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

Nov 14, 2015

Plug in $\left(a + h\right)$ for $x$.

Explanation:

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

$\textcolor{b l u e}{f \left(a + h\right) = 3 {\left(a + h\right)}^{2} + 3 \left(a + h\right) - 2}$

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

So, in order to square $\left(a + h\right)$, we "FOIL" the following: $\left(a + h\right) \left(a + h\right)$.
We should receive ${a}^{2} + a h + a h + {h}^{2}$, which is equivalent to ${a}^{2} + 2 a h + {h}^{2}$.

Therefore, we now have $\textcolor{g r e e n}{f \left(a + h\right) = 3 \left({a}^{2} + 2 a h + {h}^{2}\right) + 3 \left(a + h\right) - 2}$
Now, we can distribute the $3$ into both terms in parentheses.
We should get $f \left(a + h\right) = 3 {a}^{2} + 6 a h + 3 {h}^{2} + 3 a + 3 h - 2$.

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.

$\textcolor{red}{f \left(a + h\right) = 3 {a}^{2} + 6 a h + 3 {h}^{2} + 3 a + 3 h - 2}$