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    Struggling to make sense of this past paper question (Question 12 (ii) onwards).

    http://vle.woodhouse.ac.uk/topicdocs...010January.pdf

    I want to understand why this works - I've seen the mark scheme and know the answers, but would like someone to explain it to me in "writing" terms :rolleyes:

    Thanks
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    Its an x^2 parabola. So its symmetric. If we know the two points of same y, the max/min y will be inbetween those two points.

    The height is just putting in the x-vlaue.
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    (Original post by jamescbsmith)
    Struggling to make sense of this past paper question (Question 12 (ii) onwards).
    Quadratics are symmetrical about their vertex

    So that is how you can easily do (ii)

    Then, of course finding d is obvious

    So you use d as x to find y, then y will be more than 4 so that is (iii) finished

    For (iv) you know y so you can find x then use symmetry again to give x+width+x = 10 to find the width
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    (Original post by TenOfThem)
    Quadratics are symmetrical about their vertex

    So that is how you can easily do (ii)

    So you use d as x to find y, then y will be more than 4 so that is (iii) finished

    For (iv) you know y so you can find x then use symmetry again to give x+width+x = 10 to find the width
    Could you elaborate on what you mean by symmetry please? (gosh I feel silly asking)
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    (Original post by jamescbsmith)
    Could you elaborate on what you mean by symmetry please? (gosh I feel silly asking)
    There is a line of reflection symmetry through the vertex
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    (Original post by TenOfThem)
    There is a line of reflection symmetry through the vertex
    Ah I think I understand - so for (ii) all I really had to do is divide 10 by 2 to get x=5 as x^2 is a parabola and there is symmetry about the vertex!
    Thanks to you both!
 
 
 
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