Thread: 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)

Originally Posted by craigm

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



Should be expanded to give



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



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



As the result is zero, this implies a stationary point.

Originally Posted by craigm

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)



Now if we evaluate at x=1 we get

Originally Posted by pickslides

Your Equation



Should be expanded to give



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



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



As the result is zero, this implies a stationary point.

Sorry craigm, I did not actually find the derivative to your function

Your Equation



Should be expanded to give



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



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