How do I graph the polynomial below? f(x)=(x-1)^3(x+4)^2(x-2)

f(x)=(x-1)^3(x+4)^2(x-2)
    To graph this Polynomial, you should create an Input output table. I would include negative and positive numbers for the independent variable, -3, -2, -1, 0, 1, 2, and 3. I would substitute each in for x and follow the order of operations. Record your answers as an ordered pair. 
Graph the ordered pairs and connect your dots. 

Here is a view showing the local maximum:

We are asked to graph f(x)=(x-1)^3(x+4)^2(x-2) :
First we will use only techniques from Algebra. There are a few key attributes of graphs of polynomials we are interested in: end behavior (how the graph behaves as x grows without bound or tends to negative infinity), intercepts, and the number of turning points (local maxima or minima).
Note that f(x) is a 6th degree polynomial. (If you multiplied the factors the highest degree would be 6.) Since the leading coefficient is positive 1, as x tends to positive or negative infinity the graph tends to positive infinity. (If we zoomed sufficiently far out, the graph would essentially be a "U" shape -- near zero there would be some "shape" as the graph squiggles around.)
A sixth degree polynomial has at most 5 turning points.
The y-intercept is found by setting x=0; f(0)=32. The function is given in factored form so we can easily see the x-intercepts: x=2, x=1, and x=-4. The zero at x=-4 is of multiplicity 2 -- that means that the graph touches the x-axis at x=-4 without going through (changing sign.) The zero at x=1 has multiplicity 3, so the graph changes sign at x=1.
With the intercepts and the end behavior described, we can compute some points and connect them with a smooth curve.
** With calculus: The first derivative is:
f'(x)=3(x-1)^2(x+4)[2x^2+x-8]
The critical points are x=1 (multiplicity 2 -- an inflection point in this case), x=-4 (a local minimum), 1/4+-sqrt(65)/4 (local max when subtracting, global min when adding.)
 
The graph:

(This view shows the major points of interest except the local max.)