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.)

# 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.