Question 427134
Determine graphically the number of real zeros and the number of imaginary zeros of the polynomial function f(x) = x^3 - 3x^2 + 3x - 9.
..
f(x)=x^3-3x^2+3x-9

If you are allowed to use a graphing calculator, you will see that the function has only one real zero, x=3. 
(see the graph below). After this, divide the function,x^3-3x^2+3x-9, by (x-3) by long division or synthetic division. You will then get a quotient,(x^2+3), which gives you two imaginary zeros.

ans:
one real zero=3
two imaginary zeros=+-sqrt(-3) or +-sqrt(3)i

..

 {{{ graph( 300, 300, -5, 5, -10, 10, x^3-3x^2+3x-9) }}}

..

After reviewing my above solution, I realized the function could be factored:
x^3-3x^2+3x-9=x^2(x-3)+3(x-3)=(x-3)(x^2+3)
This would give you one real zero, 3 and two +-sqrt(-3), imaginary zeros, same as above.
This is the preferred method as you do not require a calculator for the solution.