Graph x^2<=-6x

$x^{2} \leq - 6 x$

Step 1

Add $6 x$ to both sides of the inequality.

$x^{2} + 6 x \leq 0$

Step 2

Convert the inequality to an equation.

$x^{2} + 6 x = 0$

Step 3

Factor $x$ out of $x^{2} + 6 x$ .

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

$x \cdot x + 6 x = 0$

Factor $x$ out of $6 x$ .

$x \cdot x + x \cdot 6 = 0$

Factor $x$ out of $x \cdot x + x \cdot 6$ .

$x (x + 6) = 0$

Step 4

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

$x = 0$

$x + 6 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

The final solution is all the values that make $x (x + 6) = 0$ true.

$x = 0, - 6$

Step 8

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 0$

$x > 0$

Step 9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

${(- 8)}^{2} \leq - 6 \cdot - 8$

The left side $64$ is greater than the right side $48$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < 0$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{2} \leq - 6 \cdot - 3$

The left side $9$ is less than the right side $18$ , which means that the given statement is always true.

True

Test a value on the interval $x > 0$ to see if it makes the inequality true.

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

${(2)}^{2} \leq - 6 \cdot 2$

The left side $4$ is greater than the right side $- 12$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 6$ False

$- 6 < x < 0$ True

$x > 0$ False

$x < - 6$ False

$- 6 < x < 0$ True

$x > 0$ False

Step 10

The solution consists of all of the true intervals.

$- 6 \leq x \leq 0$

Step 11

Graph x^2<=-6x

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