Algebra 1
posted by Kai on .
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?

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. 
Thank you so much, this really helped! :)

You're welcome :)

dont do it