equation of the tangent to the following curves? example:
$f(x)=x^2  (2,4)$
  $= x^2$ 
  at x= 2 
$f(x) 2^2=4$ 
$y=mx+c$ 
$y=4x+c$ 
$(2,4)$ 
$4=4\cdot2+c$ 
$4-8=c$ 
$c=-4$ 
$y=4x-4$ 
but below question 
I don't know how to solve? 
$F(x)=x^2-2x+5$   at $x= -1$
the answer is $y=-4x+4$
thanks in advance! :)
 A: Added: concerning the first part of your question, you should edit it. As a minimum, something like the following.
Example: $f(x)=x^2$. At $x=2$, $f(x)=2^2=4$. The equation of the tangent at   $(2,4)$ is
$$y=mx+c,$$
where $m$ is $f'(2)=4$. And so,  $y=4x+c$, and $$4=4\cdot 2+c\Leftrightarrow 4-8=c\Leftrightarrow  c=-4.$$ Therefore, $y=4x-4$. 

Using your notation, the equation of a straight line is
$$y=mx+c\tag{1},$$
where $m$ is its slope. If this line passes through the point $P(a,b)
$, then the coordinates $a,b$ must satisfy the equation
$b=ma+c\tag{2}.$
Subtracting $(2)$ from $(1)$, we get
$$y-b=m(x-a),\tag{3}$$
which is equivalent to
$$y=mx-ma+b.\tag{3'}$$
The slope of the tangent line to the graph of the function $f(x)$ at the point $
(a,b)=(a,f(a))$ is equal to the derivative of the function at $x=a$, i.e. $m=f^{\prime }(a)$ (Wikipedia link). Therefore, the equation of the
tangent is given by (see sketch)
$$y=f^{\prime }(a)x-f^{\prime }(a)a+f(a).\tag{4}$$

For the second function $F(x)=x^{2}-2x+5$, the equation of the tangent at $(-1,F(-1))=(-1,(-1)^{2}-2(-1)+5)=(-1,8)$ is
$$y=F^{\prime }(-1)x-F^{\prime }(-1)\left( -1\right) +8.$$
The derivative $F^{\prime }(x)=2x-2$ and $F^{\prime }(-1)=2\left( -1\right)
-2=-4$. Consequently, the equation of the tangent is
$$y=-4x-(-4)\left( -1\right) +8,$$
which is equivalent to
$$y=-4x+4.\tag{5}$$
