Question 1183251
in a 24 hour day, the low is 80 degrees and the high is 90 degrees.
the low occurs at 6:00 am.
you don't say when the high occurs, however, if it is modeled by a sinusoidal function, it would have to be either sine or cosine.
tangent won't do because tangent doesn't have a low or a high, i.e. the low or high in the tangent function is plus or minus infinity.
i'll use sine function.
we want the period to be 24 hours.
that makes the frequency equal to 360 / 24 = 15.
the general form is y = a * sin(b * (x - c)) + d
a is the amplitude.
b is the frequency.
c is the horizontal shift.
d is the vertical shift.
the amplitude will be plus or minus 5 from the horizontal center line.
the vertical shift will make the horizontal center line 85 units above the x-axis.
so far, the equation looks like it will be:
y = 5 * sin(15 * (x-c)) + 85.
with a 0 horizontal shift, the equation becomes:
y = 5 * sin(15 * (x - 0)) + 85.
that looks like this:


<img src = "http://theo.x10hosting.com/2021/072401.jpg" >


i saw that, if i made the amplitude -5 instead of +5, that the low point would be at 6:00 am.
the equation would be y = -5 * sin(15 * (x - 0)) + 85.
that looks like this:


<img src = "http://theo.x10hosting.com/2021/072402.jpg" >


i then saw that, if i left the amplitude as +5, but shifted the graph to the left 12 units, that i would also get the low point at 6:00 am.
that equation would be y = 5 * sin(15 * (x + 12)) + 85.
that looks like this:


<img src = "http://theo.x10hosting.com/2021/072403.jpg" >


i could also have used the cosine function.
it give the same graph, with the same amplitude and frequency and vertical shift, but the horizontal shift would have to be different.


one possible equation would be:
y = -5 * cos(15 * (x - 6)) + 85
that looks like this:


<img src = "http://theo.x10hosting.com/2021/072404.jpg" >