Graph -x^2-4x+32>=0

$- x^{2} - 4 x + 32 \geq 0$

Convert the inequality to an equation.

$- x^{2} - 4 x + 32 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 4 x + 32$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 4 x + 32 = 0$

Factor $- 1$ out of $- 4 x$ .

$- (x^{2}) - (4 x) + 32 = 0$

Rewrite $32$ as $- 1 (- 32)$ .

$- (x^{2}) - (4 x) - 1 \cdot - 32 = 0$

Factor $- 1$ out of $- (x^{2}) - (4 x)$ .

$- (x^{2} + 4 x) - 1 \cdot - 32 = 0$

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (- 32)$ .

$- (x^{2} + 4 x - 32) = 0$

Factor.

Factor $x^{2} + 4 x - 32$ 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 $- 32$ and whose sum is $4$ .

$- 4, 8$

Write the factored form using these integers.

$- ((x - 4) (x + 8)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x + 8) = 0$

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

$x - 4 = 0$

$x + 8 = 0$

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 $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $- (x - 4) (x + 8) = 0$ true.

$x = 4, - 8$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 4$

$x > 4$

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 $x < - 8$ to see if it makes the inequality true.

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$- {(- 10)}^{2} - 4 \cdot - 10 + 32 \geq 0$

The left side $- 28$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 8 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$- {(0)}^{2} - 4 \cdot 0 + 32 \geq 0$

The left side $32$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$- {(6)}^{2} - 4 \cdot 6 + 32 \geq 0$

The left side $- 28$ is less than the right side $0$ , which means that the given statement is false.

False

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

$x < - 8$ False

$- 8 < x < 4$ True

$x > 4$ False

$x < - 8$ False

$- 8 < x < 4$ True

$x > 4$ False

The solution consists of all of the true intervals.

$- 8 \leq x \leq 4$

Graph -x^2-4x+32>=0

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