Quadratic Functions

Question

The height  (in feet) of a ball thrown by a child is

where  is the horizontal distance in feet from the point at which the ball is thrown.

(a) How high is the ball when it leaves the child's hand?  ___  feet

(b) What is the maximum height of the ball? ____   feet(

c) How far from the child does the ball strike the ground? Round your answers to the nearest 0.01.  ____  feet

Answer 1

Hi, Cassidy!

So in a quadratic function you have a vertex which when a < 0 is called the maximum. To find this point you first have to find the Axis of symmetry which is -b/2a.

Your equation is -1/16x²+ 6x + 3 =y

a) let x =0 and solve for y. That will be the height it leaves the child’s hand at horizontal distance 0

-1/16(0) + 6(0) +3 = y

0+0+3=y

3=y

b) so if ax² + bx + c = y then, a= -1/16 b= 6

Plug in those values into x= -b/2a ——> x= -6/(2(-1/16) )

x= -6 / -2/16 = -6 /-1/8 = -6 (-8/1) = 48.

x=48 means it went horizontallly 48 ft and reached the maximum height

Now you have to find the max height at time x

y=( -1/16) (48)² + 6(48) +3

y= -144 + 288 +3

y=147

Your vertex is (48,147)

so the max height is 147 ft.

c) To find this value, you must let y=0 this give you the x intercepts.

0= -1/16x² + 6x +3

Solve using the quadratic formula

x= (-b +- sqrt( b²-4ac) )/2a

x= (-6 +- sqrt ((6)² -4(-1/16)(3)) )/2(-1/16)

X=-6 + sqrt ( 36+ .75) / -1/8 X= -6 - sqrt ( 36+ .75) / -1/8

x= -.4974226 x= 96.49742261

Since x represents distance, you pick the positive answer

x= 96.50 ft