Find Decreasing Intervals for y=(x+1)(7-x): Quadratic Function Analysis

To find the intervals where the function $y = (x+1)(7-x)$ is decreasing, we begin by expanding the function:
$y = (x+1)(7-x) = 7x - x^2 + 7 - x = -x^2 + 6x + 7$.

The function is now in the form $y = -x^2 + 6x + 7$.
This is a quadratic function, opening downward because the coefficient of $x^2$ is negative.

Let's find the critical points by taking the derivative and setting it to zero.
The derivative of $y$ is$\ y' = \frac{d}{dx}(-x^2 + 6x + 7) = -2x + 6$.
Solving $-2x + 6 = 0$ gives $x = 3$.

The vertex, $x = 3$, is where the function changes from increasing to decreasing.

To determine the interval where the function is decreasing, consider the derivative:$<br> -2x + 6 < 0$.
Solving gives $-2x < -6$, resulting in $x > 3$.

Therefore, the function is decreasing for $x > 3$.

The correct answer is: $x > 3$