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    the graph y=f(x) is translated by 1 unit in the positive x direction and 4 units in the positive y direction to a graph of y=g(x)

    find an expression for g(x) in the form ax^2 + bx +c

    any ideas how I would do this?
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    What about the rest of the question
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    (Original post by TenOfThem)
    What about the rest of the question
    we need it?
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    (Original post by madfish)
    we need it?
    What's the original equation of the graph prior to transformation? i.e. what's f(x) equal to?
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    (Original post by Nistar)
    What's the original equation of the graph prior to transformation? i.e. what's f(x) equal to?
    (1-2x)(1+2x) i.e 1-4x^2
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    (Original post by madfish)
    we need it?
    that's half a question, you need to know f(x) to do it...

    otherwise let f(x)=dx^2+ex+f and then c=f+4 (as an example, it actually isn't for your question) or whatever the translation was etc
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    I'm assuming f(x) was x^2?

    f(x) = x^2
    g(x) = (x-1)^2 + 4

    And then expand to get g(x) in the form wanted in the question.

    If f(x) is not x^2 then steps should be similar to above.
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    (Original post by madfish)
    (1-2x)(1+2x) i.e 1-4x^2
    So 1-4x2 needs to be translated 1 unit to the positive x direction and 4 units in the positive y direction.

    The x-direction unit of 1 will change to a negative within the equation to represent that translation and needs to be added onto x itself. i.e. a translation of y=x by a units in the positive x-direction will be y=(x-a)

    Then the translation by 4 units in the positive y-direction will just be added onto the end of the equation i.e. a translation of y=x by a units in the positive y-direction would be y=(x) + a

    Work it out from there
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    (Original post by SherlockHolmes)
    I'm assuming f(x) was x^2?

    f(x) = x^2
    g(x) = (x-1)^2 + 4

    And then expand to get g(x) in the form wanted in the question.

    If f(x) is not x^2 then steps should be similar to above.
    sorry f(x)=(1-2x)(1+2x) i.e 1-4x^2
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    (Original post by Nistar)
    So 1-4x2 needs to be translated 1 unit to the positive x direction and 4 units in the positive y direction.

    The x-direction unit of 1 will change to a negative within the equation to represent that translation and needs to be added onto x itself. i.e. a translation of y=x by a units in the positive x-direction will be y=(x-a)

    Then the translation by 4 units in the positive y-direction will just be added onto the end of the equation i.e. a translation of y=x by a units in the positive y-direction would be y=(x) + a

    Work it out from there
    Its the fact it's in the form 1-4x^2 is giving me trouble.. how would I change that equation to correspond with the translations?
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    This may help:

    b - a = -a + b

    1-4x^2 = -4x^2 + 1
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    (Original post by SherlockHolmes)
    This may help:

    b - a = -a + b

    1-4x^2 = -4x^2 + 1
    sorry to sound thick but I still don't get it ;(
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    (Original post by madfish)
    Its the fact it's in the form 1-4x^2 is giving me trouble.. how would I change that equation to correspond with the translations?
    Well if you've got 1-4x2, and following what I said in my earlier reply:
    g(x)= 5-4(x-1)2

    As subtracting 1 for the translation by 1 unit in the positive x-direction gives 1 - 4(x-1)2 , and translation by 4 units in the positive y-direction then gives 5 - 4(x-1)2

    1-4x2 is also equal to -4x2 + 1 if that helps to make it clearer.
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    f(x) is translated by 1 unit in the positive x direction and 4 units in the positive y direction.

    If f(x) is translated by 1 unit in the positive x direction, then f(x) becomes f(x-1)

    If f(x) is translated 4 units in the positive y direction, then f(x) becomes f(x) + 4

    Combining the two, f(x) becomes f(x-1) + 4
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    replace x with x - 1 in f(x)... then add 4

    job done
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    (Original post by the bear)
    replace x with x - 1 in f(x)... then add 4

    job done
    haha laughed at this a little , cheers bro
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    (Original post by madfish)
    haha laughed at this a little , cheers bro
    It's better to try and understand it too... TSR won't be with you in the exam to give you answers.
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    Giving you the answer isn't going to help you.

    Say you have some function f(x) and at x=3 it has the value f(3) = 7

    If the graph has been translated by one unit in the positive x-direction, then at which value of x do we have that f(x) = 7? Given that the graph has been essentially moved to the right by one, we see it no longer happens at x=3 but instead happens at x=4, and all values gets shifted by one. How can this be expressed in terms of f(x)? For simplicity, let's denote the translation by one unit in the positive x-direction by the function F(x), so using the example we've already done

    f(3) = 7, \ \ F(4) = 7

    We can put this more generally

    f(x) = y, \ \ F(x+1) = y

    So, you should be able to tell, we can express F(x) in terms of f(x). Specifically

    F((x-1)+1) =F(x) = f(x-1)

    Now, translation in the y-direction is easy. Let's go back to the original example of if we have f(x) and at x=3 it has the value f(3) = 7

    Translating the graph up by four units, it's easy to notice that the value at x=3 is not going to be 7 but 7+4 = 11

    If we're translating the graph up by 4 units, then, denoting this translation as a new function G(x) we get that

    G(x) = f(x) + 4

    Because all y-values are going to increase by 4.

    If we're combining these two translations we can express this as

    G \cdot F(x) = f(x-1) + 4 = F \cdot G(x)

    The significance of showing G \cdot F(x) = F \cdot G(x) is that it doesn't matter which transformation you do first, it doesn't make a difference.
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    (Original post by Noble.)
    Giving you the answer isn't going to help you.

    Say you have some function f(x) and at x=3 it has the value f(3) = 7

    If the graph has been translated by one unit in the positive x-direction, then at which value of x do we have that f(x) = 7? Given that the graph has been essentially moved to the right by one, we see it no longer happens at x=3 but instead happens at x=4, and all values gets shifted by one. How can this be expressed in terms of f(x)? For simplicity, let's denote the translation by one unit in the positive x-direction by the function F(x), so using the example we've already done

    f(3) = 7, \ \ F(4) = 7

    We can put this more generally

    f(x) = y, \ \ F(x+1) = y

    So, you should be able to tell, we can express F(x) in terms of f(x). Specifically

    F((x-1)+1) =F(x) = f(x-1)

    Now, translation in the y-direction is easy. Let's go back to the original example of if we have f(x) and at x=3 it has the value f(3) = 7

    Translating the graph up by four units, it's easy to notice that the value at x=3 is not going to be 7 but 7+4 = 11

    If we're translating the graph up by 4 units, then, denoting this translation as a new function G(x) we get that

    G(x) = f(x) + 4

    Because all y-values are going to increase by 4.

    If we're combining these two translations we can express this as

    G \cdot F(x) = f(x-1) + 4 = F \cdot G(x)

    The significance of showing G \cdot F(x) = F \cdot G(x) is that it doesn't matter which transformation you do first, it doesn't make a difference.
    thanks for that

    but... If I do 1- 4(x-1)^2 + 4 this expands to -4x^2 - 2x + 4

    which is the wrong answer... what am I doing wrong?
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    (Original post by madfish)
    thanks for that

    but... If I do 1- 4(x-1)^2 + 4 this expands to -4x^2 - 2x + 4

    which is the wrong answer... what am I doing wrong?
    Expanding incorrectly

    5 - 4(x-1)^2 is NOT -4x^2 - 2x + 4
 
 
 
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