Find the Symmetry f(x)=(4-x^2)/(x^2-3x-4)

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

Step 1

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.

Step 2

Simplify.

Simplify the numerator.

Rewrite $4$ as $2^{2}$ .

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

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

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

Factor $x^{2} - 3 x - 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

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

Step 3

Find $f (- x)$ .

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

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

Remove parentheses.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

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

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

Rewrite $- 4$ as $- 1 (4)$ .

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

Factor $- 1$ out of $- (x) - 1 (4)$ .

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

Rewrite $- (x + 4)$ as $- 1 (x + 4)$ .

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

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

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

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

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

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

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

Simplify the expression.

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

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x + 4$ by $1$ .

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

Step 4

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

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

Since $\frac{(2 - x) (2 + x)}{(x + 4) (x - 1)}$ $\neq$ $\frac{(2 + x) (2 - x)}{(x - 4) (x + 1)}$ , the function is not even.

The function is not even

Step 5

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

Multiply $- 1$ by $\frac{(2 + x) (2 - x)}{(x - 4) (x + 1)}$ .

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

Since $\frac{(2 - x) (2 + x)}{(x + 4) (x - 1)}$ $\neq$ $- \frac{(2 + x) (2 - x)}{(x - 4) (x + 1)}$ , the function is not odd.

The function is not odd

Step 6

The function is neither odd nor even

Step 7

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

No origin symmetry

Step 8

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

No y-axis symmetry

Step 9

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

Function is not symmetric

Step 10

Find the Symmetry f(x)=(4-x^2)/(x^2-3x-4)

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