Find Intervals of Increase and Decrease: y = -1/3x² + 2x - 4

To find the intervals where the function $y = -\frac{1}{3}x^2 + 2x - 4$ is increasing or decreasing, we must first compute its derivative.

The derivative of the function with respect to $x$ is:

$y' = \frac{d}{dx} \left(-\frac{1}{3}x^2 + 2x - 4\right) = -\frac{2}{3}x + 2$

Next, we find the critical points by setting the derivative equal to zero:

$-\frac{2}{3}x + 2 = 0$

Solve for $x$:

$-\frac{2}{3}x = -2$

$x = 3$

The function has a critical point at $x = 3$. Since this is a quadratic function that opens downwards (as indicated by the negative coefficient of $x^2$), it is a parabola with a maximum at $x = 3$. This shows that the function is increasing on the interval $(-\infty, 3)$ and decreasing on the interval $(3, \infty)$.

Therefore, the intervals of increase and decrease of the function are:

$\nearrow: x < 3$ (increasing)

$\searrow: x > 3$ (decreasing)

Thus, the solution corresponds to:

$\searrow: x > 3 \\ \nearrow: x < 3$