Find the Domain and Range y=(6-3x)/(x^2-5x+6)

$y = \frac{6 - 3 x}{x^{2} - 5 x + 6}$

Step 1

Set the denominator in $\frac{6 - 3 x}{x^{2} - 5 x + 6}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x + 6 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

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

$x - 3 = 0$

$x - 2 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 3

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

Interval Notation:

$(- \infty, 2) \cup (2, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 2, 3}$

Step 4

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

Interval Notation:

$(- \infty, 0) \cup (0, 3) \cup (3, \infty)$

Set-Builder Notation:

${y | y \neq 0, 3}$

Step 5

Determine the domain and range.

Domain: $(- \infty, 2) \cup (2, 3) \cup (3, \infty), {x | x \neq 2, 3}$

Range: $(- \infty, 0) \cup (0, 3) \cup (3, \infty), {y | y \neq 0, 3}$

Step 6

Find the Domain and Range y=(6-3x)/(x^2-5x+6)

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