Find the x and y Intercepts 6x^3-9x-x^5

$6 x^{3} - 9 x - x^{5}$

Step 1

Write $6 x^{3} - 9 x - x^{5}$ as an equation.

$y = 6 x^{3} - 9 x - x^{5}$

Step 2

Find the x-intercepts.

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

$0 = 6 x^{3} - 9 x - x^{5}$

Solve the equation.

Rewrite the equation as $6 x^{3} - 9 x - x^{5} = 0$ .

$6 x^{3} - 9 x - x^{5} = 0$

Factor the left side of the equation.

Factor $- x$ out of $6 x^{3} - 9 x - x^{5}$ .

Reorder the expression.

Move $- 9 x$ .

$6 x^{3} - x^{5} - 9 x = 0$

Reorder $6 x^{3}$ and $- x^{5}$ .

$- x^{5} + 6 x^{3} - 9 x = 0$

Factor $- x$ out of $- x^{5}$ .

$- x \cdot x^{4} + 6 x^{3} - 9 x = 0$

Factor $- x$ out of $6 x^{3}$ .

$- x \cdot x^{4} - x (- 6 x^{2}) - 9 x = 0$

Factor $- x$ out of $- 9 x$ .

$- x \cdot x^{4} - x (- 6 x^{2}) - x \cdot 9 = 0$

Factor $- x$ out of $- x (x^{4}) - x (- 6 x^{2})$ .

$- x (x^{4} - 6 x^{2}) - x \cdot 9 = 0$

Factor $- x$ out of $- x (x^{4} - 6 x^{2}) - x (9)$ .

$- x (x^{4} - 6 x^{2} + 9) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- x ({(x^{2})}^{2} - 6 x^{2} + 9) = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- x (u^{2} - 6 u + 9) = 0$

Factor using the perfect square rule.

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

$- x (u^{2} - 6 u + 3^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

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

Replace all occurrences of $u$ with $x^{2}$ .

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

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set ${(x^{2} - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x^{2} - 3)}^{2}$ equal to $0$ .

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

Solve ${(x^{2} - 3)}^{2} = 0$ for $x$ .

Set the $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{3}, - \sqrt{3}$

The final solution is all the values that make $- x {(x^{2} - 3)}^{2} = 0$ true.

$x = 0, \sqrt{3}, - \sqrt{3}$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (\sqrt{3}, 0), (- \sqrt{3}, 0)$

Step 3

Find the y-intercepts.

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

$y = 6 {(0)}^{3} - 9 \cdot 0 - {(0)}^{5}$

Solve the equation.

Remove parentheses.

$y = 6 \cdot 0^{3} - 9 \cdot 0 - {(0)}^{5}$

Remove parentheses.

$y = 6 \cdot 0^{3} - 9 \cdot 0 - 0^{5}$

Remove parentheses.

$y = 6 {(0)}^{3} - 9 \cdot 0 - {(0)}^{5}$

Simplify $6 {(0)}^{3} - 9 \cdot 0 - {(0)}^{5}$ .

Simplify each term.

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

$y = 6 \cdot 0 - 9 \cdot 0 - {(0)}^{5}$

Multiply $6$ by $0$ .

$y = 0 - 9 \cdot 0 - {(0)}^{5}$

Multiply $- 9$ by $0$ .

$y = 0 + 0 - {(0)}^{5}$

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

$y = 0 + 0 - 0$

Multiply $- 1$ 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 4

List the intersections.

x-intercept(s): $(0, 0), (\sqrt{3}, 0), (- \sqrt{3}, 0)$

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

Step 5

Find the x and y Intercepts 6x^3-9x-x^5

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