plotting "key points" (max, min) of a phase-shifted sin function

[math_knight]

y = 3 sin 8(x+4) +5

Amplitude =3
Midline = 5
Max, Min = 5+|3|, 5-|3|
Period = 2pi/8 = 1/4 pi

So far, if you exclude the phase shift, not too bad to draw around origin. The key points (min, max, intercepts of midline) occur 4 times in one period starting with pi/2B, pi/B, 3pi/2B, and ending with 2pi/B

This comes out to 1/16 pi, 2/16 pi, 3/16 pi and 4/16 pi

Being a sin, it would intercept the Y-axis at 0, hit a maxima at 1/16 pi, hit zero again at 2/16 and so on....the problem is with the horizontal shift

The new points after shifting are -63/16 pi , - 62/16 pi , - 61/16 pi, - 60/16 pi right?

However the book's solution starts the graph drawing around the origin...nowhere near - 4. It goes one period to the left and one to the right, domain [-1/4 pi, 1/4 pi]

Is there a convenient way to draw a function that is shifted horizontally so far, but draw it starting at the origin? The book really didn't go into much detail. Using factors of the period maybe? Dividing the shift by the period? 4 / 1/4 pi? And then multiply that number by the key points (max, min) to see where max min occurs? Any help would be appreciated.

edit:a slightly different way to ask this is how do you find the minima and maxima occurring nearest to x = 0. The book states that maxima and minima, 8 and 2 occur at x=.12 and x=.516 that is WAY right of the shift of -4
edit 2: I understand that the graph goes on forever and repeats, but algebraically how do you get those max and min points?

 

[Dr.Peterson]

y = 3 sin 8(x+4) +5

Amplitude =3
Midline = 5
Max, Min = 5+|3|, 5-|3|
Period = 2pi/8 = 1/4 pi

So far, if you exclude the phase shift, not too bad to draw around origin. The key points (min, max, intercepts of midline) occur 4 times in one period starting with pi/2B, pi/B, 3pi/2B, and ending with 2pi/B

This comes out to 1/16 pi, 2/16 pi, 3/16 pi and 4/16 pi

Being a sin, it would intercept the Y-axis at 0, hit a maxima at 1/16 pi, hit zero again at 2/16 and so on....the problem is with the horizontal shift

The new points after shifting are -63/16 pi , - 62/16 pi , - 61/16 pi, - 60/16 pi right?

However the book's solution starts the graph drawing around the origin...nowhere near - 4. It goes one period to the left and one to the right, domain [-1/4 pi, 1/4 pi]

Is there a convenient way to draw a function that is shifted horizontally so far, but draw it starting at the origin? The book really didn't go into much detail. Using factors of the period maybe? Dividing the shift by the period? 4 / 1/4 pi? And then multiply that number by the key points (max, min) to see where max min occurs? Any help would be appreciated.

edit:a slightly different way to ask this is how do you find the minima and maxima occurring nearest to x = 0. The book states that maxima and minima, 8 and 2 occur at x=.12 and x=.516 that is WAY right of the shift of -4
edit 2: I understand that the graph goes on forever and repeats, but algebraically how do you get those max and min points?

First issue: You seem to think the shift is -4pi, not -4. I would start the first cycle at -4, the max at -4 + pi/16, and so on. Am I right?

Second issue: If the book shows cycles starting at -pi/4 and at 0, it can't be right! At x=0, y = 3 sin 8(0+4) +5 = 3 sin(32) + 5 = 6.65428, which is not on the midline! And at x = -pi/4, y = 3 sin 8(-pi/4 + 4) +5 = 4.78117. What you describe is two cycles before shifting. But the max and min you state do work, so maybe I'm just misinterpreting what you say about the graph.

Check that you copied the problem correctly, and looked at the right answer.

After we have the right answer, we can talk about how to graph it near zero. But the basic idea will be to find a cycle near zero, by finding how many cycles we need to go from the natural starting point.

 

[math_knight]

Hey, Dr. Peterson, thank you. I figured it out actually. I know he shift is -4 (not in terms of pi) that's why was so difficult...the period is in terms of pi but the shift isn't.

