Question

which inequality is graphed on the coordinate plane?

\\(y < 4x + 2\\)
\\(y > 4x + 2\\)
\\(y \le \frac{1}{4}x + 2\\)
\\(y \ge 4x + 2\\)

Explanation:

Identify the boundary line properties

The boundary line is solid, which means the inequality must use either \(\ge\) or \(\le\).
The line passes through the y-intercept \((0, 2)\) and another clear grid point \((-1, -2)\).
The slope \(m\) is calculated as:

$$ m = \frac{-2 - 2}{-1 - 0} = 4 $$

Thus, the boundary line equation is:

$$ y = 4x + 2 $$

Determine the shaded region

The shaded region is to the left and above the boundary line.
Testing a point in the shaded region, such as \((-2, 0)\):

$$ 0 \ge 4(-2) + 2 \implies 0 \ge -6 \quad (\text{True}) $$

Testing a point not in the shaded region, such as \((2, 0)\):

$$ 0 \ge 4(2) + 2 \implies 0 \ge 10 \quad (\text{False}) $$

Therefore, the inequality is:

$$ y \ge 4x + 2 $$

Answer:

• (A) \(y < 4x + 2\)
• (B) \(y > 4x + 2\)
• (C) \(y \le \frac{1}{4}x + 2\)
• (D) \(y \ge 4x + 2\) (Correct answer)