How do you evaluate #h(-2)# given #h(x)=3x^2+2-2x#?

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Answer 1 · 2018-03-19T22:29:27

See a solution process below:

For each occurrence of #color(red)(x)# in #h(x)# substitute #color(red)(-2)# and evaluate the expression:

#h(color(red)(x)) = 3color(red)(x)^2 + 2 - 2color(red)(x)# becomes:

#h(color(red)(-2)) = (3 * (color(red)(-2))^2) + 2 - (2 * color(red)(-2))#

#h(color(red)(-2)) = (3 * 4) + 2 - (-4)#

#h(color(red)(-2)) = 12 + 2 + 4#

#h(color(red)(-2)) = 14 + 4#

#h(color(red)(-2)) = 18#