Find Increasing Intervals for y = (x-4)(-x+6): Quadratic Function Analysis

To solve this problem, we'll follow these steps:

• Step 1: Expand the quadratic function $y=(x-4)(-x+6)$ into standard form.
• Step 2: Calculate the derivative of the function to determine critical points.
• Step 3: Analyze the derivative to determine where the function increases or decreases.

Now, let's work through each step:

Step 1: Expand the function.

We have the function $y = (x-4)(-x+6)$.

Expanding gives us:

$y = x \cdot (-x + 6) - 4 \cdot (-x + 6) = -x^2 + 6x + 4x - 24 = -x^2 + 10x - 24$.

Step 2: Calculate the derivative.

The derivative of $y = -x^2 + 10x - 24$ is:

$y' = \frac{d}{dx}(-x^2 + 10x - 24) = -2x + 10$.

Step 3: Solve $y' = 0$ to find critical points.

Set $-2x + 10 = 0$ and solve for $x$:

$-2x + 10 = 0 \Rightarrow -2x = -10 \Rightarrow x = 5$.

This critical point divides the domain into two intervals: $x < 5$ and $x > 5$.

Analyze each interval using the derivative:

For $x < 5$, choose $x = 0$, $y' = -2(0) + 10 = 10$ (positive).

For $x > 5$, choose $x = 6$, $y' = -2(6) + 10 = -12 + 10 = -2$ (negative).

Since $y' > 0$ for $x < 5$ and $y' < 0$ for $x > 5$, the function is increasing on $x < 5$.

Therefore, the interval where the function is increasing is $x < 5$.