Answer:
[tex]y\ge -\frac{2}{3}x+1[/tex] represents the graph. So, option A is true.
Step-by-step explanation:
Considering the linear inequality
[tex]y\ge -\frac{2}{3}x+1[/tex]
as
[tex]\mathrm{Domain\:of\:}\:-\frac{2}{3}x+1\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}[/tex]
and
[tex]\mathrm{Range\:of\:}-\frac{2}{3}x+1:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<f\left(x\right)<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}[/tex]
[tex]x-\mathrm{axis\:interception\:points\:of\:}-\frac{2}{3}x+1:\quad \left(\frac{3}{2},\:0\right)[/tex]
[tex]-\frac{2}{3}x+1=0:\quad x=\frac{3}{2}[/tex]
[tex]\left(\frac{3}{2},\:0\right)[/tex]
[tex]y-\mathrm{axis\:interception\:point\:of\:}-\frac{2}{3}x+1:\quad \left(0,\:1\right)[/tex]
[tex]y=-\frac{2}{3}\cdot \:0+1[/tex]
[tex]y=-0+1[/tex]
[tex]y=1[/tex]
Thus,
[tex]\mathrm{X\:Intercepts}:\:\left(\frac{3}{2},\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:1\right)[/tex]
[tex]\mathrm{Slope\:of\:}-\frac{2}{3}x+1:\quad m=-\frac{2}{3}[/tex]
Also, (-3, 3) holds true.
as
[tex]3\ge \:-\frac{2}{3}\cdot \left(-3\right)+1[/tex]
[tex]3\ge \:\frac{2}{3}\cdot \:\:3+1[/tex]
[tex]3\ge \:2+1[/tex]
[tex]3\ge \:3[/tex]
WHICH IS TRUE!
Therefore,
[tex]y\ge -\frac{2}{3}x+1[/tex] represents the graph. So, option A is true.
