Find Intervals of Increase and Decrease: Analyzing y = (2x - 16)²

To solve this problem, let's follow through the steps:

• Step 1: Differentiate the function $y=(2x-16)^2$ with respect to $x$.

The derivative of $y$ is found using the chain rule. The outer function is $u^2$ with $u = (2x-16)$. Thus, the derivative is:

$y' = 2(2x-16) \cdot \frac{d}{dx}(2x-16)$

This simplifies to:

$y' = 2(2x-16) \cdot 2 = 4(2x-16) = 8x - 64$

• Step 2: Find the critical points by setting the derivative to zero.

Set $8x - 64 = 0$:

$8x = 64$

$x = 8$, so $x = 8$ is a critical point.

• Step 3: Analyze each interval determined by this critical point to determine where $y$ is increasing or decreasing.

The number line is split into two intervals by the critical point $x = 8$: $(-∞, 8)$ and $(8, ∞)$.

For $x < 8$ (e.g., $x = 7$):

Evaluate $y'(7) = 8(7) - 64 = 56 - 64 = -8$, so $y'(x) < 0$. Thus, $y$ is decreasing for $x < 8$.

For $x > 8$ (e.g., $x = 9$):

Evaluate $y'(9) = 8(9) - 64 = 72 - 64 = 8$, so $y'(x) > 0$. Thus, $y$ is increasing for $x > 8$.

Therefore, the function $y = (2x-16)^2$ is:
Decreasing on the interval $(-∞, 8)$ and Increasing on the interval $(8, ∞)$.

This corresponds to the correct answer choice:

$\searrow:x<8\\\nearrow:x>8$