Question - Graphical Representation of a Linear Inequality Solution

Simplify the given inequality \( (3x + 1)^2 \geq (3x + 1)(3) \)

Expand both sides: \( 9x^2 + 6x + 1 \geq 9x + 3 \)

Rearrange terms to set the inequality to zero: \( 9x^2 + 6x + 1 - 9x - 3 \geq 0 \)

Simplify: \( 9x^2 - 3x - 2 \geq 0 \)

Factor the quadratic inequality: \( (3x + 1)(3x - 2) \geq 0 \)

Identify critical points where the inequality can change signs by setting each factor equal to zero: \( 3x + 1 = 0 \) and \( 3x - 2 = 0 \)

Solve for x to find the critical points: \( x = -\frac{1}{3} \) and \( x = \frac{2}{3} \)

Test intervals to determine where the inequality holds; use test points \( x < -\frac{1}{3} \), \( -\frac{1}{3} < x < \frac{2}{3} \), and \( x > \frac{2}{3} \)

Choose test points such as \( x = -1 \), \( x = 0 \), and \( x = 1 \); verify that the inequality is satisfied for \( x < -\frac{1}{3} \) and \( x > \frac{2}{3} \) but not in the middle

The solution is \( x \in (-\infty, -\frac{1}{3}] \) union \( [\frac{2}{3}, \infty) \)

Therefore, the correct graph is D.