Solve the Quadratic Inequality: y = -1/7x^2 - 23/7x > 0

Look at the following function:

$y=-\frac{1}{7}x^2-2\frac{3}{7}x$

Determine for which values of $x$ the following is true:

$f(x) > 0$

First, we need to find the roots of the quadratic equation:

The quadratic is given by:
$y = -\frac{1}{7}x^2 - \frac{17}{7}x$

Setting $y = 0$ to find the $x$-intercepts (roots):
$-\frac{1}{7}x^2 - \frac{17}{7}x = 0$

Factor out the common factor, $-\frac{1}{7}x$:
$-\frac{1}{7}x(x + 17) = 0$

This gives the roots:
$x = 0$ and $x + 17 = 0 \rightarrow x = -17$

These roots divide the number line into intervals. We need to determine where $y > 0$. Because the coefficient of $x^2$ is negative, the parabola opens downward. The function will be positive between the roots.

Thus, we test the interval:
$(-17, 0)$

Since the parabola opens downward, the function $y > 0$ is true in the interval $-17 < x < 0$.

Therefore, the solution to the problem is $-17 < x < 0$, which corresponds to choice 3.