7.

It must be (b). The given curve has a vertical tangent somewhere in the first quadrant. This eliminates (a), for which $${dy\over dx}$$ is always finite. The curve has negative slope at some points in the first quadrant (with y large), and positive slope at others (with y small). The line separating the two regions could not be x = y (which eliminates (c)), although it could be x y = 1 as in (b). Another way to eliminate (c) is to consider the second quadrant (x < 0, y > 0), where (c) would have $${dy\over dx} > 0$$. Some students considered a limit as $x \to -\infty$ and $y \to +\infty$. In (b), since $x y \to -\infty$, we would have $${dy\over dx} \to -1$$ (which fits the picture), while in (c) with $x \approx - y$ we would get $${dy\over dx} \approx \frac{-y^2}{-2 y} = \frac{y}{2} \to +\infty$$.