Find Intervals of Increase and Decrease for y = (2x - 2.25)²

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

• Step 1: Identify a simpler form of the equation and its derivative.
• Step 2: Find critical points where the derivative is zero.
• Step 3: Determine signs of the derivative on intervals around critical points.

Let's proceed with these steps:

Step 1: Our function is already in vertex form: $y = \left(2x - 2\frac{1}{4}\right)^2$. It's important to note that this term can be simplified as $y = (2x - 2.25)^2$, identifying the vertex at $x = 1.125$, which is $x = 1\frac{1}{8}$.

Step 2: Differentiate the function with respect to $x$. For $y = (2x - 2.25)^2$, use the chain rule:
$y' = 2(2x - 2.25) \cdot 2 = 4(2x - 2.25) = 8x - 9$.

Step 3: Set the derivative to zero to find critical points:
$8x - 9 = 0$ leads to $x = \frac{9}{8} = 1.125$ or $x = 1\frac{1}{8}$.

The function decreases to the left of this point and increases to the right. Specifically:

• If $x < 1\frac{1}{8}$, $8x - 9 < 0$, so $y$ is decreasing.
• If $x > 1\frac{1}{8}$, $8x - 9 > 0$, so $y$ is increasing.

Therefore, the intervals of increase and decrease are:

$\nearrow:x<1\frac{1}{8}$ (Increasing: to the left of the vertex),

$\searrow:x>1\frac{1}{8}$ (Decreasing: to the right of the vertex).