Graph y = -3sin(pi*x + 2) for one cycle

T

Timcago

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Apr 13, 2006
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Sketch the graph of the following function (one cycle only). Give the value of the amplitude, the phase shift, and the period.

. . .y = -3sin(pi*x + 2)

. . .pi*x + 2 = 2*pi -> x = (2*pi - 2/pi)
. . .pi*x + 2 = 0 -> x = -2/pi

. . .Period: (2*pi - 2/pi) - -2/pi = 2

. . .Period: 2, Amplitude: 3, Phase shift: -2/pi

I have no idea how to scale or how to graph this thing. Can anyone help?

As a side question, does y = (2/3*x - pi/3) have a amplitude or phase shift?
 
Does y = (2/3*x - pi/3) have a amplitude or phase shift?
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This is the equation of a line. It has neither amplitude nor phase shift.
 
thankyou, that settles that part

As for the tough question :shock:. Hopefully someone knows how to graph that.
 
I always irritated people with this sort of thing.

1) Look at it and decide what shape to start - sine or cosine.
2) Look at the leading coeficient. If it is negative, simply draw it upside down.
3) NOW worry about vertical shift. Label the center horizontal line for vertical shift instead of the x-axis.
4) NOW worry about amplitude. Label the upper and lower limits other than +1 and -1, if the amplitude is not 1.
5) NOW worry about phase shift. Instead of labeling the y-axis, label that line for the phase shift.
6) Finally, find the x-axis and y-axis and label them.

The ONLY problem with this methodology is that everything looks pretty much the same, visually. You have to check the markings to tell the difference.
 
The did not help :(

The problem is making an accurate scale and how the heck am i supposed to do that when my phase shift is -2/Pi
 
Timcago said:
The did not help
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If the step-by-step instructions were too far from anything you've learned to be helpful, then perhaps you could show/list/describe what you have done, so we can try to help you with whatever method you're supposed to be using. Thank you.

Timcago said:
The problem is making an accurate scale and how the heck am i supposed to do that when my phase shift is -2/Pi
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How would a shift change the scale? Just draw the curve as usual, and then re-draw the x-axis scale markings, shifted by the appropriate amount.

Eliz.
 
The period is 2. Which means that the distance from the start of the graph to the end is 2.

The phase shift is -2/Pi

So do i start with -2/Pi and go 2 places over on the number line so that my scale starts are -2/Pi and ends at 1.36?

When i do that and graph it on my t-89 according to that scale all i get is a straight line.

When i increase the scale from -2/Pi to 120 i can see the 2 maximums and the 3 zeros.

Am i to graph a straight line on my paper?
 
Since sine curves are not customarily straight lines, and since this one certainly isn't, it would not likely be correct to graph a straight line. Since we cannot see what you're doing on your calculator, it is difficult to reply to that.

If I may venture to say, it might be better if you learned how to do the graphing yourself (using steps such as were suggested earlier), instead of hoping the calculator gives you the right picture to copy down.

Eliz.
 
Period: 2, Amplitude: 3, Phase shift: -2/pi
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tkhunny wrote:

1) Look at it and decide what shape to start - sine or cosine.
2) Look at the leading coeficient. If it is negative, simply draw it upside down.
3) NOW worry about vertical shift. Label the center horizontal line for vertical shift instead of the x-axis.
4) NOW worry about amplitude. Label the upper and lower limits other than +1 and -1, if the amplitude is not 1.
5) NOW worry about phase shift. Instead of labeling the y-axis, label that line for the phase shift.
6) Finally, find the x-axis and y-axis and label them.
Click to expand...

Hi, timcago,

Let’s follow tk’s advice. 1) First, we have a sine curve. The “simple” sine curve starts at (0,0) and proceeds upward and to the right, into the 1st quadrant, up to (pi/2, 1) down thru the x-axis at (pi, 0), down to (3pi/2, -1), and back up to (2pi, 0). This completes one period, correct?

2) Since the leading coefficient is negative, flip this curve upside down (a reflection thru the x-axis).

3) There is no vertical shift/translation in this problem, so the x-axis remains the axis of symmetry.

4) The amplitude is 3, so stretch the high and low points of the curve from their positions at 1 and –1 to 3 and –3. (You now have the following points plotted: (0,0) (pi/2, -3) (pi, 0) (3pi/2, 3) and (2pi, 0).)

5) The phase shift is 2/pi TO THE RIGHT. Slide your whole graph to the right, along the x-axis.

6) Finally, adjust the x-scale for the period, 2 instead of 2pi. You now have the following points: (2/pi, 0) (.5 + 2/pi, -3) (1 + 2/pi, 0) (1.5 + 2/pi, 3) and (2 + 2/pi, 0).

Note: On your calculator, make sure you’re working in radians, not degrees. Set your window: x from 0 to 3, and y from –4 to 4, say. That should give you something to look at – but don’t rely on your calculator; learn the steps necessary to analyze what your equation should look like.

Hope that helps.
 
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