Find Intervals of Increase and Decrease for y = (x - 3/4)²

The function given is $y = \left(x - \frac{3}{4}\right)^2$. This is a quadratic function with its vertex (or minimum point) at $x = \frac{3}{4}$.

To find where the function is increasing or decreasing, follow these steps:

• Step 1: Differentiate $y$ with respect to $x$.

$y' = \frac{d}{dx}\left[\left(x - \frac{3}{4}\right)^2\right] = 2\left(x - \frac{3}{4}\right)$

• Step 2: Find the critical points by setting the derivative equal to zero.

$2\left(x - \frac{3}{4}\right) = 0 \quad \Rightarrow \quad x = \frac{3}{4}$

• Step 3: Determine the sign of $y'$ in the intervals divided by the critical point.

For $x < \frac{3}{4}$, $\left(x - \frac{3}{4}\right) < 0$ and hence $y' < 0$, indicating decreasing behavior.

For $x > \frac{3}{4}$, $\left(x - \frac{3}{4}\right) > 0$ and hence $y' > 0$, indicating increasing behavior.

Thus, the function decreases on the interval $(-\infty, \frac{3}{4})$, and increases on the interval $(\frac{3}{4}, \infty)$.

Consequently, the intervals of increase and decrease are:

$\searrow: x > \frac{3}{4} \\ \nearrow: x < \frac{3}{4}$