SOLUTION: This question has already been asked but I don't understand what numbers you plug in to the equation f(x)=ax^2+bx+c. Thank you for your help. A suspension bridge with weight unif

Algebra.Com
Question 187836: This question has already been asked but I don't understand what numbers you plug in to the equation f(x)=ax^2+bx+c. Thank you for your help.
A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 100 meters from the center. (Assume that the road is level.): A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 100 meters from the center. (Assume that the road is level.)

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart.
You have points (-200,75),(0,0),(200,75
------------------------------------------------
The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. (Assume that the road is level.)
---
Use the three points in the equation f(x) = ax^2 + bx + c to generate three
equations to solve for a,b, and c.
---
(-200,75): a(-200)^2 + b(-200) + c = 75
(0,0)....: a(0)^2 + b(0) + c = 0
(200,75).: a(200)^2 + b(200) + c = 75
-------------------------------
Rearrange after noticing that c=0 in the 2nd equation:
200^2a -200b = 75
200^2a +200b = 75
--------------------------
Subtract 1st Eq. from 2nd to get:
400b = 0
b = 0
------------------
Substitutute into 200^2a -200b = 75 to solve for "a":
200^2a = 75
a = 75/(200^2)
---------------------
Equation of the parabola: f(x) = [75/200^2]x^2
--------------------
Then f(100) = [75/200^2][100^2] = 18.75 meters
=======================================================
======================
Cheers,
Stan H.

RELATED QUESTIONS

I don't understand this question: y=-11x2+bx+c has vertex (−2,5) a=11 b=? (answered by ikleyn)
(a) Let f(x): (-inf, 0) U (0 inf) --> R be defined by f(x) = x- (1/x) Show that f(x) has (answered by ikleyn)
I notice that this question has already been asked but I am still having a hard time... (answered by Maggy)
The product of two consecutive numbers is 72. Use a quadratic equation to find the two... (answered by phoihe001,Alan3354)
I can not understand how to factor trinomials in ax^2+bx+c format. I have been studying... (answered by mananth)
hey! I am having a lot of trouble with my year 12 maths and was wondering if anyone was (answered by Fombitz,rothauserc)
Use the quadratic equation to solve 6x^2+x=0. I know I have to use ax^2+bx+c=0. What... (answered by Alan3354)
Appreciate if you could explain a step by step in detail how to get the answer (C).... (answered by ikleyn,math_tutor2020,greenestamps)
Quadratic function in standard form whose graph passes though the given points. (1,-2)... (answered by stanbon)