# Alevel maths trasnformations questions helpp

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#1
Can someone tell me if these answers are correct. Ty in advance.

(a) Give full details of a sequence of two transformations needed to tranform the graph y = |x-2| to the graph of y = |2x-6|

Stretch parallel to x-axis sf 1/2 and then translation 2 units negative x-direction??

(b) Give details of the transformations thet map the curve y = e^x to the curve y = e^2x+3

Stretch in y-direction, sf 1/2 and then translation 2 units positive x-direction???

(I cant find mark scheme for these questions so pls help)
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1 month ago
#2
(Original post by ccchimpo)
Can someone tell me if these answers are correct. Ty in advance.

(a) Give full details of a sequence of two transformations needed to tranform the graph y = |x-2| to the graph of y = |2x-6|

Stretch parallel to x-axis sf 1/2 and then translation 2 units negative x-direction??

(b) Give details of the transformations thet map the curve y = e^x to the curve y = e^2x+3

Stretch in y-direction, sf 1/2 and then translation 2 units positive x-direction???

(I cant find mark scheme for these questions so pls help)
For both, it would help to see your explanation. Also sketching (in desmos) helps to understand/verify your thoughts
https://www.desmos.com/calculator/acjgotedqv
Ive tried to put in the two functions for a) to get you started

For b) is tthe exponent (2x+3) or just 2x?
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
For both, it would help to see your explanation. Also sketching (in desmos) helps to understand/verify your thoughts
https://www.desmos.com/calculator/acjgotedqv
Ive tried to put in the two functions for a) to get you started

For b) is tthe exponent (2x+3) or just 2x?
The exponent is (2x+3)
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#4
(Original post by mqb2766)
For both, it would help to see your explanation. Also sketching (in desmos) helps to understand/verify your thoughts
https://www.desmos.com/calculator/acjgotedqv
Ive tried to put in the two functions for a) to get you started

For b) is tthe exponent (2x+3) or just 2x?
Ok ty.
I checked with desmos for (a) by transforming the coordinates in the original equation and I got the coordinates for the transformation so I'm guessing the sequence of transformations I chose were right.

For (b) though, I thought the sequence of transformations was stretch parallel to x-axis sf 1/2 then translation 3 units in negative x-direction but the graph doesn't seem to match up with that and idk where I went wrong.
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1 month ago
#5
(Original post by ccchimpo)
Ok ty.
I checked with desmos for (a) by transforming the coordinates in the original equation and I got the coordinates for the transformation so I'm guessing the sequence of transformations I chose were right.

For (b) though, I thought the sequence of transformations was stretch parallel to x-axis sf 1/2 then translation 3 units in negative x-direction but the graph doesn't seem to match up with that and idk where I went wrong.
In desmos, if you click in the top right you can get an url to share. For the first one, the original curve had a minimum at x=2. The new curve has a minimum at x=3. If you mutliply the first one by 2 (scale by 1/2) then move two in the negative x direction, Im not sure the minima match. So not sure what you checked.
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#6
(Original post by mqb2766)
In desmos, if you click in the top right you can get an url to share. For the first one, the original curve had a minimum at x=2. The new curve has a minimum at x=3. If you mutliply the first one by 2 (scale by 1/2) then move two in the negative x direction, Im not sure the minima match. So not sure what you checked.
I think it matches. The original curve y = |2x-2| has minimum (2,0). The transformed curve has (3,0).
A stretch parallel to x-axis scale factor 1/2 so multiply 2 x 1/2 = 1 to get y = |2x-4|
Translate 2 units positive x direction to get y = |2x - 6| so 1 + 2 = 3. Giving you (3,0)

In the original post I meant to put positive x-direction instead of negative x-direction
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1 month ago
#7
(Original post by ccchimpo)
I think it matches. The original curve y = |2x-2| has minimum (2,0). The transformed curve has (3,0).
A stretch parallel to x-axis scale factor 1/2 so multiply 2 x 1/2 = 1 to get y = |2x-4|
Translate 2 units positive x direction to get y = |2x - 6| so 1 + 2 = 3. Giving you (3,0)

In the original post I meant to put positive x-direction instead of negative x-direction
When you do the x-axis stretch why would the -2 change to -4? Stretching changes the location of the minimum. It doesn't with your work.
Last edited by mqb2766; 1 month ago
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#8
(Original post by mqb2766)
When you do the x-axis stretch why would the -2 change to -4? Stretching changes the location of the minimum. It doesn't with your work.
so would it be stretch parallel to x-axis sf 1/2 to get from y = |x-2| to y = y = |2x-2| then translation 4 units positive x direction to get y = |2x - 6| ?
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