College Algebra, Chapter 3, 3.1, Section 3.1, Problem 20

Evaluate the function $f(x) = x^2 + 2x$ at $f(0), \quad f(3), \quad f(-3), \quad f(a), \quad f(-x), \quad f\left( \frac{1}{a} \right)$
For $f(0)$,

$
\begin{equation}
\begin{aligned}
f(0) &= (0)^2 + 2(0) && \text{Replace } x \text{ by } 0\\
\\
&= 0
\end{aligned}
\end{equation}
$


For $f(3)$,

$
\begin{equation}
\begin{aligned}
f(3) &= (3)^2 + 2(3) && \text{Replace } x \text{ by } 3\\
\\
&= 9 + 6 && \text{Simplify}\\
\\
&= 15
\end{aligned}
\end{equation}
$

For $f(-3)$,

$
\begin{equation}
\begin{aligned}
f(3) &= (-3)^2 + 2(-3) && \text{Replace } x \text{ by } -3\\
\\
&= 9 - 6 && \text{Simplify}\\
\\
&= 3
\end{aligned}
\end{equation}
$

For $f(a)$,

$
\begin{equation}
\begin{aligned}
f(a) &= (a)^2+ 2(a) && \text{Replace } x \text{ by } a\\
\\
&= a^2 + 2a
\end{aligned}
\end{equation}
$

For $f(-x)$,

$
\begin{equation}
\begin{aligned}
f(-x) &= (-x)^2 + 2(-x) && \text{Replace } x \text{ by } -x\\
\\
&= x^2 - 2x
\end{aligned}
\end{equation}
$



For $f\left( \frac{1}{a} \right)$,

$
\begin{equation}
\begin{aligned}
f\left( \frac{1}{a} \right) &= \left( \frac{1}{a} \right)^2 + 2\left( \frac{1}{a} \right) && \text{Replace } x \text{ by } \frac{1}{a}\\
\\
&= \frac{1}{a^2} + \frac{2}{a}
\end{aligned}
\end{equation}
$