How to graph y < x 2 + 4 x ?

First consider the equation ${\displaystyle y=x^2+4x}$. This can be factored as ${\displaystyle y=x(x+4)}$.
Note that this has zeros when ${\displaystyle y=0}$, which are when ${\displaystyle x=0}$ and ${\displaystyle x+4=0\Rightarrow x=-4}$.
The vertex of this parabola will be halfway between the zeros, at ${\displaystyle x=-2}$. Note that ${\displaystyle y(-2)={(-2)}^2+4(-2)=-4}$.
Consequently, it is simple to create a parabola using points ${\displaystyle (-4,0),(-2,-4),(0,0)}$:
graph{x^2+4x [-14.09, 8.41, -5.855, 5.395]}
We want to find when ${\displaystyle y<x^2+4x}$, which will be all points lying below the parabola. Since ${\displaystyle y<}$ the parabola and not ${\displaystyle y\le }$ the parabola, the line of the graph will be dotted:
graph{y < x^2+4x [-14.09, 8.41, -5.855, 5.395]}