Help with a translating a graph please :) Watch

maruchan
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The curve y=x2 is translated by the vector(4 3) and then reflected in the line y=-1?

I have got up to y=(x-4)2 +3 but I don't get how to reflect it without sketching?

Is there a formula?

Thanks for the help
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yankang.qi
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A quick way you can do this is you can see that its a reflection in a horizontal line. This means the x value of the turning point does not change. The y value of the turning point does change however, the distance from the line is 4, so it'll be 4 away from the line on the other side of the line. Making the new turning point (4,-5). Then since the graph has been flipped, it is not an x2 graph anymore, but a -x2 graph. So you'll need to be a minus in front of the (x-4)2.
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maruchan
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(Original post by yankang.qi)
A quick way you can do this is you can see that its a reflection in a horizontal line. This means the x value of the turning point does not change. The y value of the turning point does change however, the distance from the line is 4, so it'll be 4 away from the line on the other side of the line. Making the new turning point (4,-5). Then since the graph has been flipped, it is not an x2 graph anymore, but a -x2 graph. So you'll need to be a minus in front of the (x-4)2.
Ah thank you! I didn't think of that.
Do you think you can explain this to me in the written solutions it says that a reflection in y=-1 is equal to the translation 2(y+1) why is that?
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yankang.qi
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(Original post by maruchan)
Ah thank you! I didn't think of that.
Do you think you can explain this to me in the written solutions it says that a reflection in y=-1 is equal to the translation 2(y+1) why is that?
hmm I'm not sure what it means by translation 2(y+1)
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maruchan
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https://www.physicsandmathstutor.com...olutions-2016/

It's on these solutions Section 1 Question 89.

I completely don't understand what it means by that

(Original post by yankang.qi)
hmm I'm not sure what it means by translation 2(y+1)
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yankang.qi
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(Original post by maruchan)
https://www.physicsandmathstutor.com...olutions-2016/

It's on these solutions Section 1 Question 89.

I completely don't understand what it means by that
I have no idea why they are doing it, I get how they are doing it but it seems a bit pointless. I would just ignore it because I don't think it'll be useful at all for GCSE or A level. If you are GCSE the translation questions won't be very hard and that just seems like a very roundabout method. At A level the transformation questions won't need that either as it'll be either harder and require some other method or be as easy as the GCSE ones for few marks.
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maruchan
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https://www.physicsandmathstutor.com...olutions-2016/

It's on these solutions Section 1 Question 89.

I completely don't understand what it means by that

(Original post by yankang.qi)
hmm I'm not sure what it means by translation 2(y+1)
(Original post by yankang.qi)
I have no idea why they are doing it, I get how they are doing it but it seems a bit pointless. I would just ignore it because I don't think it'll be useful at all for GCSE or A level. If you are GCSE the translation questions won't be very hard and that just seems like a very roundabout method. At A level the transformation questions won't need that either as it'll be either harder and require some other method or be as easy as the GCSE ones for few marks.
Alright thank you! I'm at A-level yr 13 and that completely confused me ! thank you very much !
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yankang.qi
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(Original post by maruchan)
https://www.physicsandmathstutor.com...olutions-2016/

It's on these solutions Section 1 Question 89.

I completely don't understand what it means by that




Alright thank you! I'm at A-level yr 13 and that completely confused me ! thank you very much !
No problem
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RDKGames
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(Original post by yankang.qi)
hmm I'm not sure what it means by translation 2(y+1)
(Original post by maruchan)
Ah thank you! I didn't think of that.
Do you think you can explain this to me in the written solutions it says that a reflection in y=-1 is equal to the translation 2(y+1) why is that?
Because when you reflect a point in some line, the distance between the point and the reflection line must be the exact same as the distance between the line of refl. and the reflection point.

Every single coordinate (x,y) on that parabola y = (x-4)^2 + 3 has a perpendicular distance of |y-(-1)|, i.e. |y+1| to the line of reflection. Since y>0 for all y on this curve, then y+1 > 0 and we say that |y+1| = y+1, hence every single coord on this parabola has a distance of y+1 to the line of reflection.
So when you reflect these points in this line, they will go a distance of y+1 down to the reflection line, and then ANOTHER distance of y+1 to the reflection spot.

Hence, every single coordinate (x,y) is translated down by a distance of 2(y+1), and so we get a reflection.

Coming back to the parabola, well I'm sure you know that translating y=f(x) down by A is the same as writing y=f(x)-A. Same thing here, and we get that Y = [(x-4)^2 + 3] - 2(y+1), where Y is the new y coordinate of the parabola. The old one, y, is precisely just (x-4)^2 + 3 and so 2(y+1) is simply 2(x-4)^2 + 8. Hence why they have the new curve as

Y = (x-4)^2 + 3 - [2(x-4)^2 + 8].


I wouldn't worry if you can't wrap your head around this. A much simpler approach would be the following:

In order to reflect in the line y=-1, let's just shift everything by 1 unit so that our line of reflection coincides with the x-axis. Now reflect everything in the x-axis. Then shift everything back down by 1 unit. Hopefully it makes sense to you and it's much less to digest.
Last edited by RDKGames; 3 weeks ago
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maruchan
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(Original post by RDKGames)
Because when you reflect a point in some line, the distance between the point and the reflection line must be the exact same as the distance between the line of refl. and the reflection point.

Every single coordinate (x,y) on that parabola y = (x-4)^2 + 3 has a perpendicular distance of |y-(-1)|, i.e. |y+1| to the line of reflection. Since y>0 for all y on this curve, then y+1 > 0 and we say that |y+1| = y+1, hence every single coord on this parabola has a distance of y+1 to the line of reflection.
So when you reflect these points in this line, they will go a distance of y+1 down to the reflection line, and then ANOTHER distance of y+1 to the reflection spot.

Hence, every single coordinate (x,y) is translated down by a distance of 2(y+1), and so we get a reflection.

Coming back to the parabola, well I'm sure you know that translating y=f(x) down by A is the same as writing y=f(x)-A. Same thing here, and we get that Y = [(x-4)^2 + 3] - 2(y+1), where Y is the new y coordinate of the parabola. The old one, y, is precisely just (x-4)^2 + 3 and so 2(y+1) is simply 2(x-4)^2 + 8. Hence why they have the new curve as

Y = (x-4)^2 + 3 - [2(x-4)^2 + 8].


I wouldn't worry if you can't wrap your head around this. A much simpler approach would be the following:

In order to reflect in the line y=-1, let's just shift everything by 1 unit so that our line of reflection coincides with the x-axis. Now reflect everything in the x-axis. Then shift everything back down by 1 unit. Hopefully it makes sense to you and it's much less to digest.
I completely understand that now. Thank you very much, you're a star! I was wondering what it meant by that but I would do it the simpler away first . Are there any other rules I should now for reflections for the x axis? Thank you again
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