Question

4200 ft. apart with height of 546 ft above roadway...cable is 15ft above roadway...I have vertex at (0,15) and (2100, 546) , (-2100, 546) but can't get equation to work

Answer 1

gonna have to chase the question am i...... stay put! lol

Answer 2

those are some awfully big number, this could get messy :)

Answer 3

yes I've worked it out several times

Answer 4

your values are good...

(2100^2)a +(2100)b + c = 546

(-2100^2)a -(2100)b + c = 546

0a +0b + c = 15 we see the c = 15 here right?

Answer 5

(2100^2)a +(2100)b + c-c = 546 -15

(2100^2)a +(2100)b = 531

(-2100^2)a -(2100)b + c-c = 546-15

(2100^2)a -(2100)b = 531
---------------------------

531 -2100b
a = -----------
(2100)^2

substitute this into the next equation right?

(2100^2)a +(2100)b = 531
------------------------------

Answer 6

(2100^2)(531 -2100b)
-------------------- -(2100)b = 531
(2100)^2



531 -2100b - 2100b = 531

it says b = 0 :)

Answer 7

which is what you would expect it to be since we are on the y axis already and just raised by 15 right?

Answer 8

so we are using y-y1=c(x-x1)2
yes, y-intercept is (0, 15)

Answer 9

531
a = ----------
(2100)^2

\[y = \frac{531}{2100^2} x^2 + 15\]

Answer 10

see if this fits :)

Answer 11

when x = 2100 we get: 531 + 15 = 546 right?

Answer 12

come on..you know im right....tell me i did good :)

Answer 13

I was squaring 2100 and got lost in math...
YES you're right...AWESOME!!!!
THANKS ALOT!

Answer 14

I was comparing my work to yours...all different attempts
, lol

Answer 15

lol...... thanx :) you had it tight to begin with, dont let the numbers scare you ;)

Answer 16

Thanks!

Answer 17

you have time for another