Function transformations - what went wrong here... 
Looking at (b) only,

$$ A(2,4), \quad B(4,0), \quad  C(8,4) $$
If I follow that answer in words，translation by the vector (3,0)
I know the graph will shift to the RIGHT by 3 units, so using the coordinates,
we get
$$ A(5,4), \quad B(7,0), \quad  C(11,4) $$
then if we reflect to the $y$-axis, we get
$$ A(-5,4), \quad B(-7,0), \quad  C(-11,4) $$.
However, if "shift to the RIGHT by 3 units", using function notation,
we should use
$$
f(x) \mapsto f(x-3)
$$
Using this we should get
$$ A(-1,4), \quad B(1,0), \quad  C(5,4) $$,
then we reflect again,
$$ A(1,4), \quad B(-1,0), \quad  C(-5,4) $$,
which is the correct answer. (obviously, the $x$ coordinate for $c$ is wrong)
WHY is the first method wrong????
Just moving the points to the right by adding $x$ coordinate 3,
then reflect gets the wrong answers, this is just confusing......
 A: You have got the order of operations mixed up. Indeed, $f(3-x) = f(-(x-3))$. However, this implies that

*

*you replace $x$ by $-x$, or graphically: reflect around the $y$-axis.

*you replace $x$ by $(x-3)$, or graphically: shift $3$ units to the right.

If we follow your order of operations, we get $f(x) \rightarrow f(x-3) \rightarrow f(-x-3)$.
It is also possible to solve this question without rewriting the function.

*

*replace $x$ by $x+3$, or graphically: shift $3$ units to the left.

*replace $x$ by $-x$, or graphically: reflect around the $y$-axis.

Both results yield $f(3-x)$.
A: The description underlined in purple is actually the two transformations in the wrong order.
Translation by $\binom{3}{0}$ results in $f(x-3)$.
Now following this with a reflection in the $y$ axis results in $f(-x-3)$.
This is because in order to perform the reflection, every $x$ in the previous expression must be replaced by $-x$. This is not the same as replacing every $x-3$ with $-(x-3)$.
On the other hand, when you write $f(3-x)=f(-(x-3))$, consider firstly a reflection in the $y$ axis followed by a translation by $\binom{3}{0}$
As a sequence, $f(x)\rightarrow f(-x)\rightarrow f(-(x-3))$ and sure enough this gives the correct graph (except for the error with point $C$ of course).
