algebra 2 graphing parabola

Question

A ball is thrown directly upward from a height of 4 ft with an initial velocity of 20 ft/sec. The function s(t)=−16t2+20t+4 gives the height of the​ ball, in​ feet, t seconds after it has been thrown. Determine the time at which the ball reaches its maximum height and find the maximum height.

Answer 1

h't=-32t +20 = 0

t = 20/32= 5/8 second to reach max height

max h= h(5/8)= -16(5/8)^2 +20(5/8) +4

= -16(25/64) + 25/2 + 4

= --25/4 +25/2 +4

= 25/4 + 4

= (25 + 16)/4

= 41/4

= 10 1/4 feet

but you seem to be in an algebra class not calculus

so alternative method is rewrite the equation in vertex form

h= -16t^2 +20t + 4

factor out -16

-16(t^2 - 5t/4 ) + 4

complete the square

= -16(t^2 -5t/4 + (5/8)^2) + 4 + 16(5/8)^2

= -16(t-5/8)^2 + 4 + 25/4

= -16(t-5/8)^2 + 41/4

where vertex = (5/8, 41/4)

41/4 = max, 5/8 = time at max

Answer 2

The graph of this function is a downward opening parabola, so the max height occurs at the vertex. One way to find the vertex is to realize that the vertex lies on the axis of symmetry and the equation for the axis of symmetry is x = -b/2a. For this problem a = -16 and b = 20. So x = -20/-32 = 5/8. So after 5/8 seconds the ball will be at its max. To find the actual height of the ball at that time evaluate s(5/8).

s(5/8) = -16(5/8)²+20(5/8) + 4.

Evaluation left for you.

You can check your work here:

desmos.com/calculator/al9drf7pcx