Find Increasing Intervals for y = (4x+8)(-x+2): Quadratic Function Analysis

To find where the function $y = (4x + 8)(-x + 2)$ is increasing, we first need to express it as a standard quadratic function.

First, expand the product:

$y = (4x + 8)(-x + 2)$

$= 4x(-x) + 4x(2) + 8(-x) + 8(2)$

$= -4x^2 + 8x - 8x + 16$

Simplify this to: $y = -4x^2 + 16$

Now, differentiate $y$ with respect to $x$:

$y' = \frac{d}{dx} (-4x^2 + 16)$ = $-8x$

The function is increasing where its derivative is positive:

$-8x > 0$

Solving the inequality, we have:

$x < 0$

Therefore, the function is increasing on the interval $x < 0$.

The correct choice is therefore $x < 0$.