Before we begin, let's recall two important definitions.

A function $f$ is an even function when $f (- x) = f (x)$ for all $x$ in its domain.
A function $f$ is an odd function when $f (- x) = - f (x)$ for all $x$ in its domain.

Let's see how the graphs of these types of functions look.

Consider the graphs above. We can see that for even functions, if

(x, y)

is on the graph, then

(- x, y)

is also on the graph. Meanwhile, for an odd function, if

(x, y)

is on the graph, then

(- x, - y)

is also on the graph. Now, consider the given function.

g (x) = x^{2} + 5 x + 1

Let's calculate

g (- x) .

g (- x) = (- x)^{2} + 5 (- x) + 1

$(- a)^{2} = a^{2}$

g (- x) = x^{2} + 5 (- x) + 1

g (- x) = x^{2} - 5 x + 1

Next, let's calculate

- g (x) .

- g (x) = - (x^{2} + 5 x + 1)

Distribute $- 1$

- g (x) = - x^{2} - 5 x - 1

Finally, let's think about what these results tell us.

$g (x)$	$g (- x)$	$- g (x)$
$x^{2} + 5 x + 1$	$x^{2} - 5 x + 1$	$- x^{2} - 5 x - 1$

Since $g (x) \neq = g (- x)$ and $g (- x) \neq = - g (x),$ the function $g$ is neither an even nor an odd function.

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