Question

How do you use the first and second derivatives to sketch `y= -(x-2) (x+2) (x-4)`?

Answer 1

`y=-(x-2)(x+2)(x-4)`
graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}

Explanation:

Fist we find the roots of the equation, ie the values of `x` st `y=0`:

`y=0 => -(x-2)(x+2)(x-4)=0`
`:. (x-2)(x+2)(x-4)=0`
`:. x=2, x=-2, x=4`

Next we find the critical points, ie the values for which `y'=0`, we will need to multiply out as a cubic in order to this:

`y=-(x-2)(x+2)(x-4)`
`:. y=-(x-2)(x^2-4x+2x-8)`
`:. y=-(x-2)(x^2-2x-8)`
`:. y=-(x^3-2x^2-8x-2x^2+4x+16)`
`:. y=-(x^3-4x^2-4x+16)`
`:. y=-x^3+4x^2+4x-16`

So, differentiating wrt `x` gives:
`:. y'=-3x^2+8x+4`
`y'=0 => -3x^2+8x+4 = 0`
`:. 3x^2-8x-4 = 0`
`:. 3x^2-8x-4 = 0`

This quadratic does not factorise, so we use the quadratic formula:
`x=(-(-8)+-sqrt((-8)^2-4(3)(-4)))/(2(3))`
`:. x=(8+-sqrt(64+48))/(6)`
`:. x=(8+-sqrt(112))/(6)`
`:. x=-0.43, 3.10` (2dp)

, or equally we can solve by completing the square:
`3x^2-8x-4 = 0`
`:. 3(x^2-8/3x-4/3) = 0`
`:. (x^2-8/3x-4/3) = 0`
`:. (x-8/6)^2 -(8/6)^2-4/3 = 0`
`:. (x-8/6)^2 -64/36-4/3 = 0`
`:. (x-8/6)^2 = 64/36+48/36`
`:. (x-8/6)^2 = 112/36`
`:. (x-8/6) = +-sqrt(112/36)`
`:. x=8/6 +-sqrt(112/36)`
again leading to `x=-0.43, 3.10` (2dp)

The coordinates of these points are then given by using the original formula for `y` to give:
`x=-0.43 => y=-16.9` (1dp)
`x=3.10 => y=5.0` (1dp)

These values of `x` correspond to the critical points (or turning points). we now use the second derivative to determine the nature (max, min or point of inflection) of these points.

from earlier, `y'=-3x^2+8x+4`
`:. y''=-6x+8`
`x=-0.43 => y''=10.5 > 0` (1dp)
`x=3.10 => y''=-10.6<0` (1dp)

So we can now determine the nature of the turning points which are:
`(-0.43,-16.9)` minimum
`(3.10, 5.0)` maximum

We now have enough to sketch the curve:
graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}