Find Intervals of Increase and Decrease for y = -x² + (3/4)x - 2

To find the intervals of increase and decrease for the function $y = -x^2 + \frac{3}{4}x - 2$, we proceed as follows:

• Step 1: Find the derivative of the function.
  The derivative $\frac{dy}{dx}$ of $y = -x^2 + \frac{3}{4}x - 2$ is calculated as:
$\frac{dy}{dx} = \frac{d}{dx}(-x^2 + \frac{3}{4}x - 2) = -2x + \frac{3}{4}$
• Step 2: Find the critical points by setting the derivative to zero.
  Setting the derivative equal to zero gives us:
$-2x + \frac{3}{4} = 0$
• Solve for $x$:
$-2x = -\frac{3}{4} \quad \Rightarrow \quad x = \frac{3}{8}$
• Step 3: Test intervals around the critical point $x = \frac{3}{8}$.
  Choosing a test point from each interval:
• For $x < \frac{3}{8}$, choose $x = 0$:
$\frac{dy}{dx} = -2(0) + \frac{3}{4} = \frac{3}{4} > 0$ The function is increasing in this interval.
• For $x > \frac{3}{8}$, choose $x = 1$:
$\frac{dy}{dx} = -2(1) + \frac{3}{4} = -2 + \frac{3}{4} = -\frac{5}{4} < 0$ The function is decreasing in this interval.
• Step 4: Summarize the intervals.
  The function $y = -x^2 + \frac{3}{4}x - 2$ is increasing on the interval $x < \frac{3}{8}$ and decreasing on the interval $x > \frac{3}{8}$.

Thus, the intervals of increase and decrease for the function are $\searrow:x > \frac{3}{8}~~|~\nearrow:x < \frac{3}{8}$.