Find the x and y Intercepts y=x^3+x^2-12x

$y = x^{3} + x^{2} - 12 x$

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

Factor the left side of the equation.

Factor $x$ out of $x^{3} + x^{2} - 12 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} + x^{2} - 12 x = 0$

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

$x \cdot x^{2} + x \cdot x - 12 x = 0$

Factor $x$ out of $- 12 x$ .

$x \cdot x^{2} + x \cdot x + x \cdot - 12 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot x$ .

$x (x^{2} + x) + x \cdot - 12 = 0$

Factor $x$ out of $x (x^{2} + x) + x \cdot - 12$ .

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

Factor.

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$x ((x - 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$x (x - 3) (x + 4) = 0$

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

$x = 0$

$x - 3 = 0$

$x + 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

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 $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $x (x - 3) (x + 4) = 0$ true.

$x = 0, 3, - 4$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (3, 0), (- 4, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

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

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

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

Multiply $- 12$ by $0$ .

$y = 0 + 0 + 0$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(0, 0), (3, 0), (- 4, 0)$

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

Step 4

Find the x and y Intercepts y=x^3+x^2-12x

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