Question 1113885
<br>
{{{x^3-2x^2-4x+8 = x^2(x-2)-4(x-2) = (x^2-4)(x-2)}}}<br>
So
{{{(x^3-2x^2-4x+8)/(x-2) = x^2-4}}}
as long as x-2 is not 0.<br>
So the given function is the same as the function x^2-4 except at x=2, where the given function is undefined.<br>
So there is a hole in the graph of the given function at x=2.  At x=2, the value of the function x^2-4 is 0, so the hole in the graph of the given function is at (2,0).<br>
The zeros of the given function are the zeros of x^2-4, which are 2 and -2.<br>
The graph of x^2-4 is a parabola opening upward; the function value is negative between the two zeros of the function and positive "outside" the zeros:
positive on (-infinity, -2) and (2, infinity);
negative on (-2,2).<br>
The vertex of x^2-4 is at (0,-4); the function is decreasing on (-infinity, 0) and increasing on (0, infinity).