Graph of $\log(3-x)$

Question

So if $y=\log(3-x) = \log(-x+3)$ then you reflect $\log(x)$ in the $y$ axis to get $\log(-x)$.

Then because it is $+3$ inside brackets you then shift to the left by $3$ giving an asymptote of $x=-3$ and the graph crossing the $x$ axis at $(-4,0)$.

However this does not work. The answer shows the $+3$ in the bracket shifting the curve to the right by $3$ giving an asymptote of $x=3$ and the curve crossing the $x$ axis at $(2,0)$.

Why does it do this? Can anyone please explain?

Answer 1

There is a small set of algebraic operations that correspond to geometric transformations:

When we have the graph of a function $y = f(x)$...

Shifting:

• The substitution $x \mapsto x - h$ shifts a graph $h$ units to the right (that'd be left, if $h$ is negative)

• The substitution $y \mapsto y - k$ shifts a graph $k$ units down (up if $k$ is negative)

Reflecting:

• The substitution $x \mapsto -x$ reflects the graph across the $y$-axis; a "left/right flip"

• The substitution $y \mapsto -y$ reflects the graph across the $x$-axis; an "up/down flip"

Now, here's the key thing: These transformations have to be written exactly like this, only replacing $x$ or $y$ with something.

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So, when we break down $y = \ln(3 - x)$ as you have...

$$ y = \ln(x) \xrightarrow{x\ \mapsto\ -x} y = \ln(-x) \longrightarrow y = \ln(-x + 3) $$

the last transformation, $-x \mapsto -x + 3$, is not one of our basic transformations: We are adding $3$ to $-x$, not $x$. As written, we simply can't recognize this as corresponding to any of our basic transformations. But, if we think about it a little differently...

$$ y = \ln(x) \xrightarrow{{x}\ \mapsto\ -x} y = \ln(-\color{red}{x}) \xrightarrow{\color{red}{x}\ \mapsto\ \color{red}{x - 3}} y = \ln\bigl(-(\color{red}{x - 3})\bigr) = \ln(-x + 3) $$

which we recognize as the sequence of transformations 1) Flip the graph left/right, and 2) Shift to the right 3 units.

There is an alternative:

$$ y = \ln(x) \xrightarrow{x\ \mapsto\ x + 3} y = \ln(x + 3) \xrightarrow{x\ \mapsto\ -x} y = \ln(-x + 3) $$

so we see the transformation can also be achieved by 1) Shifting the graph $3$ units left, then 2) Flipping left and right.

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So, long story short: To recognize a graph as the transformation of another graph, you have to figure out how to only use substitutions like "add this to $x$", or "make $x$ negative", not adding things to $-x$, or $2x$, etc.

Answer 2

Note that

• $3-x$ put the origin in x=3 (right translation) and reverse axis direction
• thus the vertical asympthote is at $x=3$ and the root is at $x=2$

Answer 3

Start with $y=\log(x)$. To shift this left three units, replace "$x$" with "$x+3$". Now you have $y=\log(x+3)$.

Now reflect over the $y$-axis. To do this, replace "$x$" with "$-x$". Now you have $y=\log(-x+3)$.

The order that the horizontal graph transformations happen is opposite from what you might think by the order of operations. If you first do "$x\mapsto-x$ and then do $x\mapsto x+3$, you get $\log(x)\mapsto\log(-x)\mapsto\log(-(x+3))$ which is not what you set out with.