Question

1. which is the graph of $f(x) = 3|x|$? a. graph, b. graph, c. graph, d. graph

Explanation:

Step1: Recall the parent function of absolute value

The parent function of absolute value is \( y = |x| \), which has a V - shape with the vertex at the origin \((0,0)\) and the slope of the right - hand side (for \(x\geq0\)) is \(1\) and the slope of the left - hand side (for \(x < 0\)) is \(- 1\).

Step2: Analyze the transformation of the function \(y = 3|x|\)

For a function of the form \(y=a|x|\), when \(a>0\), the graph is a vertical stretch of the parent function \(y = |x|\) by a factor of \(a\). In the function \(f(x)=3|x|\), \(a = 3>0\). So the graph of \(y = 3|x|\) is a vertical stretch of \(y = |x|\) by a factor of \(3\). This means that for the same \(x\) - value, the \(y\) - value of \(y = 3|x|\) is 3 times the \(y\) - value of \(y=|x|\). The vertex of \(y = 3|x|\) is still at the origin \((0,0)\) because there is no horizontal or vertical shift (the function is of the form \(y=a|x - h|+k\) with \(h = 0\) and \(k = 0\)). Also, the slope of the right - hand side (for \(x\geq0\)) is \(3\) and the slope of the left - hand side (for \(x < 0\)) is \(-3\), which means the graph is steeper than the graph of \(y = |x|\).

Now let's analyze each option:

• Option A: The vertex of the graph is not at the origin, so this is not the graph of \(y = 3|x|\).
• Option B: The graph has its vertex at the origin \((0,0)\) and is steeper than the graph of \(y = |x|\) (since the slope is \(3\) for \(x\geq0\) and \(- 3\) for \(x < 0\)), which matches the graph of \(y=3|x|\).
• Option C: The vertex of the graph is not at the origin, so this is not the graph of \(y = 3|x|\).
• Option D: The graph is opening downwards, but for \(y = 3|x|\), since \(a=3>0\), the graph should open upwards, so this is not the graph of \(y = 3|x|\).

Answer:

B. The graph with vertex at the origin and opening upwards with a steeper slope (the second graph)