## anonymous 5 years ago 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

1. amistre64

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

2. amistre64

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

3. anonymous

yes I've worked it out several times

4. amistre64

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?

5. amistre64

(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 ------------------------------

6. amistre64

(2100^2)(531 -2100b) -------------------- -(2100)b = 531 (2100)^2 531 -2100b - 2100b = 531 it says b = 0 :)

7. amistre64

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

8. anonymous

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

9. amistre64

531 a = ---------- (2100)^2 $y = \frac{531}{2100^2} x^2 + 15$

10. amistre64

see if this fits :)

11. amistre64

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

12. amistre64

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

13. anonymous

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

14. anonymous

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

15. amistre64

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

16. anonymous

Thanks!

17. anonymous

you have time for another

Find more explanations on OpenStudy