Find Decreasing Intervals for y = (x-4)(-x+6): Quadratic Function Analysis

The function given is $y = (x-4)(-x+6)$. To analyze its behavior, we first convert this into a standard quadratic form by expanding:

$y = (x-4)(-x+6) = -x^2 + 6x + 4x - 24 = -x^2 + 10x - 24$.

The derivative with respect to $x$ of the function $y = -x^2 + 10x - 24$ is $\frac{dy}{dx} = -2x + 10$.

To find critical points, we set the derivative equal to zero:

$-2x + 10 = 0$

Solving for $x$, we find:

$-2x = -10$

$x = 5$.

Next, we test intervals around the critical point $x = 5$:

• For $x < 5$, choose a test point like $x = 0$: $-2(0) + 10 = 10 > 0$, so the derivative is positive, indicating the function is increasing.
• For $x > 5$, choose a test point like $x = 6$: $-2(6) + 10 = -2 < 0$, so the derivative is negative, indicating the function is decreasing.

Therefore, the function $y = (x-4)(-x+6)$ is decreasing for $x > 5$.

Thus, the correct answer is $x > 5$.