Determine When the Quadratic Function 4x² + 8x Is Less Than Zero

Look at the following function:

$y=4x^2+8x$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

To solve for which values of $x$ the function $y = 4x^2 + 8x$ is negative:

Step 1: Start by factoring the quadratic expression:
$y = 4x^2 + 8x = 4x(x + 2)$ .
Step 2: Set each factor to zero to find the roots:
$4x = 0$ or $x + 2 = 0$ , giving roots $x = 0$ and $x = -2$ .
Step 3: Analyze the intervals determined by these roots:
- For $x < -2$ , pick a test point like $x = -3$ ; plug into $4x(x + 2)$ :
$4(-3)((-3) + 2) = 4(-3)(-1) = 12 > 0$
- For $-2 < x < 0$ , pick a test point like $x = -1$ ; plug into $4x(x + 2)$ :
$4(-1)((-1) + 2) = 4(-1)(1) = -4 < 0$
- For $x > 0$ , pick a test point like $x = 1$ ; plug into $4x(x + 2)$ :
$4(1)(1 + 2) = 4(1)(3) = 12 > 0$ .

Hence, the quadratic is negative in the interval $-2 < x < 0$ .

The correct answer is therefore $-2 < x < 0$ .

$-2 < x < 0$

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