Find the x and y Intercepts r(x)=(x^3-3x^2-4x)/(x-4)

$r (x) = \frac{x^{3} - 3 x^{2} - 4 x}{x - 4}$

Step 1

Find the x-intercepts.

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

$0 = \frac{x^{3} - 3 x^{2} - 4 x}{x - 4}$

Solve the equation.

Set the numerator equal to zero.

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

Solve the equation for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 3 x^{2} - 4 x$ .

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

$x \cdot x^{2} - 3 x^{2} - 4 x = 0$

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

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

Factor $x$ out of $- 4 x$ .

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

Factor $x$ out of $x \cdot x^{2} + x (- 3 x)$ .

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

Factor $x$ out of $x (x^{2} - 3 x) + x \cdot - 4$ .

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

Factor.

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$x ((x - 4) (x + 1)) = 0$

Remove unnecessary parentheses.

$x (x - 4) (x + 1) = 0$

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

$x = 0$

$x - 4 = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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.

$x = - 1$

The final solution is all the values that make $x (x - 4) (x + 1) = 0$ true.

$x = 0, 4, - 1$

Exclude the solutions that do not make $0 = \frac{x^{3} - 3 x^{2} - 4 x}{x - 4}$ true.

$x = 0, - 1$

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Simplify the numerator.

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

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

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

$y = \frac{0 - 3 \cdot 0 - 4 \cdot 0}{0 - 4}$

Multiply $- 3$ by $0$ .

$y = \frac{0 + 0 - 4 \cdot 0}{0 - 4}$

Multiply $- 4$ by $0$ .

$y = \frac{0 + 0 + 0}{0 - 4}$

Add $0 + 0$ and $0$ .

$y = \frac{0 + 0}{0 - 4}$

Add $0$ and $0$ .

$y = \frac{0}{0 - 4}$

Simplify the expression.

Subtract $4$ from $0$ .

$y = \frac{0}{- 4}$

Divide $0$ by $- 4$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

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

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

Step 4

Find the x and y Intercepts r(x)=(x^3-3x^2-4x)/(x-4)

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