Function transformations - what went wrong here...

Question

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......

Answer 1

You have got the order of operations mixed up. Indeed, $f(3-x) = f(-(x-3))$. However, this implies that

1. you replace $x$ by $-x$, or graphically: reflect around the $y$-axis.
2. 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.

1. replace $x$ by $x+3$, or graphically: shift $3$ units to the left.
2. replace $x$ by $-x$, or graphically: reflect around the $y$-axis.

Both results yield $f(3-x)$.

Answer 2

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).