kill(all)$

van=facebook.com/LearningCalculus$

f(x):=expand((x-2)*x*(x+2))$ '(f(x))=f(x);

fp(x):=''(diff(f(x),x));

           The slope of the tangent line passing through [2,0] is fp(2) = 8

           What is the y-intercept of the tangent line passing through [2,0]?
           We rearrange the definition of slope m to obtain: y1 - y0 = m (x1 - x0)
           We know that x0 = 0 for y intercept so we just need y0.

e: first(solve(y1-y0=m*x1,y0));

           We know the line passes through [2,0] so we use that for x1, y1 and m = fp(2)

result1: subst(0,y1,subst(2,x1,subst(fp(2),m,e)));

           Our first line equation is therefore this:

y=fp(2)*x+rhs(result1);

            The second line is harder. It also passes through 2,0 but it is tanget at some location x < 0.  
            A trial solution is obtained by inspection where
                         the m value is -1 and
                         the x  value is -1 and
                         the y  value is  3.
            so we ask does y = -x + 2 pass through [2,0] and [-1, 3] lie  and on the curve.
            Since the answer to both questions is yes, we have our second line equation:

y = fp(-1)*x+2;