Another way to think about this is to look at signs in each interval.
You have f(x)= x (x+2)^2 (x-1)^4= (x- 0)(x- (-2))^2(x- 1)^4.
I have written each as "x- " because "x- a" is negative if x< a, positive if x< a.
Here the terms are 0 at x= -2, 0, and 1. If x is any number less than -2, then it is less than all three so each of x, x+ 2, and x- 1 is negative. The even power, of course, is positive anyway so we have (-)(+)(+)= - for x< -2. If -2< x< 0, the x+ 2 term is now positive but since it was to an even power it was positive any way so we still have (-)(+)(+)= -. The graph goes up, touches the x-axis at (-2, 0) then goes back down again. It must turn some where (do you know how find minimum points? That may require calculus) to come back up to 0. For 0< x< 1, the x term is now positive so we have (+)(+)(+)= +. The graph continues up until it turns back to go to 0 at x= 1. For 1< x, all three terms are still positive: (+)(+)(+)= + so the graph must turn back up.