Math Help - find the slope of the curve:

1. find the slope of the curve:

hi i have no idea where to start on this.

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

2. Originally Posted by craigm
hi i have no idea where to start on this.

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

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

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.

3. Originally Posted by craigm
hi i have no idea where to start on this.

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

$y'(1)=(3)(1+3+2)+(1-1)(2+3)=18$

4. Originally Posted by pickslides

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

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

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

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