Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

Graph transformation

Announcements Posted on
Take our short survey, £100 of Amazon vouchers to be won! 23-09-2016
    • Thread Starter
    Offline

    2
    ReputationRep:
    What is the graph transformation f(1-x)? I'm trying to graph f(x)=2ln(1-x), and I got stretch in y axis factor 2, shift left 1 unit and reflect in y axis. However, the answers say it is a shift right one unit. I checked on an online graph maker and they also show a shift right. Can anyone explain why the 1 causes a shift to the right and not the left? I thought it had something to do with the fact it has been reflected but was not sure...
    • Thread Starter
    Offline

    2
    ReputationRep:
    Actually I thought about it, I think it depends on the order in which you do the transformations.... Does anyone know what the correct order of doing transformations is? Like how we learn BODMAS, is there a kind of sequence that can be used to correctly do graph transformations?
    Offline

    3
    ReputationRep:
    (Original post by Uni12345678)
    What is the graph transformation f(1-x)? I'm trying to graph f(x)=2ln(1-x), and I got stretch in y axis factor 2, shift left 1 unit and reflect in y axis. However, the answers say it is a shift right one unit. I checked on an online graph maker and they also show a shift right. Can anyone explain why the 1 causes a shift to the right and not the left? I thought it had something to do with the fact it has been reflected but was not sure...
    Depends on the order, but there is no stretch in f(1-x) from f(x)

    Firstly it would be reflected in x=0 (the y-axis) - f(x) \rightarrow f(-x)

    Then it would be shifted by vector [1,0] - f(-x) \rightarrow f(-(x-1))=f(1-x)
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Depends on the order, but there is no stretch in f(1-x) from f(x)

    Firstly it would be reflected in x=0 (the y-axis) - f(x) \rightarrow f(-x)

    Then it would be shifted by vector [1,0] - f(-x) \rightarrow f(-(x-1))=f(1-x)
    Oh thank you! I get it! So when considering a transformation that involves moving along the x axis, the form must be including a positive x value as you have shown, yes?
    Offline

    3
    ReputationRep:
    (Original post by Uni12345678)
    Oh thank you! I get it! So when considering a transformation that involves moving along the x axis, the form must be including a positive x value as you have shown, yes?
    What do you mean?

    Translating y=f(x) by vector [a,b] would make it y-b=f(x-a)
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    What do you mean?

    Translating f(x) by vector [a,0] would make it f(x-a)
    So when we learn the rules like f(x-1) is shift to right etc... The x must be positive for the transformation to be valid; if x is negative we can't just reflect and then move by 1 unit to the right because of -1, it must be to the right because the x is negative so we factorise to change it to. Positive by taking out negative 1 and that would make the transformation go to the left as it would create a positive 1...

    Okay it makes sense to me, I'm sorry it's complicated ahhaha
    Offline

    3
    ReputationRep:
    For transforming  f(x) to  f(ax+b) there are always two different sets of transformations.
    The first is a translation by vector  \begin{pmatrix} -b \\ 0 \end{pmatrix} followed by a stretch parallel to the x axis SF 1/a.
    It is also a stretch parallel to x axis SF 1/a followed by a translation by vector  \begin{pmatrix} -\frac{b}{a} \\ 0 \end{pmatrix} .
    These are the only order they can be in. Of course if a is negative then there will be a reflection in the y axis.
    Offline

    3
    ReputationRep:
    (Original post by Uni12345678)
    So when we learn the rules like f(x-1) is shift to right etc... The x must be positive for the transformation to be valid; if x is negative we can't just reflect and then move by 1 unit to the right because of -1, it must be to the right because the x is negative so we factorise to change it to. Positive by taking out negative 1 and that would make the transformation go to the left as it would create a positive 1...

    Okay it makes sense to me, I'm sorry it's complicated ahhaha
    The best advice I can give you is to put brackets around the x, and do any transformations concerning it inside that bracket. I will redo the first example with brackets to clear up any confusion and hopefully make it clearer:

    Reflection in the y-axis: f[(x)] \rightarrow f[(-x)]=f[-x]

    Translate by vector [1,0]: f[-(x)] \rightarrow f[-(x-1)]=f[1-x]

    Then for the banter we can choose to translate it by another vector [-8,0]: f[1-(x)] \rightarrow f[1-(x+8)]=f[-7-x]

    And then stretch it in along the x-axis by scale factor 1/2: f[-7-(x)] \rightarrow f[-7-(2x)]=f[-7-2x]
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    The best advice I can give you is to put brackets around the x, and do any transformations concerning it inside that bracket. I will redo the first example with brackets to clear up any confusion and hopefully make it clearer:

    Reflection in the y-axis: f[(x)] \rightarrow f[(-x)]=f[-x]

    Translate by vector [1,0]: f[-(x)] \rightarrow f[-(x-1)]=f[1-x]

    Then for the banter we can choose to translate it by another vector [-8,0]: f[1-(x)] \rightarrow f[1-(x+8)]=f[-7-x]

    And then stretch it in along the x-axis by scale factor 1/2: f[-7-(x)] \rightarrow f[-7-(2x)]=f[-7-2x]
    Thank you! You've been very helpful, also your banter is brilliant 😂

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: September 18, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Who will be the next permanent England boss?
Help with your A-levels

All the essentials

Rosette

Essay expert

Learn to write like a pro with our ultimate essay guide.

Uni match

Uni match

Our tool will help you find the perfect course for you

Study planner

Create a study plan

Get your head around what you need to do and when with the study planner tool.

Study planner

Resources by subject

Everything from mind maps to class notes.

Books

Essential advice

11 things A-level students wish they'd known before starting their course.

Hands typing

Degrees without fees

Discover more about degree-level apprenticeships.

A student doing homework

Study tips from A* students

Students who got top grades in their A-levels share their secrets

Study help links and info

Can you help? Study help unanswered threadsRules and posting guidelines

Sponsored content:

HEAR

HEAR

Find out how a Higher Education Achievement Report can help you prove your achievements.

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.