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    The line y=4x+k is a tangent to the curve y=x^2+4x+3 at the point P

    Find the value of k
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    Solve as a simultaneous equation, y = y method, as you're looking for them to meet.

    This will then involve factorising - can you identify P from there? How do you know it's a tangent at that point, rather than just a point of intersection?
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    Scratch that. The second step only works easily when k is already known. Do the first step, then ask yourself what condition needs to be met for this to be a tangent?

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    b^2-4ac=0 must be satisfied
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    why dont you just differentiate the quadratic equation and equate it to 4. Then solve for x. Sub this value into the original quadratic and get the y value. then sub both the x and y values in the linear equation to get k.
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    Im really confuse with this question, could someone show me some working out?
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    Equate them to get x^2+3-k=0 then since there is one point, P, the discriminant=0. So b^2-4ac=0 12-4k=0 therefore k=3 and so x=0 thus y=3 k=3 P=(0,3)
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    (Original post by haken102)
    Im really confuse with this question, could someone show me some working out?
    Okay.
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    y=4x+k
    y=x^2+4x+3

    So equate them...

    x^2+4x+3 = 4x+k
    x^2 + (3-k) = 0

    So for b^2-4ac, a=1, b=0, c=(3-k)

    b^2-4ac=0 gives a tangent

    (0)^2-4(1)(3-k)=0
    -4(3-k)=0
    3-k=0
    3=k

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    why did you use the discriminant?
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    nvm, i figured it out. Thank you so much guys.
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    (Original post by haken102)
    nvm, i figured it out. Thank you so much guys.
    Next time you will be better off posting in the maths forum
 
 
 
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