Question

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

Explanation:

Step1: Recall the general form of absolute value function

The general form of an absolute value function is \( y = a|x| \), where \( a \) determines the direction and the slope. If \( a < 0 \), the graph opens downward; if \( a>0 \), it opens upward. The slope of the right - hand side (for \( x\geq0 \)) is \( a \), and the slope of the left - hand side (for \( x < 0 \)) is \( - a \). For the function \( y=-\frac{3}{2}|x| \), \( a =-\frac{3}{2}<0 \), so the graph opens downward.

Step2: Analyze the slope and key points

For \( x\geq0 \), the function is \( y =-\frac{3}{2}x \). Let's find the value of \( y \) when \( x = 2 \): \( y=-\frac{3}{2}\times2=- 3 \). For \( x = 1 \), \( y=-\frac{3}{2}\times1 =-\frac{3}{2}=-1.5 \) (but we can also check the point at \( x = 2 \) first).
Let's check the three graphs:
- First graph: When \( x = 2 \), the \( y \) - value of the marked point is \( - 3 \)? Wait, no, let's re - check. Wait, the first graph's marked point at \( x = 2 \): looking at the grid, if we calculate \( y=-\frac{3}{2}|2|=-3 \). Wait, no, maybe I misread. Wait, let