Question: The question asks to find a cubic equation given point A (0/18) slope here is 0. Point B only given the x=20 and slope is -0.6.

Please help, been trying to solve it for so long now. :(

1.) y=ax^3+bx^2+cx+d

2.) y'=3ax^2+2bx+c

-0,6=1200a+40b+1c

18=0a+0b+0c+1d

Answer: 
This is one of the most fundamental definitions in all of analysis.

It says that the increment $\Delta f:=f(a+h)-f(a)$ of the function value should in first approximation be a linear function of the increment $h$ attached at the point $a$. In other terms: We want
$$f(a+h)-f(a)=Lh +r(h)\qquad(|h|\ll1)\ ,\tag{1}$$
whereby the error $r(h)$ should be *smaller by magnitudes* than the linear term $Lh$ when $h$ is small. Now in general $|Lh|$ will be of order $|h|$ for "most" $h$. This means that we should require that
$$\lim\_{h\to0}{|r(h)|\over |h|}=0$$
in order to impart any real content to $(1)$. It turns out that this condition determines $L$ uniquely. If it can be satisfied then $f$ is called *differentiable* at $a$, and one denotes the resulting $L$ by $Df\bigr|\_a$, or similar.
The division by $\|h\|$ here is exactly analogous with the division by $h$ in the definition of the (standard) derivative of a real-valued function of a real variable:

$\dfrac{df}{dx}=\underset{h\rightarrow 0}{\lim}\dfrac{f(x+h)-f(x)}{h}$

To answer the second question, $Df|\_a$ is linear because it satisfies the linearity property, that is it commutes with addition and scalar multiplication. This is a consequence of how it is defined and is not actually the hard to prove (try using the definition to show $Df|\_a+Dg|\_a=D(f+g)|\_a$ and $D(\alpha f)|\_a=\alpha Df|\_a$ directly; hint, use some linear algebra)

To answer the final question, it is the natural analog of the familiar derivative of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ for the case of a function from $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$.