Find Decreasing Intervals for the Quadratic Function y = (x+10)(x-8)

To find where the function $y = (x+10)(x-8)$ is decreasing, let's follow these steps:

• Expand the Function:

First, let's expand the product:

$y = (x+10)(x-8) = x^2 + 2x - 80$

• Differentiate the Function:

Find the derivative $y'$ of the function:

$y' = \frac{d}{dx}(x^2 + 2x - 80) = 2x + 2$

• Find Critical Points:

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

$2x + 2 = 0$

$2x = -2$

$x = -1$

• Determine Increasing or Decreasing Intervals:

The critical point is $x = -1$. We now analyze the sign of the derivative $y' = 2x + 2$ across the interval:

- For $x < -1$, choose a test point like $x = -2$:

$y' = 2(-2) + 2 = -4 + 2 = -2$ (negative)

- For $x > -1$, choose a test point like $x = 0$:

$y' = 2(0) + 2 = 2$ (positive)

Since the derivative is negative for $x < -1$, the function is decreasing in this interval.

Conclusion: Therefore, the function is decreasing for $x < -1$.

The correct answer from the given choices is:

$x < -1$