Thread: Gradient of parabola and specific max of y with 4 unknowns

1. Gradient of parabola and specific max of y with 4 unknowns

i am needing help with a Maths equation that i cannot figure out. i need the gradient at x=45 to equal 3.21609 (y'=-25.593*cos(pi/25x - 4pi/5)). i also need the max of the parabola to equal (x,55) where x is unknown. a , b and c are not given. it is really bugging me because i have wasted about 20 pages and still cannot find it.

where y'=0, x=-B/2A and y=(-AB^2)/(4A^2)+C
therefore 55 + (AB^2)/(4A^2)=C

y = Ax^2 +BX +C
y'= 2Ax + B

therefore at x = 45

3.21609 - 90A=B

using these i altered the equations but got nowhere. is the problem there (equations) if anyone can help?

2. equation found

if it helps anyone, i set a new axis to (0,0) and solved for y=AX^2 where y= drop in my case -40.383 (14.617-55). i let its y' = 3.21609 and got x = in terms of A. used that in y=AX^2 and got A. (3.21609^2)/4A = -40.383, a's cancel. then used y=AX^2 + BX +C and y' = 2AX + B where x was 45 and A was found value. this got B. then i used y=AX^2 + BX +C where y = 55 and x = -B/2A and got C!!!

final equation = -0.064x^2 + 8.98X -259.773

3. I am sorry

this is incoherent what aer you looking for?...what is $a$ in $ax^2+bx+c$?

4. i am looking for a, b and c in parabola general equation. if you know a bit about parabola's im sure that if you read the whole thing you will understand what i am saying.

i had to get the gradient of the derivitive to equal that of another equation, and have a maximum of a particular y-value. that was all i was given. no x value for the max, no a,b or c for the general equation...nothing!

5. Originally Posted by m-sternberg
i am needing help with a Maths equation that i cannot figure out. i need the gradient at x=45 to equal 3.21609 (y'=-25.593*cos(pi/25x - 4pi/5)). i also need the max of the parabola to equal (x,55) where x is unknown. a , b and c are not given. it is really bugging me because i have wasted about 20 pages and still cannot find it.

where y'=0, x=-B/2A and y=(-AB^2)/(4A^2)+C
therefore 55 + (AB^2)/(4A^2)=C

y = Ax^2 +BX +C
y'= 2Ax + B

therefore at x = 45

3.21609 - 90A=B

using these i altered the equations but got nowhere. is the problem there (equations) if anyone can help?
I'm sorry... I also have trouble understanding your actual question. From what I can tell...

* You want the derivative of y'=-25.593*cos(pi/25x - 4pi/5) when x = 45 to equal 3.21609

* You also want the local maximum point of a parabola to be (x, 55)

The problem here is we are confused at what you are actually trying to do, could you please post the questions and given information ONLY.

Note: This may help. The general equations for a parabola are:

General form: $y=a(x-h)^2+k$
Turning Point form: $y=ax^2+bx+c$