Example: Determine the Amplitude, Period, and Phase Shift of the function then sketch the graph over at least one period.
\[ T(x) = \dfrac{{1}}{{2}}\sin\left(2x-\frac{\pi}{{4}}\right) \]Solution
Let's work out all the features of the graph using what we know about the transformations. In order to do so, we first must factor the inner portion of the function:
\[T(x) = {\color{ red }\dfrac{ 1 }{ 2 } }\sin\left({\color{ blue }2 }\left(x-{\color{ green }\frac{\pi}{{8}}}\right)\right)\]
Having done that, we can now clearly identify the following:
- Amplitude \({\color{ red }\frac{{1}}{{2}}}\)
- Phase Shift: \({\color{ green }\frac{\pi}{ 8 } }\)
- Period: \(P=\frac{2\pi}{ {\color{ blue }2} }=\pi\)
Because we know the period, we can also determine some of the zeros (for sketching the graph). The zeros will begin at \(x=0\) shifted by the phase shift \(\omega = \frac{\pi}{{8}}\) and continue every half period \(\left(\frac{\pi}{{2}}\right)\), so the zeros occur at\[ x=\frac{\pi}{{8}}, \frac{5\pi}{{8}},\frac{9\pi}{{8}},\dots\]If the zeros are every half period, then the max/min occur every quarter period (exactly halfway between the zeros!). Thus, the max/min will be at \[x=\frac{3\pi}{{8}},\frac{7\pi}{{8}},\dots\] The sign of the coefficient of \(\sin\) is positive, so there are no reflections of the graph. Based on the basic \(y=\sin(x)\) graph, we know that initially the graph goes up to the right towards a maximum. Thus, in our case, the first quarter period at \(x=\frac{3\pi}{{8}}\) is a max of \(\frac{{1}}{{2}}\) and the second quarter period \(x=\frac{ 7\pi }{ 8 }\) is a min of \(-\frac{{1}}{{2}}\). We know that they are \(\pm\frac{{1}}{{2}}\) due to the amplitude being \(\frac{{1}}{{2}}\) and because there is no vertical shift.
Since we have the zeros, amplitude, and max/min over one full period, we can now sketch the graph:
When sketching your own graph, you should label the inputs spaced out along the