Question

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

Answer 1

Please see below.

Explanation:

Let us first consider the graph of `y=x^2+4x`. This can be written in vertex form as `y=(x+2)^2-4` and hence vertex is `(-2,-4)`

and is the equation of a vertical parabola, which opens upwards and we have a minima at vertex `(-2,-4)`. Taking a few values to the left and right of `x=-2`, we can easily draw the graph of parabola, which appears as shown below:

graph{x^2+4x [-9.25, 10.75, -5.72, 4.28]}

Observe that this divides the Cartesian plane in three parts of which one is the line graph of parabola,,

the other two parts are the area lying within the cup-shaped portion inside parabola. Observe that the point `(-2,0)` lies in this area. If we put `(-2,0)` we find that `x^2+4x=4-8=-4` and hence as `y=0`, we have `y>x^2+4x`

Hence in this area we have `y>x^2+4x`.

The third part is area outside the parabola and a point in this area is `(5,0)`. Putting this we get `5^2+4*5=25+20=45>0` and hence `y<x^2+4x`

hence this area satisfies `y<x^2+4x`

Now we are given the inequality `y<=x^2+4x` and hence all points on parabola and below it satisfy this inequality and hence graph appears as

graph{y<=x^2+4x [-9.25, 10.75, -5.72, 4.28]}

Note that had we only `y< x^2+4x`, the graph would not have included points on parabola and for this parabola would have appeared as dotted. But here it appears as full line as points on parabola are included. This is because `y<=x^2+4x` includes equality.