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Math Help - Horizontal Stretch and Shrink

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Ok I think some things are starting to click in my brain. Question, the book doesn't specify to use f(1) but to just tell what would happen. You using f(1) and and the book not telling me to use f(1) is giving the same answer. Does this mean whenever I horizontally stretch or shrink, I should aim to try to find how much the graph stretches or shrinks by trying to make it equal to 1?
    no, i used f(1) as an example. it was just an illustration, nothing more. i just wanted to show you why we have to choose larger values of x (hence stretch) or choose smaller values of x (hence shrink)

    *sigh* my professor didn't explain it the way you are in the least bit. He just told us one gets closer to the x-axis and the other gets further away.
    that discussion works for vertical stretches or shrinkings.

    Can you by any chance you can just tell me how to do horizontal stretching and shrinking in a really easily understandable way?
    Vertical Stretching:

    Given a function f(x), we obtain cf(x), where c is a constant, by stretching the graph vertically by a factor of c. that is, we leave the x-values the same, and multiply all y-values by c.

    Vertical Shrinking:

    Given a function f(x), we obtain (1/c)f(x), where c is a constant, by shrinking the graph vertically by a factor of c. that is, we leave the x-values the same, and multiply all y-values by (1/c).


    Horizontal Shrinking:

    Given a function f(x), we obtain f(cx), where c is a constant, by shrinking the graph horizontally by a factor of c. that is, we leave the y-values the same, and multiply all x-values by (1/c).

    Horizontal Stretching:

    Given a function f(x), we obtain f((1/c)x), where c is a constant, by stretching the graph horizontally by a factor of c. that is, we leave the y-values the same, and multiply all x-values by c.

    so you see, it's like we want to get back to f(x) if a constant interferes. since y = f(x), if we get f(cx), we multiply x by (1/c) to get f(c(1/c)x) = f(x). so it shrinks, since we assume c >= 1, so we are in effect multiplying by something less than 1

    I feel like i'm getting more and more confused and I have my first major test on this stuff tomorrow.
    calm down. everything will be fine
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  2. #17
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    heh i've calmed down a bit so let's see if I can get this.

    Mind if we use a different problem?

    y = -f(2x) , It would horizontally shrink by 1/2 and reflect across the x-axis

    y = f(2x) - 1 , It would horizontally shrink by 1/2 and vertically shift down one unit.


    Did i get those right?

    Also if those two were supposedly written in function notation, how would it look?

    I'm curious because for instance

    y = 2f(x + 2) - 2 is the same as f(x) = 2(x + 2) - 2 where both would horizontally shift two units to the right, vertically shift 2 units down and vertically stretch by 2.

    So

    How would y = f(4x) or y = f(2x) - 1 look if it were to start with

    F(x) =

    sorry if it's a stupid question but I'm curious =/
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    heh i've calmed down a bit so let's see if I can get this.

    Mind if we use a different problem?

    y = -f(2x) , It would horizontally shrink by 1/2 and reflect across the x-axis

    y = f(2x) - 1 , It would horizontally shrink by 1/2 and vertically shift down one unit.
    correct!

    see, nothing to worry about

    Also if those two were supposedly written in function notation, how would it look?

    I'm curious because for instance

    y = 2f(x + 2) - 2 is the same as f(x) = 2(x + 2) - 2 where both would horizontally shift two units to the right, vertically shift 2 units down and vertically stretch by 2.

    So

    How would y = f(4x) or y = f(2x) - 1 look if it were to start with

    F(x) =

    sorry if it's a stupid question but I'm curious =/
    no, you can't write them like that. we can't changed the notation. the best we can do, is to actually find the formula, write it out, make the changes, and call it a new function, say g(x)


    PS. remember what i said about not mixing up y = f(x + 2) and y = x + 2, because that's what you did here. if you recall, i typed it in all capital letters and red font.
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  4. #19
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    [quote=Jhevon;66596]correct!

    see, nothing to worry about

    no, you can't write them like that. we can't changed the notation. the best we can do, is to actually find the formula, write it out, make the changes, and call it a new function, say g(x)

    MmMm.... ok so what is the differenece between

    y = f(x + 7) and f(x) = (x + 7)?

    Both are horizontally shifting 7 units to the left right?
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post

    MmMm.... ok so what is the differenece between

    y = f(x + 7) and f(x) = (x + 7)?

