Answer :
We start with the function
[tex]$$
y = 4 \sin(3\theta - 60) + 1.
$$[/tex]
This function is in the form
[tex]$$
y = A \sin(B\theta - C) + D,
$$[/tex]
where:
- The amplitude is given by [tex]\( |A| \)[/tex].
- The period is given by [tex]\( \frac{360^\circ}{B} \)[/tex] (since the angles are in degrees).
- The horizontal (phase) shift is given by [tex]\( \frac{C}{B} \)[/tex].
- The vertical shift is given by [tex]\( D \)[/tex].
Let’s identify and compute each of these step by step.
1. Amplitude
Here, [tex]\( A = 4 \)[/tex]. The amplitude is
[tex]$$
\text{Amplitude} = |4| = 4.
$$[/tex]
2. Period
We have [tex]\( B = 3 \)[/tex]. Then the period is
[tex]$$
\text{Period} = \frac{360^\circ}{3} = 120^\circ.
$$[/tex]
3. Phase Shift
The constant inside the sine function is [tex]\( C = 60 \)[/tex]. The phase shift is
[tex]$$
\text{Phase shift} = \frac{60}{3} = 20^\circ.
$$[/tex]
This means the graph is shifted to the right by [tex]\( 20^\circ \)[/tex].
4. Vertical Shift
The vertical shift is given by [tex]\( D = 1 \)[/tex], so the whole graph is moved up by 1.
5. Marking the [tex]\( x \)[/tex] Values on the Scale for One Period
Since one period spans [tex]\( 120^\circ \)[/tex], we divide it into quarters to mark the key points. The quarter of a period is
[tex]$$
\frac{120^\circ}{4} = 30^\circ.
$$[/tex]
Starting at the phase shift, the key [tex]\( x \)[/tex]-values are:
- Start of the period:
[tex]\( x = 20^\circ \)[/tex].
- First quarter period:
[tex]\( x = 20^\circ + 30^\circ = 50^\circ \)[/tex].
- Half period:
[tex]\( x = 20^\circ + 60^\circ = 80^\circ \)[/tex].
- Three quarters period:
[tex]\( x = 20^\circ + 90^\circ = 110^\circ \)[/tex].
- End of the period:
[tex]\( x = 20^\circ + 120^\circ = 140^\circ \)[/tex].
---
Summary of the Results:
- Amplitude: [tex]\(4\)[/tex]
- Period: [tex]\(120^\circ\)[/tex]
- Phase Shift: [tex]\(20^\circ\)[/tex] to the right
- Vertical Shift: [tex]\(1\)[/tex] (up)
- Key [tex]\( x \)[/tex]-values over one period:
[tex]\(20^\circ\)[/tex], [tex]\(50^\circ\)[/tex], [tex]\(80^\circ\)[/tex], [tex]\(110^\circ\)[/tex], [tex]\(140^\circ\)[/tex]
This completes the detailed step-by-step solution for the problem.
[tex]$$
y = 4 \sin(3\theta - 60) + 1.
$$[/tex]
This function is in the form
[tex]$$
y = A \sin(B\theta - C) + D,
$$[/tex]
where:
- The amplitude is given by [tex]\( |A| \)[/tex].
- The period is given by [tex]\( \frac{360^\circ}{B} \)[/tex] (since the angles are in degrees).
- The horizontal (phase) shift is given by [tex]\( \frac{C}{B} \)[/tex].
- The vertical shift is given by [tex]\( D \)[/tex].
Let’s identify and compute each of these step by step.
1. Amplitude
Here, [tex]\( A = 4 \)[/tex]. The amplitude is
[tex]$$
\text{Amplitude} = |4| = 4.
$$[/tex]
2. Period
We have [tex]\( B = 3 \)[/tex]. Then the period is
[tex]$$
\text{Period} = \frac{360^\circ}{3} = 120^\circ.
$$[/tex]
3. Phase Shift
The constant inside the sine function is [tex]\( C = 60 \)[/tex]. The phase shift is
[tex]$$
\text{Phase shift} = \frac{60}{3} = 20^\circ.
$$[/tex]
This means the graph is shifted to the right by [tex]\( 20^\circ \)[/tex].
4. Vertical Shift
The vertical shift is given by [tex]\( D = 1 \)[/tex], so the whole graph is moved up by 1.
5. Marking the [tex]\( x \)[/tex] Values on the Scale for One Period
Since one period spans [tex]\( 120^\circ \)[/tex], we divide it into quarters to mark the key points. The quarter of a period is
[tex]$$
\frac{120^\circ}{4} = 30^\circ.
$$[/tex]
Starting at the phase shift, the key [tex]\( x \)[/tex]-values are:
- Start of the period:
[tex]\( x = 20^\circ \)[/tex].
- First quarter period:
[tex]\( x = 20^\circ + 30^\circ = 50^\circ \)[/tex].
- Half period:
[tex]\( x = 20^\circ + 60^\circ = 80^\circ \)[/tex].
- Three quarters period:
[tex]\( x = 20^\circ + 90^\circ = 110^\circ \)[/tex].
- End of the period:
[tex]\( x = 20^\circ + 120^\circ = 140^\circ \)[/tex].
---
Summary of the Results:
- Amplitude: [tex]\(4\)[/tex]
- Period: [tex]\(120^\circ\)[/tex]
- Phase Shift: [tex]\(20^\circ\)[/tex] to the right
- Vertical Shift: [tex]\(1\)[/tex] (up)
- Key [tex]\( x \)[/tex]-values over one period:
[tex]\(20^\circ\)[/tex], [tex]\(50^\circ\)[/tex], [tex]\(80^\circ\)[/tex], [tex]\(110^\circ\)[/tex], [tex]\(140^\circ\)[/tex]
This completes the detailed step-by-step solution for the problem.
