Move the points so that they all lie on the graph of \( f(x)=\frac{-1}{30}(x+4)(x-3)(x+5)(x-1) \) and do not coincide.

1. Any point on the graph of \[ f(x)=\frac{-1}{30}(x+4)(x-3)(x+5)(x-1) \] must have coordinates \[ (x,\, f(x)) = \left(x,\, \frac{-1}{30}(x+4)(x-3)(x+5)(x-1)\right). \] 2. To “move the points” so that they all lie on the graph and do not coincide, choose distinct \( x \)–values. Then, reassign each point’s coordinates using the formula for \( f(x) \). 3. A convenient choice is to use the zeros of \( f(x) \), where the output is 0. Notice that the factors give zeros when: \[ x+4=0,\quad x-3=0,\quad x+5=0,\quad x-1=0, \] so \[ x=-4,\quad x=3,\quad x=-5,\quad x=1. \] The corresponding points are: \[ (-4, 0),\quad (3, 0),\quad (-5, 0),\quad (1, 0). \] 4. These four points are distinct (they do not coincide) and lie on the graph of \( f(x) \). 5. In summary, one valid solution is to move the given points so that they are relocated to: \[ (-5,0),\quad (-4,0),\quad (1,0),\quad (3,0). \] Thus, by assigning any point \((x, f(x))\) using different \(x\)–values (for example, the four values above), all the points will lie on the graph of \[ f(x)=\frac{-1}{30}(x+4)(x-3)(x+5)(x-1) \] and none will coincide.