NotNotBatman
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The graph of ln(4-x); I've thought of this in two ways; ln(-x+4) and ln -(x-4), using the former doesn't it reflect the graph in the y axis and then shift the graph 4 units in the negative x direction meaning it crossed the x axis at (-3,0)?
But when I use ln -(x-4) doesn't it reflect the graph in the y axis and shift it 4 units in the positive x direction meaning it crosses the x axis at (5,0)?

However the answer in the textbook is different. I know I can set x and y equal to zero and find out where it crosses the co-ordinate axis, but I want to work this out in terms of translations of graphs.
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Notnek
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(Original post by NotNotBatman)
The graph of ln(4-x); I've thought of this in two ways; ln(-x+4) and ln -(x-4), using the former doesn't it reflect the graph in the y axis and then shift the graph 4 units in the negative x direction...
Let's try that.

Start with f(x) = \ln x

"Reflect the graph in the y axis" - this is a transformation f(x)\rightarrow f(-x)

f(-x) = \ln (-x)

Call this new function g(x) then "shift the graph 4 units in the negative x direction" is a transformation g(x)\rightarrow g(x+4).

g(x) = \ln (-x)

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

Can you see your mistake?
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NotNotBatman
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(Original post by notnek)
Let's try that.

Start with f(x) = \ln x

"Reflect the graph in the y axis" - this is a transformation f(x)\rightarrow f(-x)

f(-x) = \ln (-x)

Call this new function g(x) then "shift the graph 4 units in the negative x direction" is a transformation g(x)\rightarrow g(x+4).

g(x) = \ln (-x)

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

Can you see your mistake?
In the answers, it shows the graph crossing at x=3, but ln(-x-4) would cross at x=5 so I'm still confused.
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Notnek
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(Original post by NotNotBatman)
In the answers, it shows the graph crossing at x=3, but ln(-x-4) would cross at x=5 so I'm still confused.
The question asks you about ln(4-x) not ln(-x-4). My last post was to show you that if you apply the two transformations that you thought were correct to ln(x), you end up with ln(-x-4) and not ln(4-x). So your reasoning was incorrect.

Does that make sense?
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NotNotBatman
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(Original post by notnek)
The question asks you about ln(4-x) not ln(-x-4). My last post was to show you that if you apply the two transformations that you thought were correct to ln(x), you end up with ln(-x-4) and not ln(4-x). So your reasoning was incorrect.

Does that make sense?
Yes, but how would I correctly apply transformations for this question?
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razzor
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(Original post by NotNotBatman)
Yes, but how would I correctly apply transformations for this question?
You need to forget about the ln(-x+4) and instead pull out the minus sign as you had already done: i.e. ln[-(x-4)]

Now you can apply the appropriate transformations: Reflexion in the y-axis and translation of 4 units in the x direction.
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NotNotBatman
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(Original post by razzor)
You need to forget about the ln(-x+4) and instead pull out the minus sign as you had already done: i.e. ln[-(x-4)]

Now you can apply the appropriate transformations: Reflexion in the y-axis and translation of 4 units in the x direction.
Is there an order for transformation of graphs?
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razzor
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(Original post by NotNotBatman)
Is there an order for transformation of graphs?
For a function of the form f(-(x+c)) where c is a constant, you should reflect first then translate.

In your scenario, you have f(-(x-4)) with f(x) = lnx.
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NotNotBatman
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(Original post by razzor)
For a function of the form f(-(x+c)) where c is a constant, you should reflect first then translate.

In your scenario, you have f(-(x-4)) with f(x) = lnx.
Thanks.
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