Analyze Quadratic Inequality: Where is y = -2x² - 16x Less Than Zero?

Look at the following function:

$y=-2x^2-16x$

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

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

To solve this problem, we need to determine when the quadratic function $y = -2x^2 - 16x$ is less than zero. Let us follow these steps:

First, identify the roots of the quadratic equation by setting $y = 0$:

$-2x^2 - 16x = 0$

Factor out the common factor:

$-2x(x + 8) = 0$

The solutions to this equation give the x-values where the function equals zero (its roots):

So, $x = 0$ or $x = -8$

Now, analyze the intervals determined by these roots:

• Interval 1: $x < -8$
• Interval 2: $-8 < x < 0$
• Interval 3: $x > 0$

The quadratic $y = -2x^2 - 16x$ is a downward-opening parabola. We know it is zero at the roots.

Let's analyze the sign of $y$ in each interval:

• In Interval 1 ($x < -8$): Choose a test point $x = -10$
  $y = -2(-10)^2 - 16(-10) = -200 + 160 = -40$ (negative).
• In Interval 2 ($-8 < x < 0$): Choose a test point $x = -4$
  $y = -2(-4)^2 - 16(-4) = -32 + 64 = 32$ (positive).
• In Interval 3 ($x > 0$): Choose a test point $x = 1$
  $y = -2(1)^2 - 16(1) = -2 - 16 = -18$ (negative).

Hence, the function is negative in Intervals 1 and 3: where $x < -8$ or $x > 0$.

Therefore, the solution to the problem is $x < -8$ or $x > 0$.

Referring to the multiple-choice options, the correct answer is: Option 2.

Thus, the solution to the problem is $x > 0$ or $x < -8$.