Reblausvtct

2023-02-19

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

Kylie Woodward

First consider the equation $y={x}^{2}+4x$. This can be factored as $y=x\left(x+4\right)$.
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 $y\left(-2\right)={\left(-2\right)}^{2}+4\left(-2\right)=-4$.
Consequently, it is simple to create a parabola using points $\left(-4,0\right),\left(-2,-4\right),\left(0,0\right)$:
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]}

Do you have a similar question?