OH GOSH PLEASE HELPPPPPP MEEEEEEEE, I HAVE 2 QUESTIONS PLEASEEEEEEE, I NEED AN ANSWER WITHIN THE NEXT 2 HOURS PLEASEEEEEEEE

Question

1.) f(x) = -16x^2 + 24x +3

A volleyball player hits a ball over the net. She hits it from a starting height of  ______ feet. The ball reaches it maximum height of  ______ feet in ______ seconds. No one from the other team is able to hit the ball, so it hits the ground  ______seconds after being hit.

2.) Use the following equation to fill in the blanks: f(x) = -12(x-1)^2+16

A basketball player shot a free throw. He started his shot from a height of  ______ feet.  The ball reached its maximum height of  _______ feet in  _______seconds.  The ball missed the hoop, so it hit the ground  _______ seconds after being shot.

Answer 1

1) The starting height would be when x=0 as the ball has not been thrown yet so starting height is 3 ft.

The maximum height occurs at the vertex of this downward facing parabola. The x coordinate of the vertex is at -b/2a or -24/-32= 3/4. When x = 3/4 y = -16(3/4)^2 +24(3/4) +3= 12. ball reaches a maximum height of 12 feet after 3/4th of a second or .75 seconds

To find when the ball hits the ground solve 0= -16x^2 + 24 x +3 ( when the ball hits the ground height is zero )

multiply the equation by -1 to get 16x^2 -24x -3+ 0. Unfortunately this does not factor so you need the quadratic formula to solve x=24+sqroot[ (-24)^2 -4 x16 x-3)]32= (24+sqroot768 )/32= 1.616co ball hits the ground 1.616 seconds after it is hit. I omitted the - part of the quadratic formula since this would give us a negative time which does not make sense

2) You can use the same technique to answer part 2 Starting height is when x =0 which gives you a starting height of 4 feet. Find the x and y coordinates of the vertex. The y coordinate is the maximum height and the x coordinate is the time when this occurs. To find when the ball hits the ground set the equation to zero and solve. I will leave these for you to do

Answer 2

starting height = 3 feet

max height = 6 feet in 3/4 of a second

-16x^2 +24x + 3 is in the quadratic form ax^2+bx+ c

where b is initial velocity, and c= initial height

set that polynomial = 0 and solve for t to find when it hits the ground

x=0.75 + (1/32)sqr(24^2+12(16)) = about 1.6 seconds