QUESTION IMAGE
Question
- which is the graph of $|\frac{1}{2}x + 2|\geq1$?
Step1: Solve the absolute - value inequality
For the inequality \(|\frac{1}{2}x + 2|\geq1\), we consider two cases:
Case 1: \(\frac{1}{2}x+2\geq1\)
Subtract 2 from both sides: \(\frac{1}{2}x\geq1 - 2=-1\)
Multiply both sides by 2: \(x\geq - 2\)
Case 2: \(\frac{1}{2}x + 2\leq - 1\)
Subtract 2 from both sides: \(\frac{1}{2}x\leq-1 - 2=-3\)
Multiply both sides by 2: \(x\leq - 6\)
Step2: Analyze the graph
The solution of the inequality \(|\frac{1}{2}x + 2|\geq1\) is \(x\leq - 6\) or \(x\geq - 2\). This means the graph should have closed - circles (because the inequality is \(\geq\)) at \(x=-6\) and \(x = - 2\) and the line should extend to the left of \(x=-6\) and to the right of \(x=-2\)
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The third option (the one with closed - circles at \(x=-6\) and \(x=-2\) and the line extending to the left of \(x = - 6\) and to the right of \(x=-2\)) is the correct graph.