Find Increasing Intervals for y = (3x+3)(9-x): Derivative Analysis

To determine where the function $y = (3x+3)(9-x)$ is increasing, we will use the following steps:

• Step 1: Expand and simplify the quadratic expression.
• Step 2: Find the derivative of the function.
• Step 3: Determine where the derivative is positive.

Let's go through these steps:

Step 1: Expand and simplify the quadratic expression:
The function given is $y = (3x + 3)(9 - x)$.

We expand this expression:
$y = 3x \cdot 9 + 3x \cdot (-x) + 3 \cdot 9 + 3 \cdot (-x)$.
This simplifies to:
$y = 27x - 3x^2 + 27 - 3x$.
Combining like terms, we get the quadratic equation:
$y = -3x^2 + 24x + 27$.

Step 2: Find the derivative of the function:
The quadratic equation found is $y = -3x^2 + 24x + 27$.
Taking the derivative, we have:
$y' = \frac{d}{dx}(-3x^2 + 24x + 27) = -6x + 24$.

Step 3: Determine where the derivative is positive:
To find where the function is increasing, solve the inequality:
$y' = -6x + 24 > 0$.
This simplifies to:
$-6x > -24$.
Dividing both sides by -6 (and remembering to reverse the inequality sign) gives:
$x < 4$.

Thus, the function is increasing on the interval where $x < 4$.

Therefore, the solution to the problem is $x < 4$..