    Both are horizontally shifting 7 units to the left right?
    f(x) = x + 7 is not being shifted, it is just a function, we're not doing anything to it, it is simply a line

    y = f(x + 7) is not necessarily a line, it can be anything. all we know, is that whatever it is, is that we're shifting it to the left 7 units
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  6. #21
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    Quote Originally Posted by Jhevon View Post
    f(x) = x + 7 is not being shifted, it is just a function, we're not doing anything to it, it is simply a line

    y = f(x + 7) is not necessarily a line, it can be anything. all we know, is that whatever it is, is that we're shifting it to the left 7 units
    ahhh ic ic


    Final question ,

    The graph f is given. Sketch the graphs of the following functions

    a) y = f( x - 2 )

    b) y = f(x) - 2

    c) y = 2f(x)

    d) y = -f(x) + 3

    I was wondering how would I go about finding the individual points for the new graph?
    Attached Thumbnails Attached Thumbnails Horizontal Stretch and Shrink-graph.jpg  
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    ahhh ic ic


    Final question ,

    The graph f is given. Sketch the graphs of the following functions

    a) y = f( x - 2 )

    b) y = f(x) - 2

    c) y = 2f(x)

    d) y = -f(x) + 3

    I was wondering how would I go about finding the individual points for the new graph?
    let's take this slow.

    you know that when we have f(x - 2) we shift 2 units, horizontally to the right. no vertical movement occurs right, just horizontal. in other words, the x's are affected, not the y's. so for example, if i move the point (0,0) 2 units to the right, what does it become?
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  8. #23
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    Quote Originally Posted by Jhevon View Post
    let's take this slow.

    you know that when we have f(x - 2) we shift 2 units, horizontally to the right. no vertical movement occurs right, just horizontal. in other words, the x's are affected, not the y's. so for example, if i move the point (0,0) 2 units to the right, what does it become?

    It'll be (2,0) . I know that for something as easy as f(x - 2), all I would have to do is horizontally move those 4 points over 2 and that would give me my new graph. My concern is, how would I do something like 2f(x) or -f(x) + 3? Is there a way to do it through calculations
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  9. #24
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    It'll be (2,0) . I know that for something as easy as f(x - 2), all I would have to do is horizontally move those 4 points over 2 and that would give me my new graph. My concern is, how would I do something like 2f(x) or -f(x) + 3? Is there a way to do it through calculations
    i wanted you to get what you actually did in terms of the coordinates. that is, you actually added 2 to the x-coordinate.

    so for f(x - 2), you would change all the (x,y) coordinates on the graph to (x + 2, y)

    now you know how the transformations shift the graphs. try to figure out what happens to the coordinates and change them accordingly. let me see what you come up with
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  10. #25
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    Quote Originally Posted by Jhevon View Post
    i wanted you to get what you actually did in terms of the coordinates. that is, you actually added 2 to the x-coordinate.

    so for f(x - 2), you would change all the (x,y) coordinates on the graph to (x + 2, y)

    now you know how the transformations shift the graphs. try to figure out what happens to the coordinates and change them accordingly. let me see what you come up with


    Oh man... i'm sorry i'm not sure what you mean. y = f(x - 2) from here where do I plug in and how do I do it to see the transformation from the picture I gave you into the new graph that f(x - 2) is supposed to give?
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  11. #26
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Oh man... i'm sorry i'm not sure what you mean. y = f(x - 2) from here where do I plug in and how do I do it to see the transformation from the picture I gave you into the new graph that f(x - 2) is supposed to give?
    umm, i told you. add 2 to all the x-coordinates and leave the y-coordinates unchanged. that will give you the points on the new graph, and just connect the dots in the same way they were connected before
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  12. #27
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    Quote Originally Posted by Jhevon View Post
    umm, i told you. add 2 to all the x-coordinates and leave the y-coordinates unchanged. that will give you the points on the new graph, and just connect the dots in the same way they were connected before



    Ohhhhh ok that was dumb of me. What about for the problems like y = 2f(x) and -f(x)?
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  13. #28
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    Quote Originally Posted by JonathanEyoon View Post
    Ohhhhh ok that was dumb of me. What about for the problems like y = 2f(x) and -f(x)?
    recall that in the usual convention, we say y = f(x), therefore any operation performed on f(x), we just do it on all the y-coordinates (leaving the x-coordinates unchanged, of course)

    so for y = 2f(x), we simply change all the points from (x,y) to (x,2y). we do a similar thing for y = -f(x)
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