Question 1036323
Let (b,0) be the point of tangency.

Now the derivative is given by f'(x) = {{{-6x^2 + 6(a+2)x-12a}}}.

Setting this to 0 to find the critical values, we get {{{x^2 -(a+2)x + 2a = (x-2)(x-a) = 0}}}.
==> x = 2 or x = a.

Case(i).  Let b = a.
==> {{{-2a^3+3(a+2)a^2-12a^2+4a^2 = 0}}}
==> {{{a^2(a-2) = 0}}}, after reduction.
==> a = 0 (double root), or a = 2.
Discard a = 0, because if it is to be the x-coordinate of the point of tangency, a has to be positive.  (Remember tangency to positive x-axis.)
==> a = 2.
==> {{{ f(x)=-2x^3+12x^2-24x+16 = -2(x-2)^3 }}}.
But this function has to be discarded as well, because even though f'(2) = 0, f"(2) = 0, and hence there is no maximum at that point, but a point of inflection.


Case (ii). Let b = 2.
==> {{{f(b=2) = -16+3(a+2)*4-24a+4a^2 = 0}}}
==> {{{4a^2-12a+8 = 0}}}
==> {{{a^2 - 3a + 2 = 0}}} ==> (a-1)(a-2) = 0 ==> a = 1 or a = 2.

Now we already know what happens when a = 2, and so we proceed letting a = 1.
==> {{{ f(x)=-2x^3+9x^2-12x+4 = (x-2)^2(-2x+1) }}}
By using the 2nd derivative test, we find that there is a relative maximum
at x = 2. (There is relative min at x = 1.)

Therefore {{{highlight(a = 1)}}}.