First of all, we compute the points of interest, i.e. the points where the curve cuts the x axis: since the expression is already factored, we have
[tex]x(x-1)(x+2) = 0 \iff x=0\ \lor\ x-1=0\ \lor\ x+2=0[/tex]
Which means that the roots are
[tex]x=0\ \lor\ x=1\ \lor\ x=-2[/tex]
Next, we can expand the function definition:
[tex]y = x(x-1)(x+2) = x^3 + x^2 - 2x[/tex]
In this form, it is much easier to compute the derivative:
[tex]y' = 3x^2+2x-2[/tex]
If we evaluate the derivative in the points of interest, we have
[tex]y'(-2) = 6,\quad y'(0)=-2,\quad y'(1)=3[/tex]
This means that we are looking for the equations of three lines, of which we know a point and the slope. The equation
[tex]y-y_0=m(x-x_0)[/tex]
is what we need. The three lines are:
[tex]y-0=6(x+2) \iff y = 6x+12[/tex] This is the tangent at x = -2
[tex]y-0=-2(x-0) \iff y = -2x[/tex] This is the tangent at x = 0
[tex]y-0=3(x-1) \iff y = 3x-3[/tex] This is the tangent at x = 1
