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Anonymous asked in Science & MathematicsMathematics · 9 years ago

Maths question, differentiation/graphs?

The line f(x)=1-6x is parallel to f(x)=x^3+ax^2-5x+1 at x=1.

Find the value of a and the equation of the tangent at the indicated point.

This question is confusing me, could someone pease explain it to me?

Thanks.

3 Answers

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  • s k
    Lv 7
    9 years ago
    Favorite Answer

    Let f(x) = 1 - 6x, g(x) = x^3 + ax^2 - 5x + 1.

    So f'(1) = g'(1).

    f'(x) = -6. So g'(1) = -6.

    g'(x) = 3x^2 + 2ax - 5, g'(1) = 3 + 2a - 5 = 2a - 2.

    2a - 2 = -6,

    a - 1 = -3,

    a = -2.

    g(1) = 1 - 2 - 5 + 1 = -5

    g'(1) = 3 - 4 - 5 = -6

    So our tangent is y + 5 = -6(x - 1), y + 5 = -6x + 6, y = -6x + 1.

  • MechEng2030
    Lv 7
    9 years ago

    f '(x) = 3x^2 + 2ax - 5 => f '(1) = -6 = 2a - 2 => a = -2

    Equation of the tangent:

    y + 5 = -6(x - 1)

    y = -6x + 1

  • ?
    Lv 7
    9 years ago

    Parallel means they have the same gradient.

    first line f'(x) = -6

    2nd line f'(x) = 3x^2 + 2ax - 5

    we know

    -6 = 3x^2 + 2ax - 5 for x=1

    -6 = 3 + 2a - 5

    -4 = 2a

    a = -2

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