This problem follows a very similar solution to the last in that we will first identify the key features of the function from the equation before beginning the sketch. To do this, we will once again start by factoring the inner portion of the function:

\[\mu(t)={\color{{orange}}7}{\color{{red}}-\dfrac{ 1 }{ 2 }}\sin\left({\color{{blue}}5\pi}\left(t-{\color{{green}}\frac{ 1 }{ 30 }}\right)\right)\]

The key features could be identified from this equation, but you can also perform one more reorganization if you so choose:

\[\mu(t)={\color{{red}}-\dfrac{ 1 }{ 2 }}\sin\left({\color{{blue}}5\pi}\left(t-{\color{{green}}\frac{ 1 }{ 30 }}\right)\right){\color{{orange}}+7}\]

While doing so is not strictly required, it may help you in identifying the key values more accurately.

• Amplitude: \( \left| {\color{ red }-\frac{ 1 }{ 2 }} \right| =\frac{ 1 }{ 2 }\)
• Phase Shift: \({\color{ green }\frac{{1}}{{30}} }\)
• Period: \(P=\frac{2\pi}{ {\color{ blue }5\pi } }=\frac{{2}}{{5}}\)
• Midline/Average Value: \({\color{ orange }+7}\)

While much of this is exactly what we need to graph, as you may have noticed from the previous problem, we also need to calculate the quarter and half period values offset from the phase shift:

• Quarter Period: \(\frac{ 2 }{ 20} = \frac{ 1 }{ 10 }\)
• Half Period: \(\frac{ 2 }{ 10 } = \frac{ 1 }{ 5}\)
• 3/4 Period: \(\frac{ 3 }{ 10 }\)

Half-Period Values (where the graph will intersect the midline):\[t=\frac{ 1 }{ 30 },\frac{ 1 }{ 30 }{\color{{red}}+\frac{ 1 }{ 5 }}, \frac{ 1 }{ 30 }{\color{{red}}+ \frac{ 2 }{ 5 }},\dots \]

\[t=\frac{{1}}{{30}},\frac{{7}}{{30}}, \frac{{13}}{{30}},\dots\]

Quarter-Period Values (where the max/min values will fall):\[t=\frac{ 1 }{ 30 }{\color{{red}}+\frac{ 1 }{ 10 }},\frac{ 1 }{ 30 }{\color{{red}} + \frac{ 3 }{ 10 }},\dots\]

\[t = \frac{ 4 }{ 30 }, \frac{ 10 }{ 30 },\dots\]

\[t=\frac{ 2 }{{15}}, \frac{ 1 }{ 3 },\dots\]

Interestingly, while we definitely "should" simplify fractions, you may find it is easier if we leave them with a common denominator (30 in this case) and then put the simplified fraction in the final graph. 

Finally, while I am only calculating the first period's worth of points, it pays to remember that this function is periodic, so any other max/min or midline intersection can be found by simply adding or subtracting the full period from one of the above values!!

Almost ready to sketch: We have our midline intersection values as well as the inputs for the max/min, but, we don't know if the first quarter value is a max or a min. If the graph is not reflected, it will be a max. However, if the graph is reflected, then the first quarter will land on a minimum. Checking back to the original function, the coefficient in front of the \(\sin\) is \(-\frac{{1}}{{2}}\), which is a reflection over the horizontal axis. Thus, the first quarter value is a minimum. The exact vertical value of the min and maxes will be the midline minus/plus the amplitude:

• Minimum: \(7 - \frac{{1}}{{2}}=6.5\)
• Maximum: \(7+\frac{{1}}{{2}}=7.5\)

Now we are ready to graph!

Step 1: Plot out the Quarter/Half Period Values

Step 2: Plot out the Midline

Step 3: Plot the Midline Intersections and Max/Min values

Step 4: Connect the Dots!