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Math Help - gradient

  1. #1
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    gradient

    hi, my question is: the equation of a curve is y = (x - 8)(x - 2) find the gradient of the curve
    a) at the point where the curve crosses the y-axis,
    b) at each of the points where the curve crosses the x-axis

    for a) i expanded the brackets then differentiated:

    x^2 - 2x - 8x + 16 \implies x^2 - 10x + 16 \implies \frac{dy}{dx} = 2x - 10 since x is zero on the y-axis the answer must be -10 and my book confirms this.

    for b) i just made it 2x - 10 = 0 \implies 2x = 10 \implies x = 5 but i don't think this is going in the right direction as the answer the book gives is "6 and -6"

    could someone show me how its done please?

    thanks, mark
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mark View Post
    hi, my question is: the equation of a curve is y = (x - 8)(x - 2) find the gradient of the curve
    a) at the point where the curve crosses the y-axis,
    b) at each of the points where the curve crosses the x-axis

    for a) i expanded the brackets then differentiated:

    x^2 - 2x - 8x + 16 \implies x^2 - 10x + 16 \implies \frac{dy}{dx} = 2x - 10 since x is zero on the y-axis the answer must be -10 and my book confirms this.

    for b) i just made it 2x - 10 = 0 \implies 2x = 10 \implies x = 5 but i don't think this is going in the right direction as the answer the book gives is "6 and -6"

    could someone show me how its done please?

    thanks, mark
    HI

    y=(x-8)(x-2)

    The curve cuts the x-axis at x=8 and x=2

    So substitute 8 and 2 respectively into the equation of gradient u got .
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