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"?

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.
