1. x-intercepts

`{: (y,=x^4-6x^2 ),( ,=x^2(x^2-6) ),( ,=x^2(x+sqrt6)(x-sqrt6) ),( ,=0) :}`

when

`x = 0`, `x=-sqrt(6)` and `x=sqrt(6)`

2. y-intercepts:

When `x = 0`, `y = 0`.

3. maxima and minima?

`{: ((dy)/(dx),=4x^3-12x ),( ,=4x(x^2-3) ),( ,=4x(x+sqrt3)(x-sqrt3) ),( ,=0) :}`

Now `(dy)/(dx)=0` when `x = 0` or `x=-sqrt(3)` and `x=sqrt(3)`

So we have max or min at `(0,0)` and `(-sqrt(3),-9)` and `(sqrt(3),-9)`.

4. Second derivative:

`(d^2y)/(dx^2)=12x^2-12`

Now `y^('') > 0` for `x = -sqrt3` so `(-sqrt3, -9)` is a local MIN

Now `y^('') < 0` for `x = 0` so `(0, 0)` is a local MAX

Now `y^('') > 0` for `x = sqrt3` so `(sqrt3, -9)` is a local MIN

5. Points of inflection:

We now use the second derivative to find points of inflection:

`(d^2y)/(dx^2)=12x^2-12`

`=12(x+1)(x-1)`

`=0`

when `x = -1` or `x = 1`

If `x < -1`, `y^('') > 0`, and for `-1 < x < 1`, we have `y^('') < 0`.

So the sign of `y^('')` has changed, so `(-1, -5)` is a point of inflection.

If `x > -1`, `y^('') > 0`,

So the sign of `y^('')` has changed, so `(1, -5)` is a point of inflection.

So we are ready to sketch the curve:

Graph quartic showing maximum, minimums and points of inflection