# Natural log graph

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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.

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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#2

(Original post by

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...

**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...

Start with

"Reflect the graph in the y axis" - this is a transformation

Call this new function then "shift the graph 4 units in the negative x direction" is a transformation .

Can you see your mistake?

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(Original post by

Let's try that.

Start with

"Reflect the graph in the y axis" - this is a transformation

Call this new function then "shift the graph 4 units in the negative x direction" is a transformation .

Can you see your mistake?

**notnek**)Let's try that.

Start with

"Reflect the graph in the y axis" - this is a transformation

Call this new function then "shift the graph 4 units in the negative x direction" is a transformation .

Can you see your mistake?

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#4

(Original post by

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.

**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.

Does that make sense?

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(Original post by

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?

**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?

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#6

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Yes, but how would I correctly apply transformations for this question?

**NotNotBatman**)Yes, but how would I correctly apply transformations for this question?

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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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.

**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.

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#8

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Is there an order for transformation of graphs?

**NotNotBatman**)Is there an order for transformation of graphs?

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

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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.

**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.

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