Finite Math Examples

$h (x) = \frac{- 5 x^{2}}{(x - 7) (x + 5)}$

Step 1

Set the denominator in $\frac{- 5 x^{2}}{(x - 7) (x + 5)}$ equal to $0$ to find where the expression is undefined.

$(x - 7) (x + 5) = 0$

Step 2

Solve for $x$ .

Tap for more steps...

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 $0$ .

$x - 7 = 0$

$x + 5 = 0$

Step 2.2

Set $x - 7$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.2.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.2.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.3

Set $x + 5$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.3.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.3.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.4

The final solution is all the values that make $(x - 7) (x + 5) = 0$ true.

$x = 7, - 5$

Step 3

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq - 5, 7}$

Step 4

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

Interval Notation:

$(- \infty, - \frac{175}{36}] \cup [0, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{175}{36}, y \geq 0}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 5) \cup (- 5, 7) \cup (7, \infty), {x | x \neq - 5, 7}$

Range: $(- \infty, - \frac{175}{36}] \cup [0, \infty), {y | y \leq - \frac{175}{36}, y \geq 0}$

Step 6