SOLUTION: Please Help Me!
1. The path of a ball thrown into the air from a height of 3 feet is given by y=-1/8x^2+x+3, where y is the height of the ball in feet at the horizontal distance
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Question 311192: Please Help Me!
1. The path of a ball thrown into the air from a height of 3 feet is given by y=-1/8x^2+x+3, where y is the height of the ball in feet at the horizontal distance of x feet from the thrower.
a. How high is the ball at its maximum height?
b. estimate the horizontal distance the ball traveled before hitting the ground.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
1. The path of a ball thrown into the air from a height of 3 feet is given by
y = x^2 + x + 3, where y is the height of the ball in feet at the
horizontal distance of x feet from the thrower.
:
a. How high is the ball at its maximum height?
Find the axis of symmetry; x=-b/(2a). In this equation: a=-1/8, b=1
x =
x =
x =
x = +4
Substitute 4 for x in the original equation to find the max height
y = -(4^2) + 1(4) + 3
y = -(16) + 1(4) + 3
y = -2 + 4 + 3
y = 5
:
Looks like this:
You can see y = 5 (max) when x=4
:
b. estimate the horizontal distance the ball traveled before hitting the ground.
Looks like slightly over 10 ft, doesn't it.
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