## kiamonstaa Y= -16x^2+190+0 A= B= C= graph when done. one year ago one year ago

1. ilikephysics2

I think you have to do the quadratic formula

2. kiamonstaa

umm i forgot what that is.... the b= something right?

3. uzumakhi

in quadratic equation term with x^2 is A term with x is B term with constant is C

4. ilikephysics2

yes -b + or - sqrt(b^2-4ac) all over at 2a

5. kiamonstaa

okay thx

6. ilikephysics2

in this case you can solve for your variables a b c

7. jazy

$b = -b \pm \frac{ \sqrt{b^2 - 4ac} }{ 2 }$

8. ilikephysics2

jazy its 2a

9. ilikephysics2

@jazy

10. jazy

I thought I put that. But yes, over 2a. (:

11. jazy

$b = -b \pm \frac{\sqrt{b^2 - 4ac}} { 2a }$

12. kiamonstaa

uughh this is sooo harrddd

13. jazy

ax^2 + b^2 + c -16x^2 + 190 +0 what is a, b, and c?

14. kiamonstaa

-16 190 0

15. kiamonstaa

oh. i got -380

16. uzumakhi

how do you get -380

17. ilikephysics2

you have the steps down, i think you know what you are doing

18. hartnn

also 'x' term is missing..

19. kiamonstaa

oh snap i didnt do the bottom hold on

20. kiamonstaa

so... id plug in the answer i got from the eqation as x?

21. kiamonstaa

i got 0.95

22. sudharsa

it right @jazy

23. jazy

I think you had it right up to the -380. You have to divide by 2(-16) or -32. What is -380 /-32?

24. ilikephysics2

you have it

25. kiamonstaa

yea thats when i got 11.87

26. kiamonstaa

then i pluged it in to the equation

27. jazy

right(: what was the 0.95 for though?

28. kiamonstaa

y= -16(11.87^2)............. the 11.87 is substituted for the x