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) ?

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Answer 1 · 2015-07-19T02:56:13

Substitute #-4# in place of #x# in #h(x) = -x^2-3x# and simplify.

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

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

#h(5) = -(5)^2-3(5) = -25-15 = -40#

and

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

So

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

# = -(16)+12#

# = -16+12#

# = -4#

Yes, it's a special case: #h(-4) = -4#

#p(x)# and #g(x)# are not involved at all in finding #h(-4)# (nor in finding #h( "any number")#)