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.
GWE
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 + 2.5bTwo 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.
Background? Well, I'm musing on ways to make the center piece of a vibratory polishing bowl.
It's a little more complicated than your analysis, but here 'tis:
In my solution, the final equation looks like
y = Px + Q + SQRT(R*X^2 + S*X + T)
and for a curve going through (0,0) (2,0.75) and (2.5,5) and tangent to the x axis at (0,0) and very nearly vertical at (2.5,5) I got my 5 constants as:
P = -12.083 Q = 35.208 R = 142.007 ;; there may be slight roundoff or truncation errors in these S = -850.868 T = 1239.627
It took me an amazing amount of time flailing at this problem (which I didn't at all need to solve; I just like doing math sometimes) before I resorted to looking up how aircraft designers used to lay out airframes, and found that back in the 1950s those guys really knew about 2nd order equations and how to whip them into shape.
GWE
Your proposed solution seems to fail the (0,0) test...
0 = 0 + Q + SQRT(0 + 0 + T)
Not quite so simple, I think.
In your first equation, if b is not = 0 then the curve won't be tangent to the x axis at 0,0.
I did a bit of looking at some form of the equation for an elipse, which could have a tangent at 0,0 and nearly vertical at 2.5,5 but so far I haven't found any major/minor sizes for an ellipse that come close to working.
I'm not a math wiz but I can't follow what "conic (2nd order function)" means in this puzzle. The only conic section I can think of that could have a point with zero slope at 0,0 and a point nearly asymtotic to vertical (parallel to the y axis) is an ellipse. Such an equation has terms with both x^2 and y^2 so Grant's required y = f(x) form seems odd.
What kind of slope is "close to vertical" at the 3rd point?
Grant, do you have nonlinear regression software that helped in this puzzle?
Dave, remember Square roots have a + and a - value. Mike in BC
Well, that does seem to be a valid explanation for the ambiguity presented. I'm unclear whether in the problem statement "nearly vertical" is supposed to imply a vertical asymptote near the point (2.5,5) ? If not then a simple parabola would seem to be a cleaner and easier solution.
I get
48*x^2 + 12*y^2 + 290*x*y -845y =0or
y=[(845-290x)-SQRT(81796*x^2-490100*x+714025)]/24
as a solution where the tangent is vertical at (2.5,5)
Oops. Yeah, you're right. It can't be a parabola.
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