Find the x and y Intercepts y=x^3+x^2-9x-9

$y = x^{3} + x^{2} - 9 x - 9$

Step 1

Find the x-intercepts.

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

$0 = x^{3} + x^{2} - 9 x - 9$

Solve the equation.

Rewrite the equation as $x^{3} + x^{2} - 9 x - 9 = 0$ .

$x^{3} + x^{2} - 9 x - 9 = 0$

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} + x^{2}) - 9 x - 9 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x + 1) - 9 (x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x + 1) (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(x + 1) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x + 3) (x - 3) = 0$

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

$x + 1 = 0$

$x + 3 = 0$

$x - 3 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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.

$x = - 3$

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.

$x = 3$

The final solution is all the values that make $(x + 1) (x + 3) (x - 3) = 0$ true.

$x = - 1, - 3, 3$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0), (- 3, 0), (3, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

$y = 0 + {(0)}^{2} - 9 \cdot 0 - 9$

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

$y = 0 + 0 - 9 \cdot 0 - 9$

Multiply $- 9$ by $0$ .

$y = 0 + 0 + 0 - 9$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 9$

Add $0$ and $0$ .

$y = 0 - 9$

Subtract $9$ from $0$ .

$y = - 9$

y-intercept(s) in point form.

y-intercept(s): $(0, - 9)$

Step 3

List the intersections.

x-intercept(s): $(- 1, 0), (- 3, 0), (3, 0)$

y-intercept(s): $(0, - 9)$

Step 4

Find the x and y Intercepts y=x^3+x^2-9x-9

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