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]}