But I figured it out, and I'll post it here...as much to help me remember it as to help someone in case they stumble on it via search. To be fair the book does draw more than one period to the left and right of y-axis, but the book actually says "for one period, the graph starts at 0 and ends at 1/4 pi." Peculiar...I thought the graph should start "at rest"...which would be mid-line, or +- amplitude.

At Y- Intercept in fact Y = 6.5 or so

In any case, to solve, I needed the Round(x) function....basically....the question was "how many periods fit in a shift?"

4 / pi/4 = 5.09, more than 5 periods fit in that -4 shift.

So round(5.09) = 5 ; And since the shift is -4 and you want to get towards zero, add five periods

-4 + 5/4 pi = -.073009 <<--That's where the graph period starts. I guess conversely you could subtract .09 periods from zero so long as you use your calc to avoid rounding errors.

Now, since the period can be divided into 4 parts to get the key points (which I did already) : 1/16th pi, 2/16th pi, 3/16th pi, 4/16th pi, add those 1/16ths pi to -.073009. Each additional one 1/16th pi is a key point. (max, min, midline intercept). And you do in fact get 8 and 2 occur at x=.12 and x=.516 at 1/6ths pi intervals.

But yeah, the graph the book drew in the solution actually stops at 1/4 pi! (Y stops at around 6.5!) And the book actually says the period starts at 0....strange...I guess you could look at it like that....starts at 6.5 (y-intercept) and ends at 6.5?? That's technically a cycle? Oh well. Nonetheless I get the idea. I think on an exam, I would stick to starting the graph at rest, shifted left, starting at x= -073009. That's close to the origin and would make it look nice. Thanks again though Dr.

edit: the book does state to start the graph at 0. I guess using Y-intercepts, I guess it's not a big deal, but it's a bit of a confusing way to look at phase shifts perhaps, and you still have to do all that work to find the start of the graph away from x = 0.

 

[Dr.Peterson]

Hey, Dr. Peterson, thank you. I figured it out actually. I know he shift is -4 (not in terms of pi) that's why was so difficult...the period is in terms of pi but the shift isn't.

But I figured it out, and I'll post it here...as much to help me remember it as to help someone in case they stumble on it via search. To be fair the book does draw more than one period to the left and right of y-axis, but the book actually says "for one period, the graph starts at 0 and ends at 1/4 pi." Peculiar...I thought the graph should start "at rest"...which would be mid-line, or +- amplitude.

At Y- Intercept in fact Y = 6.5 or so

In any case, to solve, I needed the Round(x) function....basically....the question was "how many periods fit in a shift?"

4 / pi/4 = 5.09, more than 5 periods fit in that -4 shift.

So round(5.09) = 5 ; And since the shift is -4 and you want to get towards zero, add five periods

-4 + 5/4 pi = -.073009 <<--That's where the graph period starts. I guess conversely you could subtract .09 periods from zero so long as you use your calc to avoid rounding errors.

Now, since the period can be divided into 4 parts to get the key points (which I did already) : 1/16th pi, 2/16th pi, 3/16th pi, 4/16th pi, add those 1/16ths to -.073009. Each additional one 1/16th is a key point. (max, min, midline intercept). And you do in fact get 8 and 2 occur at x=.12 and x=.516 at 1/6ths pi intervals.

But yeah, the graph the book drew in the solution actually stops at 1/4 pi! (Y stops at around 6.5!) And the book actually says the period starts at 0....strange...I guess you could look at it like that....starts at 6.5 (y-intercept) and ends at 6.5?? That's technically a cycle? Oh well. Nonetheless I get the idea. I think on an exam, I would stick to starting the graph at rest, shifted left, starting at x= -073009. That's close to the origin and would make it look nice. Thanks again though Dr.

edit: the book does state to start the graph at 0. I guess using Y-intercepts, I guess it's not a big deal, but it's a bit of a confusing way to look at phase shifts perhaps, and you still have to do all that work to find the start of the graph away from x = 0.

Good work, and nicely explained.

There is some ambiguity about the word "cycle". It can be applied, as they seem to be doing, to any interval one period long, regardless of where it starts; or it can be applied to what might be called a basic or standard cycle (I don't know if those terms are used) that is a shift of the basic cycle of a sine or cosine, which is the natural thing to use when actually drawing the graph.