Tangent of the Curve/ Calculus

I need a bit of help understanding this -

Find a point on your graph which has a postive gradient. Find this gradient by using calculus and establish the equation of the tangent at this point on the curve.

Can someone show me an example of doing this by using the function y= (x-1)(x-2)(x-3)?

- I hope this made sense :)

The gradient of a function is just the derivative.
So multiplying out you get
x³ - 6x² + 11x - 6
The derivative is
3x² - 12x + 11
So you want a value when this function is > 0
Factorising you get (and rounding off)
(x - 1.42)(x-2.58) > 0

So for this to be true, either both brackets must be positive,
or both must be negative. Therefore you can choose any
x > 2.58 or x < 1.42
So you could choose x = 3 as this has a positive gradient
So your y -value = 0
Therefore a point (3,0) has a positive gradient
Now that you have a point, all you have to do is find the tangent
to the curve at this point
Now at this point the gradient is 27 - 36 + 11 = 2
So you have y = 2x + c
0 = 2(3) + c therefore c = -6
So your tangent to the curve at (3,0) is y = 2x -6

Where did u get these numbers from?

Now at this point the gradient is 27 - 36 + 11 = 2

:confused:

The derivative of the function is 3x² - 12x + 11
Subsituting your point (3,0) you get
3(3)² - 12(3) + 11
= 3(9) - 36 + 11
= 27 - 36 + 11
= 2

Cheers for the help, it was much needed


- :)