Sketching Graphs

y = ln(4 - x)

If you reflect the graph in the y-axis first, and then shift the graph 4 units to the left you get an incorrect answer

BUT

If you shift the graph 4 units to the left first, and then reflect the graph in the y-axis you get the correct answer

Why is that? And is there an order as to which operation you must do first?

(edited 9 years ago)

Reply 1

Smaug123

Original post by Inevitable

You started out with

y = \log(x)

. You then apply a shift of 4-to-the-left, so

x \mapsto x+4

, and you get

y = \log(x+4)

. Reflecting then takes

x \mapsto -x

, yielding

y = \log(-x + 4)

.

Your other way round should give

y = \log(-(x+4))

\log(-x-4)

. It could have been made to work if you shifted to the right instead of the left, to get

y = \log(-(x-4))

.

Remember that you're always doing stuff to x directly, not to whatever's in the brackets with x.

Reply 2

Inevitable

Original post by Smaug123

You started out with

y = \log(x)

. You then apply a shift of 4-to-the-left, so

x \mapsto x+4

, and you get

y = \log(x+4)

. Reflecting then takes

x \mapsto -x

, yielding

y = \log(-x + 4)

.

Your other way round should give

y = \log(-(x+4))

\log(-x-4)

. It could have been made to work if you shifted to the right instead of the left, to get

y = \log(-(x-4))

.

Remember that you're always doing stuff to x directly, not to whatever's in the brackets with x.

In the first case you added 4 onto x, and then you multiplied the x by -1 to give ln(-x +4)

In the second case you multiplied x by -1 to give -x and then you added 4 but you used brackets this time which affected x but also the value of 4 i.e. by doing this to x directly, we should get ln(-x + 4) but you applied the multiplication of -1 to 4 this time giving ln(-(x+4). You didn't do this directly to x this time like in the first case.

Can you explain a bit further?

(edited 9 years ago)

Reply 3

Smaug123

Original post by Inevitable

It still happened to x directly in the second case. Think of it this way: let

h(x) = -x

, then

log(h(x)) \mapsto \log(h(x+4)) = \log(-x-4)

. Does that help?

Reply 4

Inevitable

Original post by Smaug123

It still happened to x directly in the second case. Think of it this way: let

h(x) = -x

, then

log(h(x)) \mapsto \log(h(x+4)) = \log(-x-4)

. Does that help?

So you're trying to imply that there is an order which you must do the operations in and it is not always the case that the translation comes before the reflection and we can find this out by doing the following:

y = ln(4-x)

- two operations to this, there is a leftwards shift and a reflection in the y-axis

-----

Case 1: shift first, then reflection

Let

f(x) = lnx

shifting 4 units to the left first:

f(x+4) = ln(x+4)

reflecting in the y-axis

f((-x) + 4) = ln((-x) + 4) = ln(4 - x)

This is correct because

ln(4-x) = ln(4-x)

as our original question

-----------

Case 2: reflection first, then shift to the left by 4 units

Let

f(x) = lnx

Reflection first in the y-axis:

f(-x) = ln(-x)

Shifting left by 4 units:

f(-(x + 4)) = ln(-(x+4)) = ln(-x - 4)

This is wrong because

ln(-x - 4) \not=ln(4-x)

BUT, this would be correct if we shifted to the right instead of to the left after the reflection so:

f(-x) = ln(-x)

shifting right by 4 units

f(-(x - 4)) = ln(-(x-4)) = ln(-x + 4)

Hence,

ln(-x + 4) = ln(4 - x)

and we get what the original curve from the beginning meaning we took the right steps this time.

---------

So the conclusion is, there is an order as to what you do first, and you can find out that order by taking these steps to see if you get the correct expression/curve as the original question had asked?

(edited 9 years ago)

Reply 5

user2020user

Original post by Inevitable

...

Do you have a question on this that I can try?

Reply 6

Smaug123

Original post by Inevitable

So the conclusion is, there is an order as to what you do first, and you can find out that order by taking these steps to see if you get the correct expression/curve as the original question had asked?