Precalculus Examples

$r (x) = \frac{4 x^{2} + 4}{x^{2} + 6 x + 9}$

Step 1

Set the denominator in $\frac{4 x^{2} + 4}{x^{2} + 6 x + 9}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 6 x + 9 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Factor using the perfect square rule.

Tap for more steps...

Step 2.1.1

Rewrite $9$ as $3^{2}$ .

$x^{2} + 6 x + 3^{2} = 0$

Step 2.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 2.1.3

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 3 + 3^{2} = 0$

Step 2.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

${(x + 3)}^{2} = 0$

Step 2.2

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.3

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3

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

Interval Notation:

$(- \infty, - 3) \cup (- 3, \infty)$

Set-Builder Notation:

${x | x \neq - 3}$

Step 4

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

Interval Notation:

$[\frac{2}{5}, \infty)$

Set-Builder Notation:

${y | y \geq \frac{2}{5}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 3) \cup (- 3, \infty), {x | x \neq - 3}$

Range: $[\frac{2}{5}, \infty), {y | y \geq \frac{2}{5}}$

Step 6