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I'm taking an online AP calculus course because my high school does not offer it. One of the questions on a practice quiz (for the online course, not a problem from an official AP practice test) is as follows:

Suppose a friend of yours gives you a graph of $y=f(x)$, and asks you to graph the function $y=-f(2(x-3))+4$. How would you go about doing this?

The choices are:

A. Start with the graph of $y=f(x)$, flip it over, squash it horizontally by a factor of 2, shift it 3 units to the right and 4 units up.

B. Start with the graph of $y=f(x)$, shift it 3 units to the right and 4 units up, then squash it horizontally by a factor of 2, and finally flip it over vertically.

C. Start with the graph of $y=f(x)$, squash it horizontally by a factor of 2, flip it over, shift it 3 units to the right and 4 units up.

D. Start with the graph of $y=f(x)$, shift it 3 units to the left and 4 units up, then squash it horizontally by a factor of 2, and finally flip it over vertically.

E. Start with the graph of $y=f(x)$, shift 4 units up, squash it horizontally by a factor of 2, flip it vertically, and finally shift it 3 units to the right.

The quiz says the correct answer is B. It gives the following "Feedback": "Remember to shift first, then stretch or squash, and then flip."

I think answer B is wrong. I think the correct answer should be C. A few drawings support my claim. I know that the order in which we carry out the transformations matters. I think B is wrong because if we shift horizontally before we squash horizontally, we actually need to shift SIX units right, not three. If we squash first, however, as in C, we only need to shift three units.

On the other hand, maybe the issue is exactly what "squash horizontally" is supposed to mean. My understanding is that when we transform $y=f(x)$ into $y=f(2x)$, the graph gets squashed only because the entire plane gets squashed: all points $(x,y)$ get moved to $(x/2,y)$, and this causes the shape of the graph to look squashed relative to the original. So we're squashing about the line $x=0$. I think the teacher is mistakenly using this phrase to mean "squash about the line $x=3$."

Who is right? Am I right that C could be the correct answer on some reasonable interpretation of "squash horizontally by a factor of 2"?

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  • $\begingroup$ "Flip it over"? Is that the language that is being used on the AP exams now? No talk of reflection through the appropriate axis? It certainly looks, to me at least, that A or C would be correct (shouldn't the "or" be exclusive on AP multi-choice parts??) but obviously not B. $\endgroup$ – Daniel W. Farlow Sep 22 '16 at 1:22
  • $\begingroup$ @DanielW.Farlow: this is a quiz problem from an online course, not a problem from an official AP exam. I get the sense my teacher doesn't know what she's talking about. The language she uses is very imprecise. $\endgroup$ – josh milligan Sep 22 '16 at 1:56
  • $\begingroup$ @DanielW.Farlow: and yes, I didn't even realize it but A gives the same result, too! I wonder what I should do... how do I tell the teacher she's wrong? $\endgroup$ – josh milligan Sep 22 '16 at 1:56
  • $\begingroup$ @joshmilligan Tell her she's wrong, but be tactful about it. "Hey Ms. ___, I think there may be an error for this question. It looks like A and C are actually both correct while B doesn't seem right." Then go on to explain your reasoning. If you are really worried about the teacher's ego or something like that, then I would try an email (which I just realized you said this is online...so that is obviously what you would do). Simply write out your thought process, and if she is half-decent, then she will be happy you wrote her. If you have problems...send her the link to this question! $\endgroup$ – Daniel W. Farlow Sep 22 '16 at 2:04
  • $\begingroup$ If only the school system were better... $\endgroup$ – Simply Beautiful Art Sep 22 '16 at 22:42
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Yes, there is an error. If you just consider $f(x) = x$ then the function given is:

$$y = -2(x-3) + 4 = -2x + 10$$

This modified function yields a line with $y$-intercept $10$.

The suggested quiz answer says:

Start with the graph of $y=f(x)$, shift it 3 units to the right and 4 units up, then squash it horizontally by a factor of 2, and finally flip it over vertically.

