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