Question

question 2 of 10
which point is a solution to the inequality shown in this graph?
text description for graph
a. (5, -5)
b. (0, -5)
c. (-3, 0)
d. (0, 0)

Explanation:

Step1: Find line equation

Use points $(-3,0)$ and $(0,4)$. Slope $m=\frac{4-0}{0-(-3)}=\frac{4}{3}$. Equation: $y=\frac{4}{3}x+4$, or $4x-3y+12=0$.

Step2: Test inequality direction

Shaded region is left of the solid line. Test $(-4,0)$: $4(-4)-3(0)+12=-4\leq0$, so inequality is $4x-3y+12\geq0$.

Step3: Test each option

• A. $(5,-5)$: $4(5)-3(-5)+12=20+15+12=47\geq0$? Yes, but check position: $(5,-5)$ is right of line, not shaded. Wait, correction: inequality is $y\leq\frac{4}{3}x+4$. Test A: $-5\leq\frac{4}{3}(5)+4\to-5\leq\frac{32}{3}$? Yes, but graph shows shaded left. Wait, recheck: for $(-4,0)$, $0\leq\frac{4}{3}(-4)+4\to0\leq-\frac{4}{3}$? No. Correct inequality: $y\geq\frac{4}{3}x+4$. Now test:
• A. $-5\geq\frac{20}{3}+4\to-5\geq\frac{32}{3}$? No.
• B. $-5\geq0+4\to-5\geq4$? No.
• C. $0\geq\frac{4}{3}(-3)+4\to0\geq-4+4\to0\geq0$? Yes (solid line includes points).
• D. $0\geq0+4\to0\geq4$? No.

Answer:

C. (-3, 0)