Find the x and y Intercepts f(x)=(18-3x-x^2)/(x^2-9)

$f (x) = \frac{18 - 3 x - x^{2}}{x^{2} - 9}$

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = \frac{18 - 3 x - x^{2}}{x^{2} - 9}$

Solve the equation.

Set the numerator equal to zero.

$18 - 3 x - x^{2} = 0$

Solve the equation for $x$ .

Factor the left side of the equation.

Factor $- 1$ out of $18 - 3 x - x^{2}$ .

Reorder the expression.

Move $18$ .

$- 3 x - x^{2} + 18 = 0$

Reorder $- 3 x$ and $- x^{2}$ .

$- x^{2} - 3 x + 18 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 3 x + 18 = 0$

Factor $- 1$ out of $- 3 x$ .

$- (x^{2}) - (3 x) + 18 = 0$

Rewrite $18$ as $- 1 (- 18)$ .

$- (x^{2}) - (3 x) - 1 \cdot - 18 = 0$

Factor $- 1$ out of $- (x^{2}) - (3 x)$ .

$- (x^{2} + 3 x) - 1 \cdot - 18 = 0$

Factor $- 1$ out of $- (x^{2} + 3 x) - 1 (- 18)$ .

$- (x^{2} + 3 x - 18) = 0$

Factor.

Factor $x^{2} + 3 x - 18$ 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 $- 18$ and whose sum is $3$ .

$- 3, 6$

Write the factored form using these integers.

$- ((x - 3) (x + 6)) = 0$

Remove unnecessary parentheses.

$- (x - 3) (x + 6) = 0$

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

$x - 3 = 0$

$x + 6 = 0$

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 $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- (x - 3) (x + 6) = 0$ true.

$x = 3, - 6$

Exclude the solutions that do not make $0 = \frac{18 - 3 x - x^{2}}{x^{2} - 9}$ true.

$x = - 6$

x-intercept(s) in point form.

x-intercept(s): $(- 6, 0)$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = \frac{18 - 3 \cdot 0 - {(0)}^{2}}{{(0)}^{2} - 9}$

Solve the equation.

Remove parentheses.

$y = \frac{18 - 3 \cdot 0 - 0^{2}}{{(0)}^{2} - 9}$

Remove parentheses.

$y = \frac{18 - 3 \cdot 0 - 0^{2}}{0^{2} - 9}$

Remove parentheses.

$y = \frac{18 - 3 \cdot 0 - {(0)}^{2}}{{(0)}^{2} - 9}$

Simplify $\frac{18 - 3 \cdot 0 - {(0)}^{2}}{{(0)}^{2} - 9}$ .

Simplify the numerator.

Multiply $- 3$ by $0$ .

$y = \frac{18 + 0 - 0^{2}}{0^{2} - 9}$

Raising $0$ to any positive power yields $0$ .

$y = \frac{18 + 0 - 0}{0^{2} - 9}$

Multiply $- 1$ by $0$ .

$y = \frac{18 + 0 + 0}{0^{2} - 9}$

Add $18 + 0$ and $0$ .

$y = \frac{18 + 0}{0^{2} - 9}$

Add $18$ and $0$ .

$y = \frac{18}{0^{2} - 9}$

Simplify the denominator.

Raising $0$ to any positive power yields $0$ .

$y = \frac{18}{0 - 9}$

Subtract $9$ from $0$ .

$y = \frac{18}{- 9}$

Divide $18$ by $- 9$ .

$y = - 2$

y-intercept(s) in point form.

y-intercept(s): $(0, - 2)$

List the intersections.

x-intercept(s): $(- 6, 0)$

y-intercept(s): $(0, - 2)$

Find the x and y Intercepts f(x)=(18-3x-x^2)/(x^2-9)

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