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]}
