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My argument with my math teacher. Am I right? Rate Topic: -----

#1 User is offline   H3 Proto 

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Posted 07 October 2009 - 01:13 PM

EDIT: Equations got messed up when I edited it. My radicals turned into ˆš.

I'm just gonna make a very rough sketch, I have to go to work soon. But I missed one on my Pre-Calculus test and this is what happened.

We were to take the graph of y=ˆš(x)

Posted Image

That's y equals the square root of x. We had to take the graph and do the following:

i. Shift it down 7 units

This makes is y=ˆš(x)-7



ii. Shift it left 5 units.

Now we're at y=ˆš(x+5)-7

Posted Image

iii. Reflect it over the y-axis

He says it should now be y=ˆš(-(x+5))-7

Posted Image

I said that if we reflect it over the y-axis it should be like this with the equation y=ˆš(-x+5)-7

Posted Image



His argument is that when when you move the graph of something the whole thing moves, including the axes. I was going by where the y-axis is, at x=0. All the points on my graph are reflected over that line. So am I right or no? If I want the point he wants me to argue with him before school tomorrow. We took about 10 minutes of class time discussing it.

This post has been edited by H3 Proto: 07 October 2009 - 06:43 PM

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#2 User is offline   Jackson 

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Posted 07 October 2009 - 01:47 PM

Axes never move. If they moved with the graph, then your "moves" would be negated. Your teacher is a moron.
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#3 User is offline   BAGELS 

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Posted 07 October 2009 - 01:50 PM

good point but I do not know the answer

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#4 User is offline   Droop 

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Posted 07 October 2009 - 01:55 PM

View PostH3 Proto, on Oct 7 2009, 01:13 PM, said:

I'm just gonna make a very rough sketch, I have to go to work soon. But I missed one on my Pre-Calculus test and this is what happened.

We were to take the graph of y=√(x)

Posted Image

That's y equals the square root of x. We had to take the graph and do the following:

i. Shift it down 7 units

This makes is y=√(x)-7



ii. Shift it left 5 units.

Now we're at y=√(x+5)-7

Posted Image

iii. Reflect it over the y-axis

He says it should now be y=√(-(x+5))-7

Posted Image

I said that if we reflect it over the y-axis it should be like this with the equation y=√(-x+5)

Posted Image



His argument is that when when you move the graph of something the whole thing moves, including the axes. I was going by where the y-axis is, at x=0. All the points on my graph are reflected over that line. So am I right or no? If I want the point he wants me to argue with him before school tomorrow. We took about 10 minutes of class time discussing it.

wrong your teacher is correct. the reflection needs to happen outside the parenthesis. because if it were inside you would take the square root of the negative and there would be reflection over the x.
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#5 User is offline   -SMITTY- 

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Posted 07 October 2009 - 02:05 PM

I think your teacher is right. when you reflect something over the Y axis, the equation goes from y=f(x) to y=f(-x). When you ended up with y=√(x+5)-7, you had to reflect it, by making the "x" value negative. It should give you -(x+5), not (-x+5) like you said. Your teachers equation is right. Do you have a graphing calculator? put it in, and his graphs are correct.
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#6 User is offline   BAGELS 

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Posted 07 October 2009 - 02:08 PM

put the equation in a graphing calculator

EDIT: i got beat^^^^^^

This post has been edited by BAGELS: 07 October 2009 - 02:08 PM


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#7 User is offline   M.O.P. 

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Posted 07 October 2009 - 02:15 PM

I printed it and will be taking it to my Math teacher who is actually quite good. Let you know what he says.
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#8 User is offline   Silent-7 

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Posted 07 October 2009 - 03:03 PM

View PostH3 Proto, on Oct 7 2009, 02:13 PM, said:

His argument is that when when you move the graph of something the whole thing moves, including the axes. I was going by where the y-axis is, at x=0. All the points on my graph are reflected over that line. So am I right or no? If I want the point he wants me to argue with him before school tomorrow. We took about 10 minutes of class time discussing it.

