Find Decreasing Intervals: Analyzing y=(2x+10)(3-x)

To find the intervals where the function $y = (2x + 10)(3 - x)$ is decreasing, we need to follow a systematic approach:

• Step 1: Expand the function to standard form.
  $y = (2x + 10)(3 - x)$
  $y = 6 - 2x - 30 + 10x$
  $y = -2x^2 + 10x + 6$
• Step 2: Differentiate the function to find its derivative.
  $y' = \frac{d}{dx}(-2x^2 + 10x + 6) = -4x + 10$
• Step 3: Determine critical points by setting the derivative to zero and solving.
  $-4x + 10 = 0$
  $4x = 10$
  $x = \frac{5}{2}$
• Step 4: Identify where the derivative is negative, to find where the function is decreasing.
  Divide the number line into intervals based on this critical point: $x < \frac{5}{2}$ and $x > \frac{5}{2}$.
  Test the intervals in the derivative:
  For $x < \frac{5}{2}$, choose $x = 0$:
  $y'(0) = -4(0) + 10 = 10 > 0$ (positive, so increasing)
  For $x > \frac{5}{2}$, choose $x = 3$:
  $y'(3) = -4(3) + 10 = -12 + 10 = -2 < 0$ (negative, so decreasing)

Therefore, the function $y = (2x + 10)(3 - x)$ is decreasing for: $x > \frac{5}{2}$.

The critical point $\frac{5}{2}$ indicates a turning point from increasing to decreasing.

Thus, the correct choice and solution is: $x > -1$