graph-the-function-by-hand-not-by-plotting-points-but-by-starting-with-the-graph-of-one-of-the-st-22

Question

Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions given in Section $1.2,$ and then applying the appropriate transformations.

$y=4 \sin 3 x$

Carlos Pinilla · Accepted Answer

So here in red, you have the function. No, thanks. What? And we want to draw the function or four. Sign of three x books. So these these three there Obviously that first thing these three there is is some multiplying the period by three. Sort of. You got four period in about to buy. That will happen, you know, by thirds. So at this point, robbery and this one, I think he&#x27;s gonna yeah, over, um, under lucky and syrup. So these blue function would be sign off three books that three is shrinking like three along being. Why direction? Sort of To fight to play disfunction will have complete three cycles. Lend up for this ring is stretching along the exhilaration to the four son eggs searching, searching along the way. Stretch. Hello. What sort of and that point was sworn? No, a pint will correspond for that point. Right there should be minus one. She&#x27;s responsible for all this. You will stay the same. He&#x27;s got ways. Uh, I should have something. What is so the period is what reply by three. So that means a shrinking is three. This drinking alone along the exertion. He&#x27;s four is stretching along the while direction so the girls will look for signs of free X should be that green grass.

Heather Zimmers · Answer

to graph this function, we&#x27;re going to start by looking at the graph of the standard function y equals sign of X, so we know that it has a period of two pi. That&#x27;s the with that one cycle, and we knew it goes up to a height of one and down to a height of negative one. So it looks roughly like this. Now what happens to it if we multiply the X by 1/2 that represents a horizontal stretch, so it&#x27;s going to be two times as wide. So we say horizontal stretch by a factor of two that changes the period. The width of one cycle is now going to be four pi instead of two pi. So if you want to draw it out that far, we can still goes up to a height of one and down to a height of negative one. That didn&#x27;t change. So now we have the graph that is stretched out wider, like so

Amrita Bhasin · Answer

this question asks us to Graff, why is sign of 1/2 X by hand? We know we&#x27;re gonna start with the parent function, which is half of axe is sign of acts and we know we get the given function. Why? By stretching the graph of f of X by a factor of two horizontally. So this, we know, is the normal sign function right now stretching the graph of f of X by a factor of two horizontally, you should get something like that.

Heather Zimmers · Answer

our coolest a graph dysfunction, but we&#x27;re going to start with the graph of one of the standard functions Y equals co Sign X and then figure out the transformations that took place. So the coastline graph has a period of two. Pi goes up to one and down to negative one and looks roughly like this. It goes on in both directions, but I&#x27;m just showing part of it. Okay, now what happens if we multiply it by two When we multiply it by two. That&#x27;s a vertical stretch, so now it&#x27;s going to be twice as tall. So now it&#x27;s going to go up to a height of two and down to a height of negative, too. We say the amplitude changed. Okay, so we can go ahead and sketch that twice is tall. All right, now what happens if we multiply that by negative one? Well, that&#x27;s going to reflect it across the X axis. Flip it upside down, you could say. So we&#x27;ll take what we just drew and make that move still going up to a height of one at a height of two down to a height of negative, too, but upside down to direct sketch. Okay? Finally, the graph that we&#x27;ve been waiting for Why equals three minus to co sign X. All we have to do now is at the three and adding the three is going to shift the whole graph of three. So now instead of the top point being at a height of two, the top point is going to be at her five a height of five. And instead of the bottom point being at a height of negative two, it&#x27;s going to be at a height of one. So let&#x27;s make room for that. So the middle line is going to be shifted up. Three. I&#x27;ll just draw in that middle line there to give us some reference. Okay, so we have this upside down graph going down to a height of one and up to a height of five. There we go.

Heather Zimmers · Answer

Our goal is to figure out what the graph of this function looks like, but we&#x27;re going to start with the graph of one of the standard functions. So let&#x27;s start with the graph of y equals sine X. Okay, so goes up to a height of one down to a height of negative one. It has a period of two pi. We can draw a couple of the cycles here, and it looks like this. That&#x27;s the basic Siggraph. Now what do we want to do to it? Well, the pie in here that represents a horizontal shrink. So it&#x27;s going to be pi times narrower than it used to be, and the old period was two pi. If we divide that by pi, the new period is going to be too. So it&#x27;s going to be roughly three times narrower, So we&#x27;ll graph y equals sign of Pi X. I&#x27;m not going to make it using the exact same scale as I used on the last one going to change my scale because now that the period is too, I&#x27;m just going to put a two out here and a negative to out there, okay, so still goes up to a height of one and down to a height of negative one. But it&#x27;s narrow. Where now? Okay, finally, we want to add one to it and adding one to any graph will shift it up one. So we want to take what we just drew and shifted up one. So it was going up to a height of one and down to a height of negative one. Now it&#x27;s going up to a height of two and down to a height of zero. So we have or Siggraph looking like this shifted up one.

Heather Zimmers · Answer

to graft. Dysfunction will start by looking at the graph of the standard function y equals tangent X. And I&#x27;m just going to grab a small bit of the tangent function. So we know that it has a vertical Assen towed at Pi over two and another one at negative pi over two in between, we have pi over four negative pi over four. And so the standard graph looks something like this. Let me put in a couple of points here. We have a height of one and a height of negative one. It was something like, Let&#x27;s try that again. Took you to make it go through the appropriate heights. Something like that. Okay, now we have those repeated over and over and over again with vertical Lassen totes in between. So there would be another one over here and so on. Okay, Now, what are we going to do to that? Well, we are going to shift it to the right for and we&#x27;re also going to shrink it vertically. Make it 1/4 is tall, so let&#x27;s go ahead and make a graph for that. Okay? Shifting it to the right. Didn&#x27;t mean to say for him and to say pi over four, shifting it to the right pi over four. Let&#x27;s do that by shifting our vertical Assen totes. Okay, The vertical Lassen tote that was previously at negative pi over two is now at negative pi over four and the vertical sm the ass and tote that was previously a pi over two is now at three pi over four. Okay, now, halfway between those, we have a zero. And then instead of having points at a height of negative one and one like we did earlier, they&#x27;re going to be at a height of 1/4 and negative 1/4 because of the vertical shrink. So it&#x27;s going to look something like this, and we would repeat that to the left and the right. There will be more of those.

Problem

Graph the function by hand, not by plotting point…

Problem 12 Easy Difficulty

Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions given in Section $1.2,$ and then applying the appropriate transformations.

$y=4 \sin 3 x$

Answer

blue graph represents $\mathrm{f}(\mathrm{x}),$ red graph represents $y=4 \sin 3 x$

( GRAPH NOT AVAILABLE )

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Biocalculus Calculus for the Life Sciences

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James Stewart

Biocalculus Calculus for the Life Sciences

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

Precalculus Review - Intro

Functions on the Real Line - Overview

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blue graph represents $\mathrm{f}(\mathrm{x}),$ red graph represents $y=4 \sin 3 x$ ( GRAPH NOT AVAILABLE )

Biocalculus Calculus for the Life Sciences

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartBiocalculus Calculus for the Life Sciences

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

blue graph represents $\mathrm{f}(\mathrm{x}),$ red graph represents $y=4 \sin 3 x$

( GRAPH NOT AVAILABLE )

James Stewart

Biocalculus Calculus for the Life Sciences