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.
Just to add a bit to the previous answers, you can check that this definition meets the "usual" requirement for differentiability if $f:\mathbb R\to \mathbb R.$ 

Take $x=a$. Then, according to the "new" definition, to find the derivative of $f$ at $a$, we seek an 

$L(a):\mathcal L(\mathbb R,\mathbb R)\to \mathcal L(\mathbb R,\mathbb R)$ such that

$\lim \_{h\to 0}\frac{f(a+h)-f(a)-L(a)h}{h}=0$. 

Now, the linear tranformations from $\mathbb R\to \mathbb R$ are of the form $fh=bh$ for some $b\in \mathbb R$, so we have now, with $f=L(a)$

$\lim \_{h\to 0}\frac{f(a+h)-f(a)-bh}{h}=0$, which simplifies to

$\lim \_{h\to 0}\left ( \frac{f(a+h)-f(a)}{h}-b \right )=0$ so that $b=f'(a)$, that is, $L(a)h=f'(a)h$