Let $\mathbb{R}^2_+=\{(x,y)\in\mathbb{R}^2|\text{ }x\geq 0,y\geq 0 \}$, $a,b,r\in \mathbb{R}$, $a,r>0$, $b\geq 0$. Put: $$A=\{(x,y)\in\mathbb{R}^2_+|\text{ }y\leq ax^2+bx-r\}$$
Is the set $A$ convex?
Let $\mathbb{R}^2_+=\{(x,y)\in\mathbb{R}^2|\text{ }x\geq 0,y\geq 0 \}$, $a,b,r\in \mathbb{R}$, $a,r>0$, $b\geq 0$. Put: $$A=\{(x,y)\in\mathbb{R}^2_+|\text{ }y\leq ax^2+bx-r\}$$
Is the set $A$ convex?
Copyright © 2021 JogjaFile Inc.
$A$ is not convex. The polynomial $y=ax^2+bx-r$ has exactly two real roots one positive and one negative. Name the positive root as $x_0$. Therefore since the points $P_1=(x_0,0)$ and $P_2=(2x_0, 3ax_0^2+bx_0)$ are in $A$ and the interconnecting line is not, so $A$ is not convex.