How do you find #h(-2n)# given #h(n)=n^2+n#?

1 Answer
smendyka
Mar 2, 2017

See the entire solution process below:

Explanation:

For each occurrence of #color(red)(n)# in the function #h(n)# replace it with #color(red)(-2n)#:

#h(color(red)(n)) = color(red)(n)^2 + color(red)(n)# becomes:

#h(color(red)(-2n)) = (color(red)(-2n))^2 + color(red)(-2n)#

#h(color(red)(-2n)) = 4n^2 - 2n#

Or

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

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