Find equation of a curve by tangent This is probaly easy.
Find the equation of the tangent to a curve $y = u (x)$ is $y =(-2a+4)(x-a)+k$, where $k$ is a constant. Given that the curve touches the $x$-axis at $x =2$, 
find the value of $k$ and the equation of the curve.
 A: Letting $y=u(x)$, its derivative is $u'(x)$, and it represents the slope of the tangent to the function $u(x)$ at the point $(x,u(x))$. Thus the line tangent to the function $u(x)$ at the point $c$, is given by the equation 
$$y-u(c) = u'(c)\cdot (x-c)\to \\y=x\cdot u'(c) +u(c) -c\cdot u'(c)=\\x\cdot (4-2a)+2a^2-4a+k$$
Now you may note that
\begin{cases}
4-2a=u'(c)\\
u(c) -c\cdot u'(c) = 2a^2-4a+k
\end{cases}
so solving the system we obtain
\begin{cases}
\int 4-2a\, \mathrm{d}c=u(c)\\
u(c) -c\cdot (4-2a) = 2a^2-4a+k
\end{cases}
\begin{cases}
c\cdot (4-2a)+b=u(c), b \text{ is a constant}\\
u(c) -c\cdot (4-2a) = 2a^2-4a+k
\end{cases}
\begin{cases}
c\cdot (4-2a)+b=u(c), b \text{ is a constant}\\
c\cdot (4-2a)+b-c\cdot (4-2a) = 2a^2-4a+k \Rightarrow b=2a^2-4a+k
\end{cases}
Finally we get to the conclusion that $u(c) = c\cdot (4-2a)+2a^2-4a+k$.
Now,$$u(2)= 2\cdot (4-2a)+2a^2-4a+k =0\Rightarrow\\8-4a+2a^2-4a=-k\\k=-2(a^2-4a+4)=-2(a-2)^2\Rightarrow \\u(x) = x\cdot (4-2a) +2a^2-4a-2a^2+8a-8\\\color{green}{u(x) = 2x\cdot (2-a)+4a-8}$$
A: I think that @Foga's answer too long.
@Harikim gives you a shortcut but here is the general approach.
First thing to note is that when you take $(2a-4)(x-a)+k$ note that this is in the tangent line form of $f'(a)(x-a)+f(a)$.
So $f'(x)=2x+4$
Then taking the integral
$$\int{f'(x)}=\int{2x-4}$$
$$f(x)=x^2-4x+c$$
Now since the curve is supposed to touch at $(2,0)$ we should first check the derivative and make sure is at zero. to make sure there are no tricks to this.
$f'(2)=0$
This we can substitute (2,0) into $y=x^2-4x+c$ 
$$0=4-8+c$$
$$c=4$$
Now if we take the tangent line formula of $y=f'(a)(x-a)+f(a)$ where $a=2$ to $y=(2a-4)(x-a)+k$. You realize $k=f(a)=f(2)$
So $k=f(2)=0$
