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Math Help - transformation of graph

  1. #1
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    transformation of graph

    hi,
    i need to put these onto a graph:
    Y=X^2
    Y=(X+3)^2
    Y=4(X-1)^2

    i can do the first one and the second but im unsure on the third one, this is what i have done so far and where i think the 3rd one will go!

    transformation of graph-so-far-1.jpg

    is this correct for number 3? if not where am i going wrong?


    thanks!
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    hi,
    i need to put these onto a graph:
    Y=X^2
    Y=(X+3)^2
    Y=4(X-1)^2

    i can do the first one and the second but im unsure on the third one, this is what i have done so far and where i think the 3rd one will go!

    Click image for larger version. 

Name:	so far 1.jpg 
Views:	15 
Size:	19.5 KB 
ID:	20874

    is this correct for number 3? if not where am i going wrong?

    thanks!
    Number 3 is wrong. There is no vertical shift. The 4 gives vertical DILATION.
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  3. #3
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    e^(i*pi)'s Avatar
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    The third one is wrong (the other two are fine). Your graph doesn't have a y intercept which implies x=0 is not valid when it is.

    Compared to the first graph you move it 1 to the right and 'squeeze it by 4'.


    It would be easiest to find where the graph crosses the axis and draw a curve to that effect.

    It crosses the x-axis when 4(x-1)^2 = 0 and it crosses the y-axis when x=0:
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  4. #4
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    would this means it goes here:

    transformation of graph-here.jpg?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by andyboy179 View Post
    is this correct for number 3? if not where am i going wrong?

    It is no correct. For example (1,0) must be a point of the graph .


    Fernando Revilla


    Edited: Sorry, I didnīt see the previous posts.
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  6. #6
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    e^(i*pi)'s Avatar
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    No - there is no vertical shift

    (-1,0), (0,-1) and (1,0) are on your graph so you can plot those points and draw the curve

    edit: this post is wrong
    Last edited by e^(i*pi); February 20th 2011 at 10:28 AM.
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    No - there is no vertical shift

    (-1,0), (0,-1) and (1,0) are on your graph so you can plot those points and draw the curve
    but that would just look like this:

    transformation of graph-here.jpg
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  8. #8
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    I did make a mistake in that post. It passes through (0,1) and (4,0) instead

    edit: see attached graph in spoiler - the green one is the graph of number 3
    Spoiler:
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  9. #9
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    so would it look like this:

    transformation of graph-here.jpg
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by andyboy179 View Post
    so would it look like this:

    Click image for larger version. 

Name:	here.jpg 
Views:	5 
Size:	10.7 KB 
ID:	20877

    Have you seen e^(i*pi)'s attachment?


    Fernando Revilla
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  11. #11
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    Quote Originally Posted by e^(i*pi) View Post
    I did make a mistake in that post. It passes through (0,1) and (4,0) instead

    edit: see attached graph in spoiler - the green one is the graph of number 3
    Spoiler:

    FernandoRevilla, i hadn't seen the attachment when i first looked, thanks!
    Last edited by andyboy179; February 20th 2011 at 11:32 AM. Reason: answered my own question!
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