Evaluate f(-x)=3x^2+2x-4

$f (- x) = 3 x^{2} + 2 x - 4$

Divide each term in $f (- x) = 3 x^{2} + 2 x - 4$ by $- 1 x$ .

$\frac{f (- x)}{- 1 x} = \frac{3 x^{2}}{- 1 x} + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{f (x)}{x} = \frac{3 x^{2}}{- 1 x} + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{f x}{x} = \frac{3 x^{2}}{- 1 x} + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Divide $f$ by $1$ .

$f = \frac{3 x^{2}}{- 1 x} + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $x^{2}$ and $x$ .

Factor $x$ out of $3 x^{2}$ .

$f = \frac{x (3 x)}{- 1 x} + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Move the negative one from the denominator of $\frac{3 x}{- 1}$ .

$f = - 1 \cdot (3 x) + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Rewrite $- 1 \cdot (3 x)$ as $- (3 x)$ .

$f = - (3 x) + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Multiply $3$ by $- 1$ .

$f = - 3 x + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$f = - 3 x + \frac{2 x}{- 1 x} + \frac{- 4}{- 1 x}$

Rewrite the expression.

$f = - 3 x + \frac{2}{- 1} + \frac{- 4}{- 1 x}$

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$f = - 3 x - 1 \cdot 2 + \frac{- 4}{- 1 x}$

Multiply $- 1$ by $2$ .

$f = - 3 x - 2 + \frac{- 4}{- 1 x}$

Dividing two negative values results in a positive value.

$f = - 3 x - 2 + \frac{4}{x}$