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Math Help - find the slope of the curve:

  1. #1
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    find the slope of the curve:

    hi i have no idea where to start on this.

    please could somebody explain

    find the slope on the line tangent to the curve
    y = (x^3 -1)(x^2+3x+2) at the point (1,0)
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  2. #2
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    Quote Originally Posted by craigm View Post
    hi i have no idea where to start on this.

    please could somebody explain

    find the slope on the line tangent to the curve
    y = (x^3 -1)(x^2+3x+2) at the point (1,0)
    Your Equation

    y= (x^3-1)(x^2+3x+2)

    Should be expanded to give

    y= x^5+3x^4+2x^3-x^2-3x-2

    Using differentiation to help find the gradient (or tangent) with respect to x

    \frac{dy}{dx} = x^5+3x^4+2x^3-x^2-3x-2

    Now substituting x=1 into the derivative will give you the gradient of the tangent.

     (1)^5+3(1)^4+2(1)^3-(1)^2-3(1)-2 = 0

    As the result is zero, this implies a stationary point.
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  3. #3
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    Quote Originally Posted by craigm View Post
    hi i have no idea where to start on this.

    please could somebody explain

    find the slope on the line tangent to the curve
    y = (x^3 -1)(x^2+3x+2) at the point (1,0)
    first we need to take the derivative.(product rule)

    y'=(3x^2)(x^2+3x+2)+(x^3-1)(2x+3)

    Now if we evaluate at x=1 we get

    y'(1)=(3)(1+3+2)+(1-1)(2+3)=18
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Your Equation

    y= (x^3-1)(x^2+3x+2)

    Should be expanded to give

    y= x^5+3x^4+2x^3-x^2-3x-2

    Using differentiation to help find the gradient (or tangent) with respect to x

    \frac{dy}{dx} = x^5+3x^4+2x^3-x^2-3x-2

    Now substituting x=1 into the derivative will give you the gradient of the tangent.

     (1)^5+3(1)^4+2(1)^3-(1)^2-3(1)-2 = 0

    As the result is zero, this implies a stationary point.
    Sorry craigm, I did not actually find the derivative to your function

    Your Equation

    y= (x^3-1)(x^2+3x+2)

    Should be expanded to give

    y= x^5+3x^4+2x^3-x^2-3x-2

    Using differentiation to help find the gradient (or tangent) with respect to x

    \frac{dy}{dx} = 5x^4+12x^3+36x^2-2x-3

    now substitute x = 1 into this equation! This should yield 18.

    My method is a little more primative to that supplied by the previous poster (thanks for the correction!) but my method is great if you have not been introduced to the product rule
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