Answer:
470.1
Step-by-step explanation:
Since this is a parabola that opens down, the maximum height is where the graph has a horizontal tangent line.
This is where the derivative=0
Since it is a parabola that opens down, we do not need to test the sides of the point where the derivative=0 as the vertex is a local maximum and the only point with a horizontal tangent.
y=[tex]-16x^2+146x+137[/tex]
y'=[tex]-32x+146[/tex]=0
-32x+146=0
x=[tex]\frac{146}{32}[/tex]=[tex]\frac{73}{16}[/tex] This is the x-coordinate of the vertex
Now plug this back into the original equation to find the maximum height.
f(73/16)=[tex]-16(\frac{73}{16})^2+146(\frac{73}{16} )[/tex]+137
=[tex]\frac{-73^2}{16} +666.125+137[/tex]
=-333.0625+666.125+137
=470.0625
To the nearest tenth this is 470.1
