Posted by Kai on Monday, February 28, 2011 at 9:25pm.
Question:
The height in feet above the ground of an arrow t seconds after it is shot can be modeled by y = 16t(squared) + 63t + 4. Can the arrow pass over a tree that is 68 feet tall? Explain.
Completely lost, could anyone help me out?

Algebra 1  Helper, Monday, February 28, 2011 at 10:31pm
y = height = 16t^2 + 63t + 4
The vertex of this parabola is called the maximum point.
Vertex = b/2a
Vertex = 63/(2(16))
Vertex = t value = 1.96875
So, the maximum value for t seconds is 1.96875
y = height = 16t^2 + 63t + 4
t = 1.96875
y = 16(1.96875)^2 + 63(1.96875) + 4
y = 16(3.875977) + 124.0313 + 4
y = 66.0157
So, the maximum height = 66.0157 feet
Use could also use derivatives to find the maximum value for t.

Algebra 1  Kai, Monday, February 28, 2011 at 11:00pm
Thank you so much, this really helped! :)

Algebra 1  Helper, Monday, February 28, 2011 at 11:02pm
You're welcome :)

Algebra 1  kkkkkkmmmmmm, Wednesday, March 23, 2016 at 6:12pm
dont do it
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