That is true. Though you forgot part of your equation... it should be y=((-x+5)^1/2)-7 instead of y=((-x+5)^1/2).
So I guess it comes down to what your teacher meant by reflecting it over the y-axis. From my understanding of that phrase, you are correct. But if there's some other interpretation, such as where you'd use -(x+5), then it's something I've never seen before and I can't really help you.


View PostDroop, on Oct 7 2009, 02:55 PM, said:

wrong your teacher is correct. the reflection needs to happen outside the parenthesis. because if it were inside you would take the square root of the negative and there would be reflection over the x.

Not true. I plugged everything into my graphing calculator and y=((-x+5)^1/2)-7 does indeed look like his drawing.

The equation y=(-(x+5)^1/2)-7 appears to be what you would get by reflecting the original graph about the y-axis, and then shifting it. That may be the key to solving your dispute. At least I hope it helps.



I'm really curious to see how this turns out now - I really think you're right here. And I'd like to think that, as an engineering major who graduated highschool Valedictorian, I know a thing or two about this math stuff. :D

This post has been edited by Silent-7: 07 October 2009 - 03:14 PM

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#9 User is offline   H3 Proto 

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Posted 07 October 2009 - 06:14 PM

I have a graphing calculator and I've put in these equations, they're right I didn't double check when I typed this but I'm pretty sure I remembered correctly. The disagreement is about whether reflecting a graph over the y-axis is interpreted as over the line x=0 (the only y-axis as I understand) or if it is reflected over x=-5 (what he is telling me). And I need to know how to explain that (if) I'm right to get my points back. It's not a big deal but my class (and teacher) gets all over my ass if I miss a problem and I've only missed one so far this year...a careless mistake.

View PostSilent-7, on Oct 7 2009, 05:03 PM, said:

The equation y=(-(x+5)^1/2)-7 appears to be what you would get by reflecting the original graph about the y-axis, and then shifting it. That may be the key to solving your dispute. At least I hope it helps.

Yes...if you do everything in the order given you get my graph. You get his if you flip the order.

This post has been edited by H3 Proto: 07 October 2009 - 06:16 PM

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#10 User is offline   JerseyPaint 

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Posted 07 October 2009 - 07:55 PM

When you moved over y down and to the left, you basically moved your center axis, so its no longer on (0,0). Thats is what I'm getting seeing right now, but I haven't done graphs in like 5 months and didn't need to study them since I was exempt from the final so "rusty" is an understatement.

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#11 User is offline   H3 Proto 

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Posted 07 October 2009 - 08:00 PM

That's what he was telling me. But the way I understood my instructions was that I was supposed to take the line y=[square root](x+5)-7 and reflect it over the y-axis, which I did.
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#12 User is offline   JerseyPaint 

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Posted 07 October 2009 - 08:34 PM

View PostH3 Proto, on Oct 7 2009, 11:00 PM, said:

That's what he was telling me. But the way I understood my instructions was that I was supposed to take the line y=[square root](x+5)-7 and reflect it over the y-axis, which I did.

The (0,0) axis point is only meant to start with and to use as a judge. That axis can change, and as I see it your axis moved to (-5, -7)

And I'm pretty sure when you reflect over the y-axis and invert x you also have to invert the changes you made to x, which was (+5), so it should be y=√-(x+5)-7

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#13 User is offline   mtaylor 

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Posted 07 October 2009 - 08:51 PM

You've just mixed up the order. You do stretches first, reflections second and translations third. Then you don't have to worry about it.

Your teacher is right.

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Posted 08 October 2009 - 06:46 AM

The teacher is ALWAYS right......:P
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#15 User is offline   Droop 

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Posted 08 October 2009 - 08:22 AM

View PostJerseyPaint, on Oct 7 2009, 08:34 PM, said:

View PostH3 Proto, on Oct 7 2009, 11:00 PM, said:

That's what he was telling me. But the way I understood my instructions was that I was supposed to take the line y=[square root](x+5)-7 and reflect it over the y-axis, which I did.

The (0,0) axis point is only meant to start with and to use as a judge. That axis can change, and as I see it your axis moved to (-5, -7)

And I'm pretty sure when you reflect over the y-axis and invert x you also have to invert the changes you made to x, which was (+5), so it should be y=√-(x+5)-7

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