Finding Increasing Intervals for y = (x+15)² + 6: Vertex Form Quadratic Analysis

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

• Step 1: Calculate the derivative of the function.
• Step 2: Determine the intervals where the derivative is positive.
• Step 3: Conclude where the function is increasing.

Now, let's work through each step in detail:

Step 1: Calculate the derivative.

Given the function $y = (x+15)^2 + 6$, calculate the derivative:

$y' = \frac{d}{dx}((x+15)^2 + 6) = \frac{d}{dx}(x+15)^2$

Using the power rule: $\frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx}$, where $u = x+15$, and $\frac{du}{dx} = 1$.

Thus, $y' = 2(x+15) \cdot 1 = 2(x+15)$.

Step 2: Determine where the derivative is positive.

Set the derivative greater than zero to find the increasing interval:

$2(x+15) > 0$.

Divide both sides by 2:

$x+15 > 0$.

Subtract 15 from both sides:

$x > -15$.

Step 3: Conclude where the function is increasing.

The interval where the function is increasing is where $x > -15$.

Therefore, the function $y = (x+15)^2 + 6$ is increasing for $x > -15$.