How do you find #h(10)# given #h(x)=-x^2-x#?

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Answer 1 · 2016-08-18T23:51:32

#h(10) = 110#

#h(color(red)(x))=-color(red)(x)^2-color(red)(x)#

#h(10)# means that x = 10. So each x is replaced by 10

#h(color(red)(10))=-color(red)(10)^2-color(red)(10)#

#h(color(red)(10)) = -100 -10 = -110#

Remember that #y = f(x)#

#y = h(x)#

#y = h(10) = -110#

means that for this function, when

#x = 10, y= -110#