Posted by alice on Saturday, December 1, 2012 at 6:59pm.
NOTE - the formula should be:
H(t) = 0.013t^2 + 0.81t + 316 + 3.5 sin 2πt
since you have t^2 involved, the amount of variation depends on the year, but will always be roughly 2*3.5 = 7
since you have sin(2pi t) the period is 2pi/2pi = 1 year
max in April, min in Sep.
Don't know what graphing tool you use, but if you go to
http://rechneronline.de/function-graphs/
and enter
0.013x^2 + 0.81x + 316 + 3.5 *sin(2*pi*x)
for your function
with 0<=x<=5 and 300<=y<=350, the graph makes this quite clear.
how did you determine the maximum and the minimum?
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