I need help doing some trigonometry problems, one of which is y = -sin(6x - pie). I need to find the domain range, x & y intercepts, and I donít know how to graph it. Can someone help me soon?
Now then, you have .
For the domain, are there any values of for which would be undefined?
For the range, remember that the sine of an angle is the -coordinate of its corresponding point on the unit circle. So then what values can the sine function take?
The -intercept should be very easy to find. When does the graph hit the -axis? The answer is: when is zero. So, substitute 0 for and you will have . Evaluate.
The -intercept is a little more tricky. Set and solve:
Where is the sine function zero? (You should know this; if not, study your unit circle!)
Finally, for graphing, remember that the graph of has four basic characteristics:
- is the amplitude, which is half of the difference between the maximum and minimum of the graph
- affects the period of the graph--the horizontal distance between the start and end of each "cycle" (i.e., how far you can move along the horizontal axis before the graph starts repeating again). The period is given by .
- is the phase-shift: how far left or right the graph is shifted horizontally (this does not affect the shape of the graph)
- is the vertical shift, similar to (3).
I'd like to complement Reckoner's post and elaborate more on the graphic part. When graphing sine functions, it is always helpful to create tables.
1. First, start by setting up a table of domain and range for the parent function
x | sin x
0 | 0
90 | 1
180 | 0
270 | -1
360 | 0
This is a one wave, and remember that the period of sin and cos is 360 (tan is 180).
2. Apply the transformations of to the parent function.
3. Apply the translations to the transformed function.
REMEMBER: you have to apply transformations first (reflection, compression, stretch) before you apply any translation.
Also, go to Sine Function and try the java applet to get a better understanding of sine waves.