Question

Help please!!

A ball is tossed into the air from a window. Its path is described by y=-16x^2 +8x+12 where y represents the height of the ball in feet above
the ground x seconds after it is tossed. What is the vertex form of the
equation? What is the maximum height of the ball?
*this is what I have...
y=-16x^2 +8x+12
Find the x-coordinate of the vertex: x=- b/2a = 1/4
vertex form: y=-16 (x- 1/4)^2
The maximum height is 12 feet.

Thanks in advance!!

Answer 1

I know I am wrong I just don't know how...

Answer 2

@wolf1728 do you mind helping me again?

Answer 3

Gee, I just might be able to help you.

Answer 4

?

Answer 5

The vertex is at x = 1/4

Answer 6

Y value of vertex is
we plug in the value of .25
y =-16x^2 +8x+12
y= -16*.25*.25 +8*.25 +12
y= -1 +2 +12
y= 13

Answer 7

So the maximum height is 13 feet... right?

Answer 8

No I thought it was 12 feet (which you also calculated).

Answer 9

I thought I was wrong, though..

Answer 10

Just going by the formula, y represents the height of the ball after x seconds. So, at x = 0 seconds is the height immediately after it is thrown.
So, when x=0, the x terms equal zero and you are just left with 12.

Answer 11

Okay, I'm kinda confused. But thank you so much! You are a big help.

Answer 12

u r welcome.
Think I will have to logout for a bit - you'll be okay right?

Answer 13

Yes, thank you. :)

Answer 14

Okay c ya later.