Is there an error in this AP Calculus quiz on function transformations? 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"?
 A: 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.
A: 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.
A: 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.
A: Suppose $f(x) = (0,0), (2,5), (4,-1)$
$y = -f(2(x-3)) - 4  = (3,-4), (4,-9), (5,-3)$

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.
A: 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.
