Find the Domain and Range f(x)=10/(x^2-7x-30)

$f (x) = \frac{10}{x^{2} - 7 x - 30}$

Step 1

Set the denominator in $\frac{10}{x^{2} - 7 x - 30}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 7 x - 30 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} - 7 x - 30$ 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 $- 30$ and whose sum is $- 7$ .

$- 10, 3$

Write the factored form using these integers.

$(x - 10) (x + 3) = 0$

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

$x - 10 = 0$

$x + 3 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 10) (x + 3) = 0$ true.

$x = 10, - 3$

Step 3

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

Interval Notation:

$(- \infty, - 3) \cup (- 3, 10) \cup (10, \infty)$

Set-Builder Notation:

${x | x \neq - 3, 10}$

Step 4

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

Interval Notation:

$(- \infty, - \frac{40}{169}] \cup (0, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{40}{169}, y > 0}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 3) \cup (- 3, 10) \cup (10, \infty), {x | x \neq - 3, 10}$

Range: $(- \infty, - \frac{40}{169}] \cup (0, \infty), {y | y \leq - \frac{40}{169}, y > 0}$

Step 6

Find the Domain and Range f(x)=10/(x^2-7x-30)

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