Question

which of the following is the graph of this absolute value function?
$y = -\frac{1}{2}|x|$

Explanation:

Step1: Analyze the parent function

The parent absolute value function is \( y = |x| \), which has a V - shape opening upwards with vertex at \((0,0)\).

Step2: Analyze the transformation

For the function \( y=-\frac{1}{2}|x| \), the negative sign in front of the absolute value means the graph of \( y = |x| \) is reflected over the \( x \) - axis (so it opens downwards), and the coefficient \(\frac{1}{2}\) is a vertical compression (it makes the graph wider or flatter compared to \( y=-|x| \)).

• Let's check the vertex: The vertex of \( y = |x| \) is at \((0,0)\), and the transformations (reflection and vertical compression) do not change the vertex, so the vertex of \( y = -\frac{1}{2}|x| \) is still at \((0,0)\).
• Let's check the direction: Since there is a negative sign, the graph opens downwards.
• Let's check the slope: For \( y = |x| \), when \( x\geq0 \), \( y=x \) (slope \( m = 1 \)), when \( x<0 \), \( y=-x \) (slope \( m=- 1 \)). For \( y=-\frac{1}{2}|x| \), when \( x\geq0 \), \( y =-\frac{1}{2}x \) (slope \( m =-\frac{1}{2} \)), when \( x<0 \), \( y=\frac{1}{2}x \) (slope \( m=\frac{1}{2} \)).

Now let's analyze the graphs:

• The first graph opens upwards, so it can't be the graph of \( y = -\frac{1}{2}|x| \) (since our function opens downwards).
• The second graph has a steeper slope (slope of \( 1 \) or \( - 1 \) for the lines) and opens downwards, but our function has a slope of \( \pm\frac{1}{2} \), so it's not the correct one.
• The third graph opens downwards, has a vertex at \((0,0)\), and the slope of the lines (for \( x\geq0 \) and \( x < 0 \)) is \( \pm\frac{1}{2} \) (we can check by taking a point, for example, when \( x = 2 \), \( y=-\frac{1}{2}\times|2|=- 1 \), which matches the point on the third graph).

Answer:

The third graph (the one with the blue lines opening downwards, vertex at (0,0), and a point at (2, - 1) or similar)