If a ball is thrown upward with a velocity of 20 feet per second, its height (in feet) after x seconds is given by the function ƒ(x) = 20x − 16x2. What is the maximum height of the ball?

Question

Answer choices

1. A.
2. 6.25 ft

1. B.
2. 0.625 ft

1. C.
2. 1.25 ft

1. D.
2. 12.5 ft

Answer 1

You can do this problem using two different techniques for finding the vertex of a quadratic equation.

1) put the problem in Standard form of a quadratic: y=ax²+bx+c

f(x)=20x-16x² could become f(x)=-16x²+20x. if you just switch the locations of the terms on the same side.

a=-16 b=20 and c=0

Use the formula for the vertex from Standard form: vertex at (-b/(2a), f(-b/(2a))

So the vertex's x-value will be -b/(2a) = -(20)/(2*-16)=-20/-32=5/8. super fast.

then put 5/8 in for x in the original equation f(x)=-16x²+20x to find the max height, 6.25 feet.

2) Alternatively, you could find the zeros of the function (the x-intercepts-when it hits the ground) by factoring. Then average the x values to find the time of the max height.

So solve 0=20x-16x²

factor out the GCF, 4x first: 0=4x(5-4x)

Then use the zero product property to solve 4x=0 and 5-4x=0. This will give you x-intercepts of 0 and 5/4.

The average of 0 and 5/4 is 5/8. That is the time of the max height. So again, substitute into the original equation to find the max height of 6.25 feet.

Answer 2

h(x) = 20x-16x^2

take the derivative and set equal to zero, solve for x

h'(x) = 20x -32x = 0

x = 20/32 = 5/8 seconds

plug that into the original equation to get max height

h(5/8) = 20(5/8) - 16(5/8)^2 = 25/2 - 16(25/64) = 12.5 - 6.25 = 6.25 feet

A is correct