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Math Help - Tangent to a curve

  1. #1
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    Tangent to a curve

    Find the value of c for which the line y=x+c is a tangent to the curve y=x^2-5x+4
    thanks!
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  2. #2
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    Quote Originally Posted by BabyMilo View Post
    thanks!
    do i dy/dx it? then dy/dx=0

    then sub x into


    to get y.

    then y=x+c
    to find c?

    thanks!
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  3. #3
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    If it is tangent to the curve, it will touch the curve only once.

    Since they touch, they must be equal.

    So x + c = x^2 - 5x + 4

    x^2 - 6x + 4 - c = 0


    Since it only touches once, the discriminant must be 0.

    So \Delta = (-6)^2 - 4(1)(4 - c) = 0

    36 - 16 + 4c = 0

    20 + 4c = 0

    4c = -20

    c = -5.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    If it is tangent to the curve, it will touch the curve only once.

    Since they touch, they must be equal.

    So x + c = x^2 - 5x + 4

    x^2 - 6x + 4 - c = 0


    Since it only touches once, the discriminant must be 0.

    So \Delta = (-6)^2 - 4(1)(4 - c) = 0

    36 - 16 + 4c = 0

    20 + 4c = 0

    4c = -20

    c = -5.
    stupid me.
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  5. #5
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    This was posted in the "Pre Calculus" section so Prove It used a method that did not require the derivative.

    Since you do mention the derivative, no, the derivative is not 0 where the line y= x+ c is tangent to it. A line is tangent to a curve where its slope is the same as the derivative. y= x+ c has slope 1 so you are looking for a place where the derivative is 1, not 0.
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  6. #6
    Super Member Bacterius's Avatar
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    Here is a method using the derivative.

    --------------------

    The derivative of the function f(x) = x^2 - 5x + 4 is f'(x) = 2x - 5 (power rule on sum of functions).

    The line y = x + c has a slope equal to 1. Thus, you are looking for the point on the curve of f(x) where the slope is equal to 1. So, you must solve 2x - 5 = 1 for x. Hmm, x = 3.

    Say a = 3 (to make it less confusing). You know that the tangent to the curve of f in a point of absciss a is equal to :

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

    Substitute : y = (x - 3) \times 1 - 2

    That is : y = x - 5. Therefore c = -5.

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  7. #7
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    Quote Originally Posted by Bacterius View Post
    Here is a method using the derivative.

    --------------------

    The derivative of the function f(x) = x^2 - 5x + 4 is f'(x) = 2x - 5 (power rule on sum of functions).

    The line y = x + c has a slope equal to 1. Thus, you are looking for the point on the curve of f(x) where the slope is equal to 1. So, you must solve 2x - 5 = 1 for x. Hmm, x = 3.

    Say a = 3 (to make it less confusing). You know that the tangent to the curve of f in a point of absciss a is equal to :

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

    Substitute : y = (x - 3) \times 1 - 2

    That is : y = x - 5. Therefore c = -5.

    Since you were already given that y= x+ c, y= 3+ c= 3^2- 5(3)+ 4= 9- 15+ 4= -2 so c= -2-3= -5.

    That seems simpler to me.
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  8. #8
    Super Member Bacterius's Avatar
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    Ah, yes, I didn't spot that
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