Find Increasing Intervals for y = (2x+10)(3-x): Quadratic Function Analysis

To solve this problem, we need to determine where the function $y = (2x + 10)(3 - x)$ is increasing by finding where its derivative is positive.

First, let's expand the function:

$y = (2x + 10)(3 - x) = 6x - 2x^2 + 30 - 10x$

$y = -2x^2 - 4x + 30$

Now, compute the derivative of $y$:

$y' = \frac{d}{dx}(-2x^2 - 4x + 30) = -4x - 4$

To find where the function is increasing, we set the derivative greater than zero:

$-4x - 4 > 0$

Solve the inequality:

• Add 4 to both sides: $-4x > 4$
• Divide by -4 (flip the inequality sign): $x < -1$

This gives us the interval where the function is increasing:

$x < -1$

Therefore, the function is increasing for $x < -1$.

Therefore, the correct answer choice is: $x < -1$.