<PRE><B>Question 725618</B>
The equation we will write will be of the form:
y = C + A*sin(B(x-D))
where
| A | is the amplitude
B is a number related to the period by the equation B = 2pi/period
C is the vertical shift
D is the phase (aka horizontal) shift.&lt;br&gt;
All we have to do is take the information from the problem and use it to figure out correct values for A, B, C and D.&lt;br&gt;
With a maximum temperature of 32 and a minimum of 25, the halfway point would be 28.5. This is the vertical shift. This is C.&lt;br&gt;
From the halfway point to the maximum (or minimum) is 3.5. This is the amplitude. This is A.&lt;br&gt;
The cycle repeats every 24 hours so the period is 24 hours. This makes
{{{B = 2pi/24 = pi/12}}}&lt;br&gt;
The normal sin function starts at the halfway point and then moves from there up to the maximum. We have to find where this should be in order for the graph to start at noon (as the problem instructs). So let&#39;s piece this together. At the y-axis, the time is noon (as the problem says it should be). One hour later, 1 pm, we have the maximum temperature. Twelve hours later, 1 am, we have the minimum temperature. 1/4 of a cycle later we will be back to the halfway point. (It is always 1/4 of a cycle from one key point to the next: from a maximum to the next halfway point, from the halfway point to the minimum, etc.) Since the period is 24 hours, 1/4 of a cycle is 6 hours. So 6 hours after the minimum we will be at the halfway point &lt;i&gt;on our way to the next maximum&lt;/i&gt;. This the the point we&#39;ve been looking for. It took us 1 hour to the maximum, 12 hours to the minimum and finally 6 hours to this point. All together it took 19 hours. This is our phase shift. This is D.&lt;br&gt;
We&#39;re finally ready to write our equation:
{{{y = 28.5 + 3.5*sin((pi/12)(x-19))}}}&lt;br&gt;
P.S. The problem says &quot;... find the equation ...&quot;. This is not possible. There is not just one equation that will work. There are literally an infinite number of equations that will work. In addition to the one we found we could use:&lt;ul&gt;&lt;li&gt;The same equation with the {{{pi/12}}} and (x-19) multiplied out.&lt;/li&gt;&lt;li&gt;The same equation with different phase shifts (D&#39;s). With a period of 24 hours we 19 plus or minus multiples of 24. For example our &quot;D&quot; could have been:&lt;ul&gt;&lt;li&gt;19+24 = 43&lt;/li&gt;&lt;li&gt;19-24 = -5&lt;/li&gt;&lt;li&gt;19+2*24 = 67&lt;/li&gt;&lt;li&gt;19-2*24 = -29&lt;/li&gt;&lt;li&gt;etc.&lt;/li&gt;&lt;/ul&gt;&lt;/li&gt;&lt;li&gt;The same equation with a negative A, -3.5, and a D of 7 (or 7 plus or minus multiples of 24)&lt;/li&gt;&lt;li&gt;Various combinations of these variations.&lt;/li&gt;&lt;/ul&gt;FWIW, here&#39;s what the graph looks like:
{{{graph(600, 600, -5, 35, -5, 35, 28.5 + 3.5*sin((pi/12)(x-19)))}}}
Note: The x values represent &quot;hours since noon&quot;.</PRE>