Finding Intervals of Increase and Decrease for y = (2x - 1/2)(x - 2¼)

To solve the problem of finding intervals of increase and decrease for the given function, follow these steps:

• Step 1: Expand the function.
  We start by expanding $y = \left(2x - \frac{1}{2}\right)\left(x - 2\frac{1}{4}\right)$:
  $y = 2x \cdot x - 2x \cdot 2\frac{1}{4} - \frac{1}{2} \cdot x + \frac{1}{2} \cdot 2\frac{1}{4}$.
  Simplifying, we get:
  $y = 2x^2 - \frac{9}{2}x + \frac{1}{2} \cdot \frac{9}{4}$.
  Thus, $y = 2x^2 - \frac{9}{2}x + \frac{9}{8}$.

• Step 2: Differentiate the function.
  Differentiate $y = 2x^2 - \frac{9}{2}x + \frac{9}{8}$ with respect to $x$:
  $\frac{dy}{dx} = 4x - \frac{9}{2}$.

• Step 3: Find the critical points.
  Set the first derivative equal to zero:
  $4x - \frac{9}{2} = 0$.
  Solving for $x$, we get $4x = \frac{9}{2}$, hence $x = \frac{9}{8}$.

• Step 4: Use the first derivative test.
  Evaluate the sign of $\frac{dy}{dx}$ around the critical point $x = \frac{9}{8}$:
  - For $x < \frac{9}{8}$, choose $x = 1$: $\frac{dy}{dx} = 4(1) - \frac{9}{2} = \frac{8}{2} - \frac{9}{2} = -\frac{1}{2}$ (negative).
  - For $x > \frac{9}{8}$, choose $x = 2$: $\frac{dy}{dx} = 4(2) - \frac{9}{2} = \frac{8}{1} - \frac{9}{2} = \frac{7}{2}$ (positive).
  Thus, the function decreases when $x < \frac{9}{8}$ and increases when $x > \frac{9}{8}$.

Conclusively, the intervals of increase and decrease are:

$\searrow: x < 1\frac{1}{4}, \nearrow: x > 1\frac{1}{4}$.