Equation of a tangent line - complex

I have a math problem from my class that I have to solve and turn back in. It is basically Regiomontanus' angle maximization problem. These are the objectives of the homework assignment:

f(x) =(x-a)(x-b)(x-c)

1. Find two values of x that make f(x) = 0

2. Find the average of the two values of x you found in part (1). What point on the graph of f has this x-value?

3. Find the equation for the tangent line of f at the point you found in part (2).

4. Find all points where the tangent line from part (3) intersects the graph y = f(x), where f is dened above.

This is what I have so far:

        Find two values of x where f(x)=0
Well, since f(x)= (x-a)(x-b)(x-c),
f(a)=(a-a)(a-b)(a-c)= 0*(a-b)*(a-c)= 0
Thus x =a is a zero.
F(b)= (b-a)(b-b)(b-c)= (b-a)*0*(b-c)= 0
Thus x = b is a zero.
The two zeroes of the graph are at x=a, x=b and if we needed a third one x=c, which we don’t.

Find the average of the two values of x you found in part 1. What point on the graph of f has this value?
Since we know that we are using x=a, and x=b for our zeroes from part 1, the average of those two points would be:
((a+b)/2  ,f((a+b)/2))   f((a+b)/2) = (((a+b)/2)-a)(((a+b)/2)-b)(((a+b)/2)-c)
Find the equation for the tangent line of f at the point you found in part 2.

For this, we need to take the derivative of f(x)=(x-a)(x-b)(x-c):

First, expand the function:

(x-a)(x-b)(x-c)= (x^2-xb-xa+ab)*(x-c)= x^3-x^2 c+xbc-x^2 a+xac+abx-abc

Then we can simplify the expression :

〖f(x)= x〗^3-x^2 (a+b+c)+x(ab+bc+ca)-abc

Then take it’s derivative:

f’(x)= (d/dx) x^3– (a+b+c)*(d/dx) x^2+(ab+bc+ca)*(d/dx)x – (d/dx)abc

(d/dx) x^3 : we need to use the power rule: x^n  nx^(n-1) : x^3  3x^(3-1)= 3x^2

(d/dx) x^2 ∶We need to use the power rule:x^n→nx^(n-1) ∶x^2→2x^(2-1)  =2x^1=2x

(d/dx)x:We need to use the power rule∶x^n→nx^(n-1) ∶x^1→1x^(1-1)=1*x^0=1

(d/dx)abc∶Since the derivitive of constants are 0,abc=0,and won^' tbe included in f'(x)

f’(x)= 3x^2– 2x(a+b+c)+(ab+bc+ac)

Now we need to plug in our x value we found in part 2.
f^' (x)=3((a+b)/2)^2-2((a+b)/2)(a+b+c)+ab+bc+ac
f^' (x)=3((a^2+2ab+b^2)/4)-(a+b)(a+b+c)+ab+bc+ac
f^' (x)=  3/4 (a^2+2ab+b^2 )-a^2+ab+ac+ab+b^2+bc+ab+bc+ac
f^' (x)=  3/4 a^2+3/2 ab+3/4 b^2+3ab+2ac+2bc-a^2+b^2
f^' (x)= -1/4 a^2+9/2 ab+7/4 b^2+2ac+2bc
We lnow that taking the derivative of f will give us the m in y = mx+b
Now we need to find the whole equation. We know what our y is and what our x is, so we  are solving for our b.
Set up the equation to find b:
(x-a)(x-b)(x-c)=-1/4 a^2+9/2 ab+7/4 b^2+2ac+2bc ((a+b)/2)+B


My issue is that I am having a hard time figuring out how I should go about finding the tangent line in part 3. It's not that I don't know how to find a tangent line, its just that this problem has me confused. This is for part 3.

I am even more confused on how I should go about part 4.

Any help is greatly apreciated, and I am sorry if I have any typing errors or lack in mathmatical knowledge, these assignments are given to us so that we can learn.

If $F(x) =\prod_{k=1}^n f_k(x)$, then
$\begin{array}\\ F'(x) &=\sum_{j=1}^n f_j'(x)\prod_{k=1, k\ne j}^n f_k(x)\\ &=\sum_{j=1}^n \frac{f_j'(x)}{f_j(x)}\prod_{k=1}^n f_k(x) \quad\text{ (if none of the } f_j(x) = 0)\\ &=\sum_{j=1}^n \frac{f_j'(x)}{f_j(x)}F(x)\\ &=F(x)\sum_{j=1}^n \frac{f_j'(x)}{f_j(x)}\\ \end{array}$
If $F(x) = (x-a)(x-b)(x-c)$, this becomes $F'(x) =(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b) = 3x^2-2(a+b+c)x+bc+ac+ab$ as you got.
This rule can be proved either from the chain rule and the derivative of the log (in the form $(\ln F(x))' =\dfrac{F'(x)}{F(x)}$) or by induction on $n$ starting (and using in the induction step) the ordinary product rule $(f(x)g(x))' =f'(x)g(x)+f(x)g'(x)$.