16+12#

#h(-4)=-4#