Find Intervals of Increase and Decrease: y = -(x + 7/8)² - 1.2

The given function is $y = -(x+\frac{7}{8})^2 - 1\frac{1}{5}$. This function is in the vertex form $y = a(x-h)^2 + k$, where $a = -1$, $h = -\frac{7}{8}$, and $k = -\frac{6}{5}$.

• The vertex of the quadratic function is $(- \frac{7}{8}, -\frac{6}{5})$.
• Since $a = -1$ which is less than 0, the parabola opens downward.
• For a parabola that opens downward, the function is increasing on the interval where $x < h$ and decreasing on the interval where $x > h$.
• Thus, the function increasing on $(- \infty, -\frac{7}{8})$ and decreasing on $(- \frac{7}{8}, \infty)$.

Therefore, the correct intervals are:
$\nearrow: x < -\frac{7}{8}$
$\searrow: x > -\frac{7}{8}$

Therefore, the final intervals of increase and decrease are:

$\searrow: x > -\frac{7}{8}$
$\nearrow: x < -\frac{7}{8}$