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Math Help - [SOLVED] Find the value of constant k in tangent

  1. #1
    Member looi76's Avatar
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    [SOLVED] Find the value of constant k in tangent

    Question:
    Find the value of the constant k for which the line y + 2x = k is a tangent to the curve y = x^2 - 6x + 14.

    Attempt:

    y = x^2 - 6x + 14
    y + 2x = k

    y = k - 2x

    = x^2 - 6x + 2x + 14 - k
    = x^2 - 4x + 14 - k
    a = 1, b = -4, c = 14

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4\times 1\times 14}}{2\times 1}

    I can't continue because Im getting a negative value inside the square root, where did I go wrong?
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  2. #2
    Moo
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    Hello,

    They're not saying that y+2x=k is the equation of the curve, but that it is tangent to the curve.

    This means that at a given point, the tangent to the curve will have y+2x=k as an equation.

    The tangent to the curve at a point of absciss a is :

    y=f'(a)(x-a)+f(a)


    Hence, calculate the derivative of y = x^2 - 6x + 14 and then... think
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  3. #3
    Member looi76's Avatar
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    Thanks Moo!

    y = x^2 - 6x + 14
    a = 1, b = -6, c = 14

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4\times1\times14}}{2\times1}

    I still can't continue!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by looi76 View Post
    Thanks Moo!

    y = x^2 - 6x + 14
    a = 1, b = -6, c = 14

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4\times1\times14}}{2\times1}

    I still can't continue!
    y'=2x-6

    start wit that
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  5. #5
    Member looi76's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    y'=2x-6

    start wit that
    Thanks Mathstud!

    You meant that I need to differentiate first.

    After I differentiate and get y = 2x - 6. What should I do next?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by looi76 View Post
    Thanks Mathstud!

    You meant that I need to differentiate first.

    After I differentiate and get y = 2x - 6. What should I do next?
    If y=-2x+k is to be a tangent line to y=x^2-6x+14..it must satify =y-f(x_0)=f'(x_0)(x-x_0) Like Moo correctly stated...therefore we get y-f(x_0)=(2(x_0)-6)(x-x_0)

    does that help?
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  7. #7
    Member looi76's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    They're not saying that y+2x=k is the equation of the curve, but that it is tangent to the curve.

    This means that at a given point, the tangent to the curve will have y+2x=k as an equation.

    The tangent to the curve at a point of absciss a is :

    y=f'(a)(x-a)+f(a)


    Hence, calculate the derivative of y = x^2 - 6x + 14 and then... think
    The deritave of y = x^2 - 6x + 14.
    = b^2 - 4ac
    = (-6)^2 - 4\times1\times14
    = -20

    Is the deritave -20 or deritave means something else?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by looi76 View Post
    The deritave of y = x^2 - 6x + 14.
    = b^2 - 4ac
    = (-6)^2 - 4\times1\times14
    = -20

    Is the deritave -20 or deritave means something else?
    derivative means differentiate f'(x)=\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h} where f'(x) is the derivative
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by looi76 View Post
    The deritave of y = x^2 - 6x + 14.
    = b^2 - 4ac
    = (-6)^2 - 4\times1\times14
    = -20

    Is the deritave -20 or deritave means something else?
    You are thinking of the discrminant in the quadractic formula
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  10. #10
    Moo
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    Ok, so let's do it step by step :

    Let f(x)=x^2 - 6x + 14

    f'(x)=2x-6 \Longrightarrow f'(a)=2a-6

    Hence the equation of the tangent line at a point is :

    y=(2a-6)(x-a)+f(a)

    y=(2a-6)(x-a)+a^2-6a+14

    y=2ax-2a^2-6x+6a+a^2-6a+14

    y=2ax-a^2-6x+14

    y+(6-2a)x=14-a^2


    So the only point where there will be a tangent line whose equation is in the form y+2x=k will be for 6-2a=2 \Longleftrightarrow a=\dots

    Hence k=\dots

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  11. #11
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    Quote Originally Posted by looi76 View Post
    Question:
    Find the value of the constant k for which the line y + 2x = k is a tangent to the curve y = x^2 - 6x + 14.
    In the case of a parabola a line is tangent if and only if it has one intersection point with the parabola.

    Thus, y=-2x+k is tangent to y=x^2-6x+14 if and only if -2x+k = x^2 - 6x + 14 has exactly one solution. That means, the quadradic x^2 - 4x + (14-k) = 0 has exactly one solution. That happens precisely when the discriminant vanishes, thus, 16 - 4(14-k) = 0.
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  12. #12
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    Re: [SOLVED] Find the value of constant k in tangent

    Quote Originally Posted by looi76 View Post
    Question:
    Find the value of the constant k for which the line y + 2x = k is a tangent to the curve y = x^2 - 6x + 14.

    Attempt:

    y = x^2 - 6x + 14
    y + 2x = k

    y = k - 2x

    = x^2 - 6x + 2x + 14 - k
    = x^2 - 4x + 14 - k
    a = 1, b = -4, c = 14

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4\times 1\times 14}}{2\times 1}

    I can't continue because Im getting a negative value inside the square root, where did I go wrong?

    Here is how you do it.. Easy and Simple

    Line: y = k-2x
    Curve: y = x-6x+14



    Workings

    k-2x = x-6x+14

    x-6x+2x+14-k = 0

    x-4x+14-k = 0

    a=1, b=-4, c=(14-k)


    [I]Use this Equation[/I] ---- b-4ac

    b-4ac
    Replace..

    =(-4)-[4(14-k)]

    =16-(56-4k)

    =16-56+4k

    4k = -16+56

    4k = 40

    k = 40/4

    k = 10
    <---- Here's your Answer!!!
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