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    "The line y=6x-7 is a tangent to the curve y=x^2+k. Find k."

    I missed this lesson, and my textbook is not at all helpful. I've been trying to do this for half an hour, but no matter what I do, I come to a dead end almost immediately.

    All I have so far is that the differentiated equation for the curve is 2x.Should I put 2x equivalent to 6x-7?

    Just point me in the right direction, please.Thanks so much!
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    Equate the 2 equations involving y. Then rearrange to form a quadratic. As the line is a tangent to the curve, the discriminant of the quadratic formed will be 0. You can use this to find k.
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    (Original post by JustJusty)
    "The line y=6x-7 is a tangent to the curve y=x^2+k. Find k."

    I missed this lesson, and my textbook is not at all helpful. I've been trying to do this for half an hour, but no matter what I do, I come to a dead end almost immediately.

    All I have so far is that the differentiated equation for the curve is 2x.Should I put 2x equivalent to 6x-7?

    Just point me in the right direction, please.Thanks so much!
    Ooooh close...

    What does the differentiated equation actually tell you about?
    Gradient.
    Therefore, the gradient of any point on y = x2 + k is 2x.

    What's the gradient of the equation y = 6x - 7 ?
    m = 6
    So, you can equate the two gradients.
    Solve for x, solve for y, solve for k.
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    This is not a question that needs any differentiation. First, lets equate the two lines together (so we can find a common point, a solution).

    So,

    x^2 + k = 6x - 7

    Now, make the equation equal to 0.

    x^2 - 6x + (7+k) = 0

    Since the straight line is a tangent to the curve, there can only be one solution. Therefore the discriminant, b^2 - 4ac = 0.

    In the equation formed above, a = 1, b= -6 and c = (7+k)

    Solve b^2 = 4ac (your only variable is k), and you should get your value of k.

    ANSWER SPOILER:

    EDIT: Whilst I was posting, two others gave the solution lol, that should reassure you of the method
    Spoiler:
    Show
    k=2
    • Thread Starter
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    (Original post by B_9710)
    Equate the 2 equations involving y. Then rearrange to form a quadratic. As the line is a tangent to the curve, the discriminant of the quadratic formed will be 0. You can use this to find k.
    Oh gosh, of course! Here I was thinking it was something horribly complicated. I forgot to take into account that a tangent basically has a double root on the curve.

    Thank you!
 
 
 
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