Find the Symmetry f(x)=(x^3-27)/(x^2-6x)

$f (x) = \frac{x^{3} - 27}{x^{2} - 6 x}$

Determine if the function is odd, even, or neither in order to find the symmetry.

1. If odd, the function is symmetric about the origin.

2. If even, the function is symmetric about the y-axis.

Find $f (- x)$ .

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Find $f (- x)$ by substituting $- x$ for all occurrence of $x$ in $f (x)$ .

$f (- x) = \frac{{(- x)}^{3} - 27}{{(- x)}^{2} - 6 (- x)}$

Simplify the numerator.

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Rewrite $27$ as $3^{3}$ .

$f (- x) = \frac{{(- x)}^{3} - 3^{3}}{{(- x)}^{2} - 6 (- x)}$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = - x$ and $b = 3$ .

$f (- x) = \frac{(- x - 3) ({(- x)}^{2} - x \cdot 3 + 3^{2})}{{(- x)}^{2} - 6 (- x)}$

Simplify.

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Apply the product rule to $- x$ .

$f (- x) = \frac{(- x - 3) ({(- 1)}^{2} x^{2} - x \cdot 3 + 3^{2})}{{(- x)}^{2} - 6 (- x)}$

Raise $- 1$ to the power of $2$ .

$f (- x) = \frac{(- x - 3) (1 x^{2} - x \cdot 3 + 3^{2})}{{(- x)}^{2} - 6 (- x)}$

Multiply $x^{2}$ by $1$ .

$f (- x) = \frac{(- x - 3) (x^{2} - x \cdot 3 + 3^{2})}{{(- x)}^{2} - 6 (- x)}$

Multiply $3$ by $- 1$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 3^{2})}{{(- x)}^{2} - 6 (- x)}$

Raise $3$ to the power of $2$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 9)}{{(- x)}^{2} - 6 (- x)}$

Simplify the denominator.

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Let $u = - x$ . Substitute $u$ for all occurrences of $- x$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 9)}{u^{2} - 6 u}$

Factor $u$ out of $u^{2} - 6 u$ .

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Factor $u$ out of $u^{2}$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 9)}{u \cdot u - 6 u}$

Factor $u$ out of $- 6 u$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 9)}{u \cdot u + u \cdot - 6}$

Factor $u$ out of $u \cdot u + u \cdot - 6$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 9)}{u (u - 6)}$

Replace all occurrences of $u$ with $- x$ .

$f (- x) = \frac{(- x - 3) (x^{2} - 3 x + 9)}{- x (- x - 6)}$

Simplify terms.

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Move the negative in front of the fraction.

$f (- x) = - \frac{(- x - 3) (x^{2} - 3 x + 9)}{x (- x - 6)}$

Factor $- 1$ out of $- x$ .

$f (- x) = - \frac{(- (x) - 3) (x^{2} - 3 x + 9)}{x (- x - 6)}$

Rewrite $- 3$ as $- 1 (3)$ .

$f (- x) = - \frac{(- (x) - 1 \cdot 3) (x^{2} - 3 x + 9)}{x (- x - 6)}$

Factor $- 1$ out of $- (x) - 1 (3)$ .

$f (- x) = - \frac{- (x + 3) (x^{2} - 3 x + 9)}{x (- x - 6)}$

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

$f (- x) = - \frac{- 1 (x + 3) (x^{2} - 3 x + 9)}{x (- x - 6)}$

Factor $- 1$ out of $- x$ .

$f (- x) = - \frac{- 1 (x + 3) (x^{2} - 3 x + 9)}{x (- (x) - 6)}$

Rewrite $- 6$ as $- 1 (6)$ .

$f (- x) = - \frac{- 1 (x + 3) (x^{2} - 3 x + 9)}{x (- (x) - 1 \cdot 6)}$

Factor $- 1$ out of $- (x) - 1 (6)$ .

$f (- x) = - \frac{- 1 (x + 3) (x^{2} - 3 x + 9)}{x (- (x + 6))}$

Rewrite $- (x + 6)$ as $- 1 (x + 6)$ .

$f (- x) = - \frac{- 1 (x + 3) (x^{2} - 3 x + 9)}{x (- 1 (x + 6))}$

Cancel the common factor.

$f (- x) = - \frac{- 1 (x + 3) (x^{2} - 3 x + 9)}{x (- 1 (x + 6))}$

Rewrite the expression.

$f (- x) = - \frac{(x + 3) (x^{2} - 3 x + 9)}{x (x + 6)}$

A function is even if $f (- x) = f (x)$ .

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Check if $f (- x) = f (x)$ .

$- \frac{(x + 3) (x^{2} - 3 x + 9)}{x (x + 6)}$ $\neq$ $\frac{x^{3} - 27}{x^{2} - 6 x}$

Since $- \frac{(x + 3) (x^{2} - 3 x + 9)}{x (x + 6)}$ $\neq$ $\frac{x^{3} - 27}{x^{2} - 6 x}$ , the function is not even.

The function is not even

A function is odd if $f (- x) = - f (x)$ .

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Find $- f (x)$ .

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Simplify the numerator.

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Rewrite $27$ as $3^{3}$ .

$- f (x) = - \frac{x^{3} - 3^{3}}{x^{2} - 6 x}$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 3$ .

$- f (x) = - \frac{(x - 3) (x^{2} + x \cdot 3 + 3^{2})}{x^{2} - 6 x}$

Simplify.

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Move $3$ to the left of $x$ .

$- f (x) = - \frac{(x - 3) (x^{2} + 3 \cdot x + 3^{2})}{x^{2} - 6 x}$

Raise $3$ to the power of $2$ .

$- f (x) = - \frac{(x - 3) (x^{2} + 3 x + 9)}{x^{2} - 6 x}$

Factor $x$ out of $x^{2} - 6 x$ .

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

$- f (x) = - \frac{(x - 3) (x^{2} + 3 x + 9)}{x \cdot x - 6 x}$

Factor $x$ out of $- 6 x$ .

$- f (x) = - \frac{(x - 3) (x^{2} + 3 x + 9)}{x \cdot x + x \cdot - 6}$

Factor $x$ out of $x \cdot x + x \cdot - 6$ .

$- f (x) = - \frac{(x - 3) (x^{2} + 3 x + 9)}{x (x - 6)}$

Since $- \frac{(x + 3) (x^{2} - 3 x + 9)}{x (x + 6)}$ $\neq$ $- \frac{(x - 3) (x^{2} + 3 x + 9)}{x (x - 6)}$ , the function is not odd.

The function is not odd

The function is neither odd nor even

Since the function is not odd, it is not symmetric about the origin.

No origin symmetry

Since the function is not even, it is not symmetric about the y-axis.

No y-axis symmetry

Since the function is neither odd nor even, there is no origin / y-axis symmetry.

Function is not symmetric

Find the Symmetry f(x)=(x^3-27)/(x^2-6x)

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