Unless you have miswritten the function, there are NO "asymptotes"- that is a polynomial.
What I would do is rearrange the factors so each term, in the form x- a, is increasing a. That is, write it as (x+ 2)x(x- 2)^2(x- 4)[/tex] (x+ 2= x- (-2) and x= x- 0).<br />
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That is a polynomial that crosses the x-axis at x= -2, x= 0, x= 2, and x= 4. To see what it looks like between those points, remember that if x< a, then x-a< 0 but if x> a then x-a> 0. Start, on the left, with x< -2. The <b>all</b> of those 5 factors (we are counting x-2 twice, of course) is negative and the product of an odd number (5) of negative numbers is negative. For x< -2, this function is negative. Its graph rises from -\infty on the left to (-2, 0). <br />
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If -2< x< 0, then x+2> 0 but all other four factors are negatve. The product of 4 negative factors is positive and it product with a positive factor is postive. The graph rises above the x-axis to some maximum, then back to (0, 0). <br />
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If 0< x< 2, then both x+ 2> 0 and x> 0. We now have two positive factors and three negative factors which is negative. The graph drops below the x-axis at (0, 0), down to some minimum, then back up to (2, 0).<br />
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If 2< x< 4, then we have four positive factors (they jumped from two to four because the x-2 factor is squared) and one negative factor. The product remains negative. So the graph does NOT cross the x-axis at (2, 0). The graph "kisses" it (is tangent to the x-axis) then drops back down to some negative minimum then back up to (4, 0). <br />
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Finally, if x> 4, all five factors are positive but the product of any number of positive factors is positive. The graph rises above the x-axis at x= 4, going up, on the right, toward positive infinity. (The graph does NOT come back down, there is NO horizontal asymptote!)
