Question

Y= -16x^2+190+0
A=
B=
C=
graph when done.

Answer 1

I think you have to do the quadratic formula

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

okay thx

Answer 6

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

Answer 7

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

Answer 8

jazy its 2a

Answer 9

@jazy

Answer 10

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

Answer 11

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

Answer 12

uughh this is sooo harrddd

Answer 13

ax^2 + b^2 + c
-16x^2 + 190 +0

what is a, b, and c?

Answer 14

-16
190
0

Answer 15

oh. i got -380

Answer 16

how do you get -380

Answer 17

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

Answer 18

also 'x' term is missing..

Answer 19

oh snap i didnt do the bottom hold on

Answer 20

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

Answer 21

i got 0.95

Answer 22

it right @jazy

Answer 23

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

Answer 24

you have it

Answer 25

yea thats when i got 11.87

Answer 26

then i pluged it in to the equation

Answer 27

right(: what was the 0.95 for though?

Answer 28

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