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    How would you find the turning point of a quadratic graph when you are only given the y intercept(0,-10) and the coordinate of one intersection on the x axis (5,0)? Any help would be appreciated
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    BTW just to add this was a question from the maths GCSE calculator paper 3 from 2017
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    (Original post by Dan_Jenkins)
    How would you find the turning point of a quadratic graph when you are only given the y intercept(0,-10) and the coordinate of one intersection on the x axis (5,0)? Any help would be appreciated
    Can you post the exact question or a picture? I looked in my copy of Edexcel Maths paper 3 2017 9-1 and I can't find it lol. I think I could solve it if it stated that a (the x squared co-efficient) = 1.

    For the case that a = 1:

    Edit: According to somebody in another thread, my solution is correct.
    Spoiler:
    Show

    If 5 is a root of the quadratic (x-5) must be a factor:
    (x-5)(x+k) = 0

    -5 and k must multiply to give -10 as -10 is the y intercept.
    -5k = -10
    k = 2
    (x-5)(x+2) = 0
    x^2 -3x -10 = 0
    By completing the square:
    (x-1.5)^2 - 2.25 - 10 = 0
    (x-1.5)^2 - 12.25 = 0

    So x= 1.5, y = -12.25
    Turning point = (1.5, -12.25)
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    (Original post by Loci Pi)
    Can you post the exact question or a picture? I looked in my copy Edexcel Maths paper 3 2017 9-1 and I can't find it lol. I think I could solve it if it stated that a (the x squared co-efficient) = 1.

    For the case that a = 1:

    If 5 is a root of the quadratic (x-5) must be a factor:
    (x-5)(x+k) = 0

    -5 and k must multiply to give -10 as -10 is the y intercept.
    -5k = -10
    k = 2
    (x-5)(x+2) = 0
    x^2 -3x -10 = 0
    By completing the square:
    (x-1.5)^2 - 2.25 - 10 = 0
    (x-1.5)^2 - 12.25 = 0

    So x= 1.5, y = -12.25
    Turning point = (1.5, -12.25)
    Please do not post full solutions on the maths forum
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    (Original post by Y11_Maths)
    Please do not post full solutions on the maths forum
    Sorry, I will spoiler it now but it was only the solution for if a = 1, which the thread starter did not specify.
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    (Original post by Y11_Maths)
    Please do not post full solutions on the maths forum
    why not?
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    (Original post by misspessimism)
    why not?
    So the thread starter can attempt the question once they are given a few hints without the answer being spoiled.
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    (Original post by Loci Pi)
    So the thread starter can attempt the question once they are given a few hints without the answer being spoiled.
    Ahh. i thought you meant legal reasons
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    (Original post by misspessimism)
    why not?
    It is against the rules in the maths forum because this is not a “do my homework” place.
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    (Original post by Dan_Jenkins)
    How would you find the turning point of a quadratic graph when you are only given the y intercept(0,-10) and the coordinate of one intersection on the x axis (5,0)? Any help would be appreciated
    https://www.thestudentroom.co.uk/sho...4#post76280584

    I've already given a solution for this problem for someone else...
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    (Original post by Y11_Maths)
    It is against the rules in the maths forum because this is not a “do my homework” place.
    Ah well thats my question out the window..
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    (Original post by Mehru1214)
    https://www.thestudentroom.co.uk/sho...4#post76280584

    I've already given a solution for this problem for someone else...
    How are you finding my paper?
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    (Original post by Y11_Maths)
    How are you finding my paper?
    It looks good, however unfortunately I had to leave it, so I'm going to do it properly tomorrow! Nice though!
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    (Original post by Loci Pi)
    Can you post the exact question or a picture? I looked in my copy of Edexcel Maths paper 3 2017 9-1 and I can't find it lol. I think I could solve it if it stated that a (the x squared co-efficient) = 1.

    For the case that a = 1:

    Edit: According to somebody in another thread, my solution is correct.
    Spoiler:
    Show


    If 5 is a root of the quadratic (x-5) must be a factor:
    (x-5)(x+k) = 0

    -5 and k must multiply to give -10 as -10 is the y intercept.
    -5k = -10
    k = 2
    (x-5)(x+2) = 0
    x^2 -3x -10 = 0
    By completing the square:
    (x-1.5)^2 - 2.25 - 10 = 0
    (x-1.5)^2 - 12.25 = 0

    So x= 1.5, y = -12.25
    Turning point = (1.5, -12.25)

    AKA me XD
    https://www.thestudentroom.co.uk/sho...4#post76280584
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    (Original post by Mehru1214)
    It looks good, however unfortunately I had to leave it, so I'm going to do it properly tomorrow! Nice though!
    Ok
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