Question

which of the following is the graph of this absolute value function?
$y = -\frac{2}{3}|x|$

Explanation:

Step1: Analyze the parent function

The parent absolute value function is \( y = |x| \), which has a V - shape with the vertex at the origin \((0,0)\) and opens upwards. For the function \( y=-\frac{2}{3}|x| \), the negative sign in front of the absolute value means the graph will open downwards (since the coefficient of \(|x|\) is negative, it reflects the graph of \( y = |x| \) over the \(x\) - axis).

Step2: Analyze the vertex and slope

The vertex of the absolute value function \( y = a|x - h|+k \) is at \((h,k)\). For \( y =-\frac{2}{3}|x|\), \(h = 0\) and \(k = 0\), so the vertex is at \((0,0)\).
The slope of the right - hand side (when \(x\geq0\)) of the absolute value function \( y = a|x| \) is \(a\). For \( y=-\frac{2}{3}|x|\), when \(x\geq0\), the function can be written as \(y =-\frac{2}{3}x\), so the slope of the right - hand side is \(-\frac{2}{3}\). When \(x = 3\), \(y=-\frac{2}{3}\times3=- 2\). Let's check the points:

• For the first graph: When \(x = 3\), \(y=-1\) (from the blue dot), which does not match \(y =-\frac{2}{3}\times3=-2\).
• For the second graph: When \(x = 2\), \(y=-1\) (from the blue dot), let's check the slope. The slope from \((0,0)\) to \((2, - 1)\) is \(\frac{-1 - 0}{2-0}=-\frac{1}{2}

eq-\frac{2}{3}\).

• For the third graph: When \(x = 3\), \(y=-2\) (from the blue dot). The slope from \((0,0)\) to \((3,-2)\) is \(\frac{-2-0}{3 - 0}=-\frac{2}{3}\), which matches the slope of the right - hand side of \(y =-\frac{2}{3}|x|\) (when \(x\geq0\), \(y =-\frac{2}{3}x\)). Also, the graph opens downwards with the vertex at \((0,0)\), which is consistent with the function \(y =-\frac{2}{3}|x|\).

Answer:

The third graph (the one with the blue dot at \((3, - 2)\) and vertex at \((0,0)\) opening downwards) is the graph of \(y =-\frac{2}{3}|x|\).