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: 
I think it's much easier to understand if you write the definition as "$Df\_{|a}$ is the unique linear map $L$ (if it exists) satisfying $f(x) = f(a) + L(x-a) + o(||x-a||)$ when $x\rightarrow a$".

Maybe it's even clearer if you write "$Df\_{|a}$ is the linear part of the unique affine map $A$ (if it exists) satisfying $f(x) = A(x) + o(||x-a||)$ when $x\rightarrow a$". 

So you can see that $A$ is the best possible affine approximation of $f$ near $a$ (because the error you make by replacing $f$ with $A$ is negligible compared to any affine map), and $Df\_{|a}$ is the linear part of this affine approximation.
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, $L(a)h=f'(a)h$