Find the x and y Intercepts 1/9x^4-4/9x^3

$\frac{1}{9} x^{4} - \frac{4}{9} x^{3}$

Step 1

Write $\frac{1}{9} x^{4} - \frac{4}{9} x^{3}$ as an equation.

$y = \frac{1}{9} x^{4} - \frac{4}{9} x^{3}$

Step 2

Find the x-intercepts.

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

$0 = \frac{1}{9} x^{4} - \frac{4}{9} x^{3}$

Solve the equation.

Rewrite the equation as $\frac{1}{9} x^{4} - \frac{4}{9} x^{3} = 0$ .

$\frac{1}{9} x^{4} - \frac{4}{9} x^{3} = 0$

Simplify each term.

Combine $\frac{1}{9}$ and $x^{4}$ .

$\frac{x^{4}}{9} - \frac{4}{9} x^{3} = 0$

Combine $x^{3}$ and $\frac{4}{9}$ .

$\frac{x^{4}}{9} - \frac{x^{3} \cdot 4}{9} = 0$

Move $4$ to the left of $x^{3}$ .

$\frac{x^{4}}{9} - \frac{4 x^{3}}{9} = 0$

Multiply each term in $\frac{x^{4}}{9} - \frac{4 x^{3}}{9} = 0$ by $9$ to eliminate the fractions.

Multiply each term in $\frac{x^{4}}{9} - \frac{4 x^{3}}{9} = 0$ by $9$ .

$\frac{x^{4}}{9} \cdot 9 - \frac{4 x^{3}}{9} \cdot 9 = 0 \cdot 9$

Simplify the left side.

Simplify each term.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{x^{4}}{9} \cdot 9 - \frac{4 x^{3}}{9} \cdot 9 = 0 \cdot 9$

Rewrite the expression.

$x^{4} - \frac{4 x^{3}}{9} \cdot 9 = 0 \cdot 9$

Cancel the common factor of $9$ .

Move the leading negative in $- \frac{4 x^{3}}{9}$ into the numerator.

$x^{4} + \frac{- 4 x^{3}}{9} \cdot 9 = 0 \cdot 9$

Cancel the common factor.

$x^{4} + \frac{- 4 x^{3}}{9} \cdot 9 = 0 \cdot 9$

Rewrite the expression.

$x^{4} - 4 x^{3} = 0 \cdot 9$

Simplify the right side.

Multiply $0$ by $9$ .

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

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

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

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

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

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

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

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

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

$x^{3} = 0$

$x - 4 = 0$

Set $x^{3}$ equal to $0$ and solve for $x$ .

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Solve $x^{3} = 0$ for $x$ .

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

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.

$x = 4$

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

$x = 0, 4$

x-intercept(s) in point form.

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

Step 3

Find the y-intercepts.

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

$y = \frac{1}{9} \cdot {(0)}^{4} - \frac{4}{9} \cdot {(0)}^{3}$

Solve the equation.

Remove parentheses.

$y = \frac{1}{9} \cdot 0^{4} - \frac{4}{9} \cdot {(0)}^{3}$

Remove parentheses.

$y = \frac{1}{9} \cdot 0^{4} - \frac{4}{9} \cdot 0^{3}$

Remove parentheses.

$y = \frac{1}{9} \cdot {(0)}^{4} - \frac{4}{9} \cdot {(0)}^{3}$

Simplify $\frac{1}{9} \cdot {(0)}^{4} - \frac{4}{9} \cdot {(0)}^{3}$ .

Simplify each term.

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

$y = \frac{1}{9} \cdot 0 - \frac{4}{9} \cdot {(0)}^{3}$

Multiply $\frac{1}{9}$ by $0$ .

$y = 0 - \frac{4}{9} \cdot {(0)}^{3}$

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

$y = 0 - \frac{4}{9} \cdot 0$

Multiply $- \frac{4}{9} \cdot 0$ .

Multiply $0$ by $- 1$ .

$y = 0 + 0 (\frac{4}{9})$

Multiply $0$ by $\frac{4}{9}$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

Step 4

List the intersections.

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

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

Step 5

Find the x and y Intercepts 1/9x^4-4/9x^3

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