Describe the graph of the function f(x) = x3 − 11x2 + 36x − 36. Include the y-intercept, x-intercepts, and the shape of the graph.
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The y-intercept is found by setting x = 0:  f(0) = (0)^3 − 11(0)^2 + 36(0) − 36.  We get f(0) = -36.  The y-intercept is thus (0, -36).

The x-intercepts are found from the roots/zeros of x^3 − 11x^2 + 36x − 36.

Synthetic division is very helpful in finding the roots of polynomials.  We list possible factors of the constant term -36 and test each in

x^3 − 11x^2 + 36x − 36 until we find one or more factors of -36 for which the remainder is zero.  -36 has the possible factors ±1, ±2, ±3, ±4, ±6, ±12, ±18, ±36.

Let's try -4.  Is this a root of x^3 − 11x^2 + 36x − 36?

-4  /    1    -11    36    -36

                4     28    256

     ------------------------------  No, -4 is not a root of x^3 − 11x^2 + 36x − 36

       1        -7     64    220       because the remainder is 220, not 0.

Eliminate the possibility x = -4.  Continue in the same way, testing each possible factor of -36 as a zero remainder.  Try x = 2.  We get:

2     /    1   -11   36   -36

                  2   -18    36

      ---------------------------

          1      -9     18    0       Since the remainder is 0, we know that 2

                                                 is a root of x^3 − 11x^2 + 36x − 36.

Similarly, test x = 3:  Using the coefficients 1  -11  36  -36, we obtain a zero remainder and thus can conclude that 3 is a root or zero of the original equation.

Thus, the roots/zeros of the original equation are {2, 3, 6}

and the corresponding factors of this equation are thus

{(x - 2), (x - 3), (x - 6).  The x-intercepts are {(0, 2), (0, 3), (0, 6)}  This function has no negative root/roots.  Thus, the entire graph lies to the right of the y-axis.  The curve increases/rises between x = 0 and x = 2 and crosses the x-axis at (0, 2).  Then it decreases/drops between x = 2 and x = 3.  Finally, the curve increases/rises again to the right of x = 6, always increasing.