Grant Erwin wrote:
This is actually a little worse than shop math. I have solved this; will confirm
the first correct answer.

Find an equation of a conic (2nd order function) which goes through the points
(0,0) (2,0.75) and (2.5,5) and is tangent to the X axis at (0,0) and is nearly
vertical at the point (2.5,5). Put it in the form y = F(x); in other words with
y alone on the left hand side of the equation.

Got any background to make this interesting? Looks pretty dull.

y= ax^2 + bx + c
0= 0 + 0 + c so c = 0
0.75 = 4a + 2b
5 = 6.25a + 2.5b

Two equations in two unknowns so put a or b in terms of the other and
plug it in and grind out an answer. Not sure if a "conic" might mean
you'd really need to start with the form y = ax^2 + bxy + cy^2 + dx + e
but I guess you'd get the same thing.