How do you evaluate the function #f (x) = 3x^2 + 3x -2# for #f (a + h)#?

1 Answer
Nov 14, 2015

Plug in #(a+h)# for #x#.

Explanation:

If the question asked for #f(2)#, all you would do is put #2# into every spot where there's an #x#. For #(a+h)#, you use the same logic, and plug #(a+h)# into every spot where there's an #x#.

#color(blue)(f(a+h)=3(a+h)^2+3(a+h)-2)#

This quickly turns into a question of algebra. Consider the first term, #3(a+h)^2#. The order of operations states that we must square #(a+h)# before we distribute the #3#.

So, in order to square #(a+h)#, we "FOIL" the following: #(a+h)(a+h)#.
We should receive #a^2+ah+ah+h^2#, which is equivalent to #a^2+2ah+h^2#.

Therefore, we now have #color(green)(f(a+h)=3(a^2+2ah+h^2)+3(a+h)-2)#
Now, we can distribute the #3# into both terms in parentheses.
We should get #f(a+h)=3a^2+6ah+3h^2+3a+3h-2#.

Now, all we have to do is combine like terms—but wait! There are none. Everything is as simplified as possible. Therefore, we have arrived at our answer.

#color(red)(f(a+h)=3a^2+6ah+3h^2+3a+3h-2)#