In the first scenario, h will be a positive number
in the second drawing h = 0
in the third scenario h will be some negative number
.02 d^2 - 0.6 d +1 will have a nadir (a low point) at -b/2a = - -0.6/(2*.02) = 15 m = d
SUstitute this value in for d .02(15^2) - 0.6(15) + 1 = h = = -3.5 so this applies to the third picture.
DO the same thing for this equation a (low point) = - b/ 2a = 15 (again)
now substitute this value into the equation and find h to see what scenario this second equation fits....... you can do it! 