Transformations

[Macha]

Hi all! I am trying to state the transformation to the question below by using the dot-dash method rather than the inspection approach. However, when checking by inspection, the translation from the x-axis should be (x+2-2)^2=x^2. Where have I gone wrong with the formal approach? Thanks!

Q: The curve y=(x+2)^2 is reflected in the y axis, translated by 2 units in the positive direction of the x-axis and then dilated by a factor 1/2 from the x-axis. Find the equation of the transformed curve.
Working thus far: x'= -x + 2 y'=y/2
x=-x'+2 y=2y'
2y'=(-x'+2+2)^2
y'= -(x'-4)^2/2

 

[Dr.Peterson]

I have no idea what you mean by "dot-dash method" or "inspection method". What you seem to be doing here is not what I have usually seen, but is a good way.

But the error I see here is that you moved a negative sign outside of the square, which effectively changes the reflection from over the y axis to over the x axis! Remove that negative sign from your final answer, and it will be correct.

Perhaps the biggest issue is where your thinking in "checking by inspection" went wrong. Can you explain that more fully?

 

[Macha]

I have no idea what you mean by "dot-dash method" or "inspection method". What you seem to be doing here is not what I have usually seen, but is a good way.

But the error I see here is that you moved a negative sign outside of the square, which effectively changes the reflection from over the y axis to over the x axis! Remove that negative sign from your final answer, and it will be correct.

Perhaps the biggest issue is where your thinking in "checking by inspection" went wrong. Can you explain that more fully?

Well if we know that it’s translated in the positive x direction by two units the function would be (x-2)^2. So when inserted back into the original function, (x+2-2)^2=x^2.

Shouldn’t it be fine to take the negative outside the bracket as I’m merely factorising?

 

[Dr.Peterson]

Shouldn’t it be fine to take the negative outside the bracket as I’m merely factorising?

Here is what you said:

Working thus far: x'= -x + 2 y'=y/2
x=-x'+2 y=2y'
2y'=(-x'+2+2)^2
y'= -(x'-4)^2/2

You are not merely factorizing, taking the negative outside the bracket; as I said, you are taking it outside the square.

Is it true that (-x)^2 = -(x)^2? No! A negative squared is positive!

So the correct answer is y'= (x'-4)^2/2.

Well if we know that it’s translated in the positive x direction by two units the function would be (x-2)^2. So when inserted back into the original function, (x+2-2)^2=x^2.

I'm not sure what you are doing here. Here's the method I am familiar with:

Q: The curve y=(x+2)^2 is reflected in the y axis, translated by 2 units in the positive direction of the x-axis and then dilated by a factor 1/2 from the x-axis. Find the equation of the transformed curve.

Reflecting in the y axis replaces x with -x: y=(-x+2)^2.
Translating 2 units right replaces x with x-2: y=(-(x-2)+2)^2 = (-x+4)^2.
Dilating by a factor of 1/2 from the x axis multiplies by 1/2: y = 1/2 (-x+4)^2.

So your answer is correct.

Now, why is your result by "inspection" wrong? I'm not sure of the method you've learned, but you seem to be either just forgetting the reflection, or doing everything in the wrong order. Perhaps you need to explain why you did each step, so I can see if there's something you're misunderstanding.

 

[Otis]

... trying to state the transformation ... using the dot-dash method ...

Hello. I spent about 10 minutes searching the Internet for variations of "dot-dash method", as it relates to function transformations, but I could not find any references using that name.

I would have used the three steps shown by Dr. Peterson in post #4. That is, I would have reflected first and then shifted horizontally. You may shift functions before reflecting them, but if you do then you need to remember to also shift the mirror while you're at it. Cheers

?

 

[Macha]

Here is what you said:

You are not merely factorizing, taking the negative outside the bracket; as I said, you are taking it outside the square.

Is it true that (-x)^2 = -(x)^2? No! A negative squared is positive!

So the correct answer is y'= (x'-4)^2/2.


I'm not sure what you are doing here. Here's the method I am familiar with:

Reflecting in the y axis replaces x with -x: y=(-x+2)^2.
Translating 2 units right replaces x with x-2: y=(-(x-2)+2)^2 = (-x+4)^2.
Dilating by a factor of 1/2 from the x axis multiplies by 1/2: y = 1/2 (-x+4)^2.

So your answer is correct.

Now, why is your result by "inspection" wrong? I'm not sure of the method you've learned, but you seem to be either just forgetting the reflection, or doing everything in the wrong order. Perhaps you need to explain why you did each step, so I can see if there's something you're misunderstanding.

I must just be thinking of taking the -1 out as a common factor as that is obligatory to identify the end point in square root graphs?

 

[Dr.Peterson]

I must just be thinking of taking the -1 out as a common factor as that is obligatory to identify the end point in square root graphs?

I'm not sure what you mean; but it is definitely not good to do things just because you expect them, without having a specific reason it would be correct.

I spent about 10 minutes searching the Internet for variations of "dot-dash method", as it relates to function transformations, but I could not find any references using that name.

I did some searching and did find that some people seem to use "dash" to refer to the "prime" symbol on x' and y'; I can't think what "dot" would refer to.

This method, though I don't think I've seen it in any textbook here (at this level, as opposed to coordinate system transformations), is an excellent one for making it very clear what is happening. I think it's a little more work than what I might call the substitution method that I demonstrated, though.

 

[Macha]

I'm not sure what you mean; but it is definitely not good to do things just because you expect them, without having a specific reason it would be correct.


I did some searching and did find that some people seem to use "dash" to refer to the "prime" symbol on x' and y'; I can't think what "dot" would refer to.

This method, though I don't think I've seen it in any textbook here (at this level, as opposed to coordinate system transformations), is an excellent one for making it very clear what is happening. I think it's a little more work than what I might call the substitution method that I demonstrated, though.

What I mean is, to find the translation of x-co-ordinate of the endpoint in a square root function, you must first factorise what is in the square root e.g(-x-4 to -(x+4)) to identify that in this case, the x-co-ordinate of the endpoint has been translated 4 units in the neg x direction.

 

[Dr.Peterson]

Okay. Yes, factoring is good.

But you recognize, I presume, that you can't move the negative outside of either a radical or a square?

The proper thing to do with [MATH](-x + 4)^2[/MATH] is to take two steps: [MATH](-x + 4)^2 = (-(x - 4))^2 = (-1)^2(x - 4)^2 = (x - 4)^2[/MATH].

And in [MATH]\sqrt{-x - 4}[/MATH], you would only go as far as [MATH]\sqrt{-x - 4} = \sqrt{-(x + 4)}[/MATH], and see that the graph has been first reflected in the y-axis, and then translated left 4 units. You can't take the sign outside the radical.