Find the x and y Intercepts f(x)=x^3+3x^2-4x-12

$f (x) = x^{3} + 3 x^{2} - 4 x - 12$

Step 1

Find the x-intercepts.

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

$0 = x^{3} + 3 x^{2} - 4 x - 12$

Solve the equation.

Rewrite the equation as $x^{3} + 3 x^{2} - 4 x - 12 = 0$ .

$x^{3} + 3 x^{2} - 4 x - 12 = 0$

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.

$(x^{3} + 3 x^{2}) - 4 x - 12 = 0$

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

$x^{2} (x + 3) - 4 (x + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x + 3) (x^{2} - 4) = 0$

Rewrite $4$ as $2^{2}$ .

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

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

$x + 3 = 0$

$x + 2 = 0$

$x - 2 = 0$

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 $x + 2$ 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 $x - 2$ 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 $(x + 3) (x + 2) (x - 2) = 0$ true.

$x = - 3, - 2, 2$

x-intercept(s) in point form.

x-intercept(s): $(- 3, 0), (- 2, 0), (2, 0)$

Step 2

Find the y-intercepts.

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

$y = {(0)}^{3} + 3 {(0)}^{2} - 4 \cdot 0 - 12$

Solve the equation.

Remove parentheses.

$y = 0^{3} + 3 {(0)}^{2} - 4 \cdot 0 - 12$

Remove parentheses.

$y = 0^{3} + 3 \cdot 0^{2} - 4 \cdot 0 - 12$

Remove parentheses.

$y = {(0)}^{3} + 3 {(0)}^{2} - 4 \cdot 0 - 12$

Simplify ${(0)}^{3} + 3 {(0)}^{2} - 4 \cdot 0 - 12$ .

Simplify each term.

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

$y = 0 + 3 {(0)}^{2} - 4 \cdot 0 - 12$

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

$y = 0 + 3 \cdot 0 - 4 \cdot 0 - 12$

Multiply $3$ by $0$ .

$y = 0 + 0 - 4 \cdot 0 - 12$

Multiply $- 4$ by $0$ .

$y = 0 + 0 + 0 - 12$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 12$

Add $0$ and $0$ .

$y = 0 - 12$

Subtract $12$ from $0$ .

$y = - 12$

y-intercept(s) in point form.

y-intercept(s): $(0, - 12)$

Step 3

List the intersections.

x-intercept(s): $(- 3, 0), (- 2, 0), (2, 0)$

y-intercept(s): $(0, - 12)$

Step 4

Find the x and y Intercepts f(x)=x^3+3x^2-4x-12

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