Find the Domain and Range y=(x^2+5x)/(x^2+7x+10)

$y = \frac{x^{2} + 5 x}{x^{2} + 7 x + 10}$

Step 1

Set the denominator in $\frac{x^{2} + 5 x}{x^{2} + 7 x + 10}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 7 x + 10 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} + 7 x + 10$ 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 $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

$(x + 2) (x + 5) = 0$

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

$x + 2 = 0$

$x + 5 = 0$

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.

$x = - 2$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x + 2) (x + 5) = 0$ true.

$x = - 2, - 5$

Step 3

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2, - 5}$

Step 4

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

Interval Notation:

$(- \infty, 1) \cup (1, \frac{5}{3}) \cup (\frac{5}{3}, \infty)$

Set-Builder Notation:

${y | y \neq 1, \frac{5}{3}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 5) \cup (- 5, - 2) \cup (- 2, \infty), {x | x \neq - 2, - 5}$

Range: $(- \infty, 1) \cup (1, \frac{5}{3}) \cup (\frac{5}{3}, \infty), {y | y \neq 1, \frac{5}{3}}$

Step 6

Find the Domain and Range y=(x^2+5x)/(x^2+7x+10)

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