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?

Question

anonymous · Answer

vertex form:
$$f(x) = a(x - h)^2 + k$$ where (h, k) is the vertex; meaning k is the max height.

you'll want to complete the square:

$$y = -16(x^2 - x/2 + 3/4) = -16[(x - 1/4)^2 + 1/4] = -16(x - 1/4)^2 - 4$$

anonymous · Answer

somethign is wrong with my answer

anonymous · Answer

i found the mistake

anonymous · Answer

i wrote +3/4 instead of -3/4 inside the bracket.

let me know if you would like me to explain any of my steps ^_^

anonymous · Answer

$$y = -16(x^2 - x/2 - 3/4) = -16[(x - 1/4)^2 - 11/16] = -16(x - 1/4)^2 + 11$$

anonymous · Answer

@Euler271 could you explain how you did it please? This all just looks like a big confusing mess to me, :/

anonymous · Answer

@ehuman @Easyaspi314

anonymous · Answer

@ybarrap

anonymous · Answer

|dw:1382108891186:dw|