Find the Domain and Range (2x-4)/(x^2+x-2)

$\frac{2 x - 4}{x^{2} + x - 2}$

Step 1

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

$x^{2} + x - 2 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} + x - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$(x - 1) (x + 2) = 0$

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

$x - 1 = 0$

$x + 2 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1, - 2}$

Step 4

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

Interval Notation:

$(- \infty, \frac{2}{9}] \cup [2, \infty)$

Set-Builder Notation:

${y | y \leq \frac{2}{9}, y \geq 2}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty), {x | x \neq 1, - 2}$

Range: $(- \infty, \frac{2}{9}] \cup [2, \infty), {y | y \leq \frac{2}{9}, y \geq 2}$

Step 6

Find the Domain and Range (2x-4)/(x^2+x-2)

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