Gradient, and $f(x,y)=x^2+4y^2$

Question

I been given the preceding function in the title my question is I calculated the gradient, and then was given this point $P(-2,0)$. What exactly is it the gradient is supposed to be a pluggable function? If it is I plug in that point and get $-4$. My professor mentioned that the value of the gradient corresponds to the steepness of the function. Any thoughts would be helpful: $$\nabla f(x,y)=2x+8y$$

Edit

As Curl mention the corrected gradient is $$\nabla f(x,y)=(2x,8y)$$

Answer 1

$F(x,y)=x^2+4y^2$ $$gradient~ f=\vec g(x,y)=\vec \nabla ~f=\vec i \frac{\partial F}{\partial x}+ \vec j \frac{\partial F}{\partial y}= 2x\vec i+4y\vec j,~~ \vec g(-2,0)=-4\vec i.$$