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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- Apr 20th 2009, 03:23 PMcraigmfind 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) - Apr 20th 2009, 05:42 PMpickslides
Your Equation

$\displaystyle y= (x^3-1)(x^2+3x+2)$

Should be expanded to give

$\displaystyle y= x^5+3x^4+2x^3-x^2-3x-2$

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

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

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

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

As the result is zero, this implies a stationary point. - Apr 20th 2009, 05:50 PMTheEmptySet
- Apr 20th 2009, 05:59 PMpickslides
Sorry craigm, I did not actually find the derivative to your function

Your Equation

$\displaystyle y= (x^3-1)(x^2+3x+2)$

Should be expanded to give

$\displaystyle y= x^5+3x^4+2x^3-x^2-3x-2$

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

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

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