Definition by tangent

Based on the information, the slope of the tangent to $(x, y)$ must be $m = \frac{\Delta y}{\Delta x} = \frac{0 - y}{(x+5)-x} = -\frac y 5.$ Thus we have the differential equation $\frac{dy}{dx} = -\frac y 5\ \ \therefore\ \ \frac {dy} y = -\frac {dx} 5\ \ \therefore\ \ \ln |y| = -\frac x 5 + c \\ \therefore\ \ f(x) = y = C\cdot e^{-x/5}.$ Plugging in the given points, $f(40) = C\cdot e^{-8} = (C\cdot e^{-4})\cdot e^{-4} = f(20)\cdot e^{-4} = 16\cdot e^{-4} \approx \boxed{0.293050}.$