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Core 1 Graph Question, someone help please? Watch

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    so a=1
    b = 2
    c= 4?
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    (Original post by liverpool2044)
    so a=1
    b = 2
    c= 4?
    No. If you plug those into the formula b^2 - 4ac what will you get? Nothing, because there is nothing to solve! You need some k terms, your a is correct however. I'll do "b" for you and you can do c.

    b = -(k+2)
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    Ahh i dont know why im so confused by this haha. I suppose theres a 2k left so 2?
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    (Original post by liverpool2044)
    Ahh i dont know why im so confused by this haha. I suppose theres a 2k left so 2?
    C is basically any term that isn't multiplied by x. So c = 2k + 1
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    (Original post by liverpool2044)
    Ahh i dont know why im so confused by this haha. I suppose theres a 2k left so 2?
    a is the sum of the coefficients of x^2
    b is the sum of the coefficients of x (x^1)
    c is the sum of all the constant terms (you could say coefficients of x^0)
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    Ah okay makes a lot of sense now, thank you.

    So its -(k+2)^2-4x1x2k-1

    Do i then expand the b^2
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    i got to

    k(k-4)+5 so the answers are k = 0 or 4?
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    (Original post by liverpool2044)
    Ah okay makes a lot of sense now, thank you.

    So its -(k+2)^2-4x1x2k-1

    Do i then expand the b^2
    Be careful with brackets (also it is 2k+1 not 2k-1)

    (-(k+2))^2 - 4*1*(2k+1) = ?

    You have to equate this to something. You are looking for a tangent, so how many points of intersection will there be with the curve?
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    (Original post by liverpool2044)
    i got to

    k(k-4)+5 so the answers are k = 0 or 4?
    That's what I got too
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    thanks! great help
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    (Original post by liverpool2044)
    thanks! great help
    no problem let's see if you understand the b^2 - 4ac now, what's a b and c in each of these:

    i) kx^2 - 4x + 3 = 0
    ii) 3x - (k+1)x^2 - kx = 0
    iii) x^2 - 3(k+1)x + k - 2 = 0
 
 
 
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