1. ## Story Problem...

An object is thrown from the origin of a coordinate system with the x-axis along the groundand the y-axis vertical. Its path, or tragectory, is given by the equation y=359x-13x^2. Find the objects maximum height.

2. Hello, epetrik!

An object is thrown from the origin of a coordinate system
with the x-axis along the ground and the y-axis vertical.
Its path, or trajectory, is given by the equation: . $y\:=\:359x-13x^2$
Find the object's maximum height.
Here is a non-Calculus solution . . .

The path is a down-opening parabola; its maximum is at its vertex.

The vertex of the parabola, $y \:=\:ax^2 + bx + c$, is at: . $x \:=\:\frac{\text{-}b}{2a}$

We have: . $a = \text{-}13,\;b = 359$
. . The vertex is at: . $x \:=\:\frac{\text{-}359}{2(\text{-}13)} \:=\:\frac{359}{26}$

Then: . $y \:=\: 359\left(\frac{359}{26}\right) - 13\left(\frac{359}{26}\right)^2 \:=\:\frac{128,\!881}{52}$

The maximum height is about: $2478.48$ units.