How do you graph `f(x)=x^4-3x^2+2x`?

I assume that you want to graph this without technology.

The graph is at the end of the solution. (I believe it's more instructive to not have it to start.)

`f(x)=x^4-3x^2+2x`

`f` is a polynomial, so, of course its domain is the set of all real numbers.

`y`-intercept `f(0)=0`
The `y`-intercept is `0`. (Or `(0, 0)` if you prefer.)

`x`- intercept(s) :
Solve `f(x)=x^4-3x^2+2x=0`

Factor out the `x`: to get `x(x^3-3x+2)=0`
By inspection or by the Rational Zero Theorem (or "Test"), `1` is a zero of `(x^3-3x+2)`.
By the Factor Theorem, `(x-1)` is a factor.
Use division (in some form) to get the quadratic factor:
`(x^3-3x+2) = (x-1)(x^2+x-2)`

The quadratic is straightforward to factor: `(x^2+x-2)=(x+2)(x-1)`

So we see that `f(x)=x(x+2)(x-1)^2` the zeros (the `x`-intercepts are: `-2, 0, 1` (`1` is a multiple zero of even multiplicity.)

(Write the intercepts: `(-2,0), (0,0)`, and `(1, 0)` if you prefer.)

Analysis of `f'`
Although we can work with `f(x)` in any form, I prefer the standard polynomial form over the 3 factor form.

`f(x)=x^4-3x^2+2x`
So `f'(x)=4x^3-6x+2` Which is never non-existent, so we only need to solve: `f'(x)=4x^3-6x+2 = 0`
By inspection or by the Rational Zero Theorem or by observing that multiple zeros of a polynomial are also zeros of the derivative, we see that `1` is again a zero, so `(x-1)` is a factor and:

`f'(x)=4x^3-6x+2 = 2(2x^3-3x+1)=2(x-1)(2x^2+2x-1)`

The quadratic factor has irrational zeros: `(-1 +- sqrt3)/2`.

Observe that `1< sqrt 3 <2`, so one of the zeros is negative and the other positive. And `(-1 + sqrt3)/2 < (-1 + 2)/2 = 1/2 <1`.

The critical numbers for `f` are Left to right
: `(-1 - sqrt3)/2, (-1 + sqrt3)/2," and " 1`.

For ease of reference, let's call the negative critical number `z_1 =(-1 - sqrt3)/2` and the positive one `z_2 = (-1 + sqrt3)/2`

We need to investigate the sign of `f'` on each of the intervals:

`(-oo,z_1)`, `(z_1, z_2,)`, `(z_2,1)`, `(1,oo)`

If you like test numbers, I'd suggest: `-10, 0, 1/2, 10`.
It may not be clear that `1/2` is in `(z_2,1)` until it is observed that:

`sqrt3 < 2` `=>` `(-1 + sqrt3)/2 < (-1 + 2)/2 = 1/2`

If you prefer, use a factor table.

Whichever method you use, you'll find that:

Increasing/Decreasing
`f'(x) < 0` on `(-oo,z_1)`, so `f` is decreasing on `(-oo,z_1)`
`f'(x) > 0` on `(z_1, z_2,)`, so `f` is increasing on `(z_1, z_2,)`
`f'(x) < 0` on `(z_2,1)`, so `f` is decreasing on `(z_2,1)`
`f'(x) > 0` on `(1,oo)`, so `f` is increasing on `(1,oo)`

Local extrema:
`f(z_1) = f((-1 - sqrt3)/2)` is a local minimum (also global)
`f(z_2) = f((-1 + sqrt3)/2)` is a local maximum
`f(1) = 0` is a local minimum.

Analysis of `f''`

`f''(x)=12x^2-6 = 6(2x^2-1)`

Whose zeros are : `+- 1/sqrt2 = +-sqrt 2/2`

Investigating the sign of `f''` on the appropriate intervals leads us to:

Concavity:
`f''(x) > 0` on `(-oo, -sqrt2 / 2)` So `f` is concave up.
`f''(x) < 0` on `(-sqrt2 / 2, sqrt2 / 2)` So `f` is concave down.
`f''(x) > 0` on `(sqrt2 / 2, oo)` So `f` is concave up.

There are two inflection points. They are:
`((-sqrt2 / 2, f(-sqrt2 / 2) )` which is `(-sqrt2 / 2 "," -5/4-sqrt2 )`
and
`((sqrt2 / 2, f(sqrt2 / 2) )` which is `(sqrt2 / 2 "," -5/4+sqrt2 )`

Now sketch the graph (It may take a couple of rough sketches first:

graph{y=x^4-3x^2+2x [-10, 10, -5, 5]}