If you follow these directions for $f(x) = x$, which has $y$-intercept $0$, then the shift 3 units to the right will create a $y$-intercept of $-3$, the shift of 4 units up will create a $y$-intercept of $1$, the horizontal squishing will not change the $y$-intercept of $1$, and the vertical flip will not move the $y$-intercept to $10$. And so the suggested answer is wrong, as you suspected.

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  • $\begingroup$ "the horizontal squishing will not change the y-intercept of 1" -- but this is part of what's at issue, right? it seems the teacher and I are using different definitions of "horizontal squashing." I see how B could be right if she interprets squashing as happening about the line $x=3$, but I thought we always use horizontal compression to mean about the line $x=0$.... $\endgroup$ – josh milligan Sep 22 '16 at 0:54
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The transformations can be split into two independent sets, the horizontal (inside $f$) and vertical (outside $f$) ones. Within each set the order of transformations must be fixed, but otherwise they can be weaved into each other.

This also means we can unweave the options for transforming $f(x)$ to $-f(2(x-3))+4$ so the horizontal ones come first. Denote the elementary transformations as follows:

  • P: "squash horizontally by a factor of 2" ($x\to2x$)
  • Q: "shift right 3 units" ($x\to x-3$)
  • R: "flip graph vertically" ($y\to-y$)
  • S: "shift up 4 units" ($y\to y+4$)

P and Q are horizontal transformations, while R and S are vertical ones.

For the transformation $x\to a(x-b)$, if we are given the elementary transformations "squash horizontally by factor $a$" and "shift right $b$ units" we should apply the squashing first; if we reverse the operations the transformation becomes $x\to ax-b$. Similarly, reversing "scale vertically by factor $a$" and "shift up $b$ units" turns $y\to ay+b$ into $y\to a(y+b)$.

From this, we see that $PQRS$ is a correct sequence for this problem. The given options' sequences are below, and I rearrange them after each arrow so that the horizontal transformations come first:

  • A: $RPQS\to PQRS$
  • B: $QSPR\to QPSR$
  • C: $PRQS\to PQRS$
  • D: $(-Q)SPR\to (-Q)PSR$
  • E: $SPRQ\to PQSR$

We see that two options, A and C, produce the required transformation. B is wrong, as you reasoned out.

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  • $\begingroup$ "scaling should always happen before translation when transforming a graph" ... no, you can translate before you scale, so long as you realize that what you translate by depends on the order. $f(2(x-3))$ can be obtained by either compressing first, then shifting 3 right, OR shifting 6 right, then compressing. at least as far as I can tell. $\endgroup$ – josh milligan Sep 22 '16 at 2:00
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I don’t think that any of us would have a question about $-f(2x)+4$ : that’s what you’d get by squashing and flipping, and then moving up by $4$. The effect of replacing $x$ by $x-3$ is to shift the squashed, flipped figure $3$ units to the right. So C looks right to me.

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Suppose $f(x) = (0,0), (2,5), (4,-1)$

$y = -f(2(x-3)) - 4 = (3,-4), (4,-9), (5,-3)$

enter image description here

That looks to me like squish (factor of 2), shift horizontally (right 3 units), flip about the x axis, shift vertically (down 4 units).

But my brain might not work the same as yours. You could say that is flip then squish then shift, and you would get the same picture. And both would be right.

or squish then flip then shift.

But the flipping definitely comes before the vertical shift. and the squish comes before the horizontal shift.

I made a transcription error. It should be $y = - f(2(x-3)) + 4$ and not -4, but I don't care to redo my graphics, and doesn't really change the message.

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Try reflecting the function across the $y=x$ line (i.e. switch the xs and ys). You should end up with this function (assuming that we only do this for a one-to-one portion of $f$s domain). In other wordds, $f$ is a piece-wise function with one-to-one pieces.

$$x = \frac{1}{\color{red}2}f^{-1}\bigg(\frac{1}{-1}(y-4)\bigg)+\color{red}3$$

It should now be obvious why the compression by a factor of $2$ needs to occur before the shift of $3$ units.

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