Sketch Graphs Ofy=Asinn(T±Ε)±Bandy=Acosn(T±Ε)±B
Learn to sketch sine and cosine graphs with a vertical shift for Queensland Year 11 Mathematical Methods (QCAA). Adding a constant lifts or lowers the whole wave without changing its shape.
You will combine amplitude, period and horizontal shift with this parameter, see the effect of the parameters, find the mean line, and work out the range as the mean plus or minus the amplitude in the QCAA course.
Every question with a fully worked solution.
- Sketch Graphs Ofy=Asinn(T±Ε)±Bandy=Acosn(T±Ε)±B - Video - Determining the Equation of a Sine and Cosine Graph Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the graph of \(y=a\sin n(t-e)+b\) or \(y=a\cos n(t-e)+b\) adds a vertical shift \(b\) to the sinusoid. The curve now oscillates about the mean line \(y=b\), with amplitude \(|a|\), period \(\dfrac{2\pi}{n}\), maximum \(b+|a|\), minimum \(b-|a|\) and range \([b-|a|,\ b+|a|]\).
The vertical shift \(b\) slides the whole wave up (or down). The curve oscillates about the horizontal mean line \(y=b\), which sits halfway between the maximum and the minimum.
The amplitude \(|a|\) is the distance from the mean line to a peak. So the maximum is \(b+|a|\) and the minimum is \(b-|a|\).
The range is therefore \([\,b-|a|,\ b+|a|\,]\). The period \(\dfrac{2\pi}{n}\) and phase shift \(e\) work exactly as before — the vertical shift does not change them.
For \(y=a\sin n(t-e)+b\) and \(y=a\cos n(t-e)+b\):
How to analyse \(y=a\sin n(t-e)+b\)
- Mean line: identify \(b\); the curve oscillates about \(y=b\).
- Amplitude: read \(|a|\); the maximum is \(b+|a|\) and the minimum is \(b-|a|\).
- Range: write \([\,b-|a|,\ b+|a|\,]\).
- Period and shift: period \(\dfrac{2\pi}{n}\); factor \(n(t-e)\) for the phase shift \(e\), then mark key points about the mean line.
Amplitude and mean:
| \(|a|\) | \(=\) | \(2\) |
| \(b\) | \(=\) | \(3\) |
Maximum — mean plus amplitude:
| \(\text{max}\) | \(=\) | \(b+|a|\) |
| \(=\) | \(3+2\) | |
| \(=\) | \(5\) |
Minimum — mean minus amplitude:
| \(\text{min}\) | \(=\) | \(b-|a|\) |
| \(=\) | \(3-2\) | |
| \(=\) | \(1\) |
Amplitude \(2\); maximum \(5\); minimum \(1\).
Amplitude and mean:
| \(|a|\) | \(=\) | \(3\) |
| \(b\) | \(=\) | \(-1\) |
Range — \([\,b-|a|,\ b+|a|\,]\):
| \(\text{range}\) | \(=\) | \([\,-1-3,\ -1+3\,]\) |
| \(=\) | \([-4,\ 2]\) |
Period — \(n=2\):
| \(\text{period}\) | \(=\) | \(\dfrac{2\pi}{2}\) |
| \(=\) | \(\pi\) |
Range \([-4,\,2]\); period \(\pi\).
Amplitude and mean:
| \(|a|\) | \(=\) | \(4\) |
| \(b\) | \(=\) | \(1\) |
Range:
| \(\text{range}\) | \(=\) | \([\,1-4,\ 1+4\,]\) |
| \(=\) | \([-3,\ 5]\) |
Shift — argument \(\left(t-\dfrac{\pi}{3}\right)\), so \(e=\dfrac{\pi}{3}\):
| \(\text{shift}\) | \(=\) | \(\dfrac{\pi}{3}\ \text{right}\) |
Amplitude \(4\), mean \(y=1\), range \([-3,\,5]\), shift \(\dfrac{\pi}{3}\) right.
Mean line — average of max and min:
| \(b\) | \(=\) | \(\dfrac{7+1}{2}\) |
| \(=\) | \(4\) |
Amplitude — half the difference:
| \(|a|\) | \(=\) | \(\dfrac{7-1}{2}\) |
| \(=\) | \(3\) |
Frequency — from the period:
| \(n\) | \(=\) | \(\dfrac{2\pi}{\pi}\) |
| \(=\) | \(2\) |
\(y=3\sin 2t+4\).
Common pitfalls
Frequently asked questions
What is the mean line of a sinusoid?
It is the horizontal line \(y=b\) halfway between the maximum and minimum; the curve oscillates about it. You find it with \(b=\dfrac{\text{max}+\text{min}}{2}\).
How do you find the maximum and minimum of y = a sin n t + b?
The maximum is \(b+|a|\) and the minimum is \(b-|a|\): start at the mean line \(y=b\) and go up or down by the amplitude \(|a|\).
What is the range of a sinusoid with a vertical shift?
The range is \([\,b-|a|,\ b+|a|\,]\) — from the minimum to the maximum, smaller value written first.
Does the vertical shift change the period?
No. The period is \(\dfrac{2\pi}{n}\) regardless of \(b\); the vertical shift only raises or lowers the curve.
How do I get the amplitude from a graph with a shift?
Take half the difference between the peak and the trough: \(|a|=\dfrac{\text{max}-\text{min}}{2}\).