Finite Math Examples

$H (x) = \frac{- 2 x^{2}}{(x - 8) (x + 8)}$

Step 1

Set the denominator in $\frac{- 2 x^{2}}{(x - 8) (x + 8)}$ equal to $0$ to find where the expression is undefined.

$(x - 8) (x + 8) = 0$

Step 2

Solve for $x$ .

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Step 2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 8 = 0$

$x + 8 = 0$

Step 2.2

Set $x - 8$ equal to $0$ and solve for $x$ .

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Step 2.2.1

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 2.2.2

Add $8$ to both sides of the equation.

$x = 8$

Step 2.3

Set $x + 8$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 2.3.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 2.4

The final solution is all the values that make $(x - 8) (x + 8) = 0$ true.

$x = 8, - 8$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 8) \cup (- 8, 8) \cup (8, \infty)$

Set-Builder Notation:

${x | x \neq 8, - 8}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - 2) \cup [0, \infty)$

Set-Builder Notation:

${y | y < - 2, y \geq 0}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 8) \cup (- 8, 8) \cup (8, \infty), {x | x \neq 8, - 8}$

Range: $(- \infty, - 2) \cup [0, \infty), {y | y < - 2, y \geq 0}$

Step 6