Question

How do you graph `y < x^2 + 4x`?

Answer 1

First consider the equation `y=x^2+4x`. This can be factored as `y=x(x+4)`.

Note that this has zeros when `y=0`, which are when `x=0` and `x+4=0=>x=-4`.

The vertex of this parabola will be halfway between the zeros, at `x=-2`. Note that `y(-2)=(-2)^2+4(-2)=-4`.

So, we can easily draw the parabola with the points `(-4,0),(-2,-4),(0,0)`:

graph{x^2+4x [-14.09, 8.41, -5.855, 5.395]}

We want to find when `y < x^2+4x`, which will be all points lying below the parabola. Since `y<` the parabola and not `ylt=` the parabola, the line of the graph will be dotted:

graph{y < x^2+4x [-14.09, 8.41, -5.855, 5.395]}