Reblausvtct

2023-02-19

How to graph $y<{x}^{2}+4x$?

Kylie Woodward

Beginner2023-02-20Added 6 answers

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 $x+4=0\Rightarrow x=-4$.

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

Consequently, it is simple to create a parabola using 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 $y\le$ the parabola, the line of the graph will be dotted:

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

Note that this has zeros when $y=0$, which are when $x=0$ and $x+4=0\Rightarrow x=-4$.

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

Consequently, it is simple to create a parabola using 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 $y\le$ the parabola, the line of the graph will be dotted:

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