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    I worked out Q(a) but not sure about (b). I got -x^2-2cx+c for part a; I might be wrong though but I assume you just expand the brackets and make it in that form. Thanks in advance!


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    This is not the answer, but your handwriting is amazing - do you use parker pens/expensive pens?
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    (Original post by KK.Violinist)
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    I worked out Q(a) but not sure about (b). I got -x^2-2cx+c for part a; I might be wrong though but I assume you just expand the brackets and make it in that form. Thanks in advance!


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    Your part (a) is slightly incorrect, you have -(x-c)(x-c) = -(x^2 - 2cx + c^2) (note the square of the c) - now just distribute the - sign into the brackets.
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    Work out all the letters in the equations then find B with making f(X) = g(X). C is the X intercept of h(X) so that should be easy and then length from that.
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    (Original post by KK.Violinist)
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    I worked out Q(a) but not sure about (b). I got -x^2-2cx+c for part a; I might be wrong though but I assume you just expand the brackets and make it in that form. Thanks in advance!


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    Note that f'(0) = gradient of line 1 - can you then write down the equation of line 1 (you know where it intersects the y-axis as well) and find where it intersects g(x)? What about finding where g(x) and f(x) intersect and then can you use g'(x) to find the gradient of \ell_2?
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    (Original post by TheYearNiner)
    This is not the answer, but your handwriting is amazing - do you use parker pens/expensive pens?
    Thanks haha, I use a standard ballpoint pen.


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    (Original post by Zacken)
    Note that f'(0) = gradient of line 1 - can you then write down the equation of line 1 (you know where it intersects the y-axis as well) and find where it intersects g(x)? What about finding where g(x) and f(x) intersect and then can you use g'(x) to find the gradient of \ell_2?
    Sorry, my minds gone blank. How would I go about finding the equation of line 1? y=mx+c where -8 is c and mx being f'(x)?


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    (Original post by KK.Violinist)
    Sorry, my minds gone blank. How would I go about finding the equation of line 1? y=mx+c where -8 is c and mx being f'(x)?


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    Differentiate, sub in the x value of the coordinate of the point given which lies on line to get the gradient; and use y-y1 =m(x-x1).

    Your handwriting is so damn good, it makes me feel like a toddler :eek:
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    (Original post by kkboyk)
    Differentiate, sub in the x value of the coordinate of the point given which lies on line to get the gradient; and use y-y1 =m(x-x1).

    Your handwriting is so damn good, it makes me feel like a toddler :eek:
    Okay, if my calculations are correct then the line is y=2x-8.

    Why is everyone talking about my handwriting!!! Haha I think it's pretty average, I've seen classmates with better handwriting xD


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    (Original post by KK.Violinist)
    Okay, if my calculations are correct then the line is y=2x-8.

    Why is everyone talking about my handwriting!!! Haha I think it's pretty average, I've seen classmates with better handwriting xD


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    Because most people doing Maths (especially university Maths students and Professors) have terrible handwriting.

    I sometimes can't read my own handwriting
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    (Original post by kkboyk)
    Because most people doing Maths (especially university Maths students and Professors) have terrible handwriting.

    I sometimes can't read my own handwriting
    Lol that's true but surprisingly my teacher has very neat handwriting that makes even mine look mediocre.

    Anyway, back on topic, now I have to find point A? I really shouldn't be doing this so late during the night. My brain keeps switching off -_-


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    (Original post by KK.Violinist)
    Lol that's true but surprisingly my teacher has very neat handwriting that makes even mine look mediocre.

    Anyway, back on topic, now I have to find point A? I really shouldn't be doing this so late during the night. My brain keeps switching off -_-


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    Simultaneous equations with line L1 and g(x).
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    (Original post by kkboyk)
    Simultaneous equations with line L1 and g(x).
    I'm slightly confused by the D. I found D=40 but not sure if I'm correct. Do I even need to find D? Sorry for my amateur replies


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    (Original post by KK.Violinist)
    I'm slightly confused by the D. I found D=40 but not sure if I'm correct. Do I even need to find D? Sorry for my amateur replies


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    What is D?
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    (Original post by kkboyk)
    What is D?
    g(x)=x^2-14x+D




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    (Original post by KK.Violinist)
    g(x)=x^2-14x+D

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    Oh its a constant.

    If you differentiate g(x) and also L1 make them equal to each other (as they both will have the same gradient at that point) you'll be able to find the x value
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    (Original post by kkboyk)
    Oh its a constant.

    If you differentiate g(x) and also L1 make them equal to each other (as they both will have the same gradient at that point) you'll be able to find the x value
    So find the second derivative for f(x) and the first for g(x) and equal to each other? If so then I got x=8


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