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Transformation of Quadratic equation

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data

    How do I simplify the transformation of this equation?
    y=-2[-1/3(x+2)]^2-4


    2. Relevant equations

    af(k(x-d))+c



    3. The attempt at a solution
    If this was a linear equation I would simply distribute the -1/3 into (x+2) and then distribute -2 into the (-1/3x -2/3) then collect -2/3 with the -4 right?
    Which would leave me 2/3x-8/3

    I tried this with the quadratic equation, which gives me 2/3x^2-8/3 but it can't be correct. I lose the reflection about the x so I know it's not right.

    I've tried searching online but find the steps for simplifying this type of equation.

    Thanks in advance
     
  2. jcsd
  3. Sep 1, 2010 #2
    From your attempt, it seems you are assuming that, inside the square bracket, you have one third of x+2, but I would have said it was one third of 1 over (x+2). In other words, one over (3x+6)
     
  4. Sep 1, 2010 #3
    and I forgot the minus sign
     
  5. Sep 1, 2010 #4

    epenguin

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    You can regard your right hand side as a difference of two squares which you should know how factorises. You should be able to see that one of the squares is of an imaginary number. Whether it is any usefulness to factorise it depends what problem this is part of.
     
  6. Sep 1, 2010 #5
    Ok so I only need to multiply the k value into the perfect square.

    a(kx-kd)^2+4

    -2(3x+6)^2-4

    Thanks for the help

    I was given transformations and asked to provide the simplified form of the equation. Nothing more, nothing less.
     
  7. Sep 3, 2010 #6

    epenguin

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    Well I don't know about that. I wouldn't call the multiplication by 3 of the bracket you have done 'simplifying'.

    I am not sure what your expression inside the [] is supposed to be - whether it corresponds to

    -2[-1/(3(x+2))]2 or -2[-(x+2)/3]2 , though it doesn't make any difference to what I said previously. Sorry but people continually post one such thing for another here so one is not sure what they mean. It is better if you can write using Latex - but that is now giving me rubbish.

    In either case your result does not seem to me right. However you are right that you can eliminate the minus sign inside the square bracket, as when squared this will become positive.

    And I am not sure what you mean by 'transformations' and 'given transformations'.

    This is the sort of equation usually given for exercises in conditions for real/nonreal roots. If y = 0 the equation has nonreal roots for reasons I already explained. But if you move the y to the RHS i.e. express the equation in the form 0 = ... then it is easy to see :biggrin::devil: without expanding how for certain values of y the equation in x will have real roots and for others nonreal.
     
  8. Sep 3, 2010 #7
    I'm doing a correspondence course, grade 11 functions in Ontario. One of the topics we cover is Transformation of functions.
    Transformations are described using this format:
    af(k(x-d) +c
    a = Controls vertical compression and stretching as well a reflection in the x-axis.
    K = controls horizontal compression and stretching as well a reflection in the y-axis.
    d = translates the graph to the right or left.
    c = translates the graph up or down.

    Here is an example of I question I might get.

    Describe the transformation that must be applied to the graph of each base function f(x) to obtain the transformed function. Write the transformed function.

    a) f(x) = 1/x, y= f(x+2) -1

    answer: the function has been shifted 2 units to the left and 1 down.
    d = -2
    c = -1
    y= 1/(x+2) -1.

    I haven't found a book at the library that shows how to re write the transformed equation based on Y = af(k(x-d))+c. The booklet I'm working from doesn't show how to get the equation of f(x) + x^2 after applying y = 2f(1/3(x-5))-1. it tells you what the transformations are but then it doesn't show that simplified equation. I"m going to try an scan the page so I post it up.

    I"m trying to learn how to use latex but it's going slow. I think my computer is bugged. every time I enter some latex I get "A = base" as my output.
    [tex] \frac {1}{2}[/tex]

    I tried entering this code: \frac {1}{2} but in my preview I get A = base
     
  9. Sep 3, 2010 #8
    edit: ok, it showed up there but in my preview latex code always shows up as "A = base"
     
  10. Sep 3, 2010 #9

    epenguin

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    I am sorry, I too am having trouble I don't understand with Tex. I used to be able to do it and don't think I am doing any different from before. I type
    -2 [\frac{-1}{3(x+2)}]^2 between tex and \tex inside [] [] and just get

    [tex] -2 [\frac{-1}{3(x+2)}]^2[/tex]

    [tex] -2 [\frac{-(x+2)}{3}]^2[/tex]

    and do not understand what is happening. Edit: the formula is better than it appeared to me but I am unable to see the result with preview while editing, can only see after have posted, rather awkward.
    re edt: it is coming better now - is either of those your formula?


    Coming to your problem it is still hard to understand what the problem is. I understand your example a) but you have not set out your problem as in this example. I would call what that example is doing 'substitution' but OK, it is doing a substitution in order to transform a function into another form. This is often needed in algebra or geometry in order to find a simpler form to work with, or to show that one form is equivalent to, for some essential properties, some other, a standard form perhaps. It is found (and is intuitive) that many geometrical properties are not altered by such transformations. I will need your scanned page(s) because I read e.g. how to get the equation of f(x) + x^2 after applying y = 2f(1/3(x-5))-1 and wonder how to get the equation of f(x) + x^2 from what after applying y = 2f(1/3(x-5))-1 to what? But I think you have to wait a time for the site control before the scans are posted.
     
    Last edited: Sep 3, 2010
  11. Sep 3, 2010 #10
    Ok here is an example using a linear equation.


    [tex]\begin{align*}
    \text {Base function}\\
    f(x) &= x\\
    \text {Transformation parameters}\\
    y &= -3[2(x + 4)] +6\\
    f(x) &= -3[2(x +4)] +6\\
    &=-3(2x +8) +6\\
    &=-6x-24 + 6\\
    \text {Transformed equation}\\
    y&=-6x-18
    \end{align*}[/tex]


    Another one, this time with a reciprocal equation, this time they don't show the steps. Just the transformed equation.

    [tex]\begin{align*}
    \text {Base function}\\
    f(x) &= \frac {1}{x}\\
    \text {Transformation parameters}\\
    y&= -\frac{3}{4}f[2(x + 3)] -7\\
    \text {Transformed equation}\\
    f(x)&=-\frac{3}{4}\left(\frac{2}{(x +3)\right)} -7
    \end{align*}[/tex]

    So based on that. What do you think the equation of this quadratic function will be?

    [tex]\begin{align*}
    \text {Base function}\\
    f(x) &= x^2\\
    \text {Transformation parameters}\\
    y&= 2f[\frac{1}{4}(x + 3)] +2\\
    \text {Transformed equation}\\
    f(x)&=2(4x + 3)^2 +2
    \end{align*}[/tex]


    By the way here's a link to an online LaTeX editor.

    http://rogercortesi.com/eqn/index.php" [Broken]

    Gota fix those equation numbers.
     
    Last edited by a moderator: May 4, 2017
  12. Sep 4, 2010 #11

    eumyang

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    Let's clean up your linear function example. The transformation parameters are
    [tex]y = -3\,f(2(x + 4)) + 6[/tex], right? When finding the transform equation, technically you should not reuse f(x) but instead write:
    [tex]\begin{aligned}
    -3\,f(2(x + 4)) + 6 &= -3(2(x +4)) + 6 \\
    &= -3(2x +8) + 6 \\
    &= -6x - 24 + 6 \\
    &=-6x-18
    \end{aligned}[/tex]

    Regarding the next example, your basic function is [tex]f(x) = \frac{1}{x}[/tex]. Your transformation parameters are
    [tex]y = -\frac{3}{4}f(2(x + 3)) -7[/tex]. Let's look at just the f part. Whatever is inside the f is what you plug into the function. So
    [tex]f(2(x + 3)) = \frac{1}{2(x + 3)}[/tex].

    There is a -3/4 outside the f, so you multiply this function by -3/4:
    [tex]-\frac{3}{4}f(2(x + 3)) = -\frac{3}{4} \cdot \frac{1}{2(x + 3)}[/tex].

    Finally, there is a -7 after the expression with f, so you subtract -7 from the function:
    [tex]-\frac{3}{4}f(2(x + 3)) - 7 = -\frac{3}{4} \cdot \frac{1}{2(x + 3)} - 7[/tex].

    This is where you stopped (and is there a typo?), but if your linear function example is any indication, I wouldn't consider this simplified. I would go ahead and multiply the two fractions:
    [tex]-\frac{3}{4} \cdot \frac{1}{2(x + 3)} - 7[/tex]
    [tex]= -\frac{3}{8(x + 3)} - 7[/tex]
    And then subtract the fractions:
    [tex]= -\frac{3}{8(x + 3)} - \frac{7(8(x + 3))}{8(x + 3)}[/tex]
    [tex]= -\frac{3}{8(x + 3)} - \frac{56x + 168}{8x + 24}[/tex]
    [tex]= -\frac{56x + 171}{8(x + 3)}[/tex]

    Where did the 1/4 go? Looks like another typo?

    Again, I wouldn't stop here. I would multiply everything out and write my answer in standard form y = ax^2 + bx + c. Distribute the 1/4, square the resulting binomial, distribute the 2, and then add the 2.
     
    Last edited by a moderator: Sep 15, 2010
  13. Sep 12, 2010 #12
    Sorry it's taken so long to post these scans but I have been busy with school and work.

    The bottom of image 1 goes with image 2 and image 4 goes with 5.
    Of particular interest is example b) on image 1. The final answer is omitted on image 2.

    This was scanned from the booklet I'm working from.

    I'm only allowed 3 attachments so I'll try to up load the next 2 in the next post.
     

    Attached Files:

  14. Sep 12, 2010 #13
    Here's image 4 and 5.
     

    Attached Files:

    Last edited: Sep 12, 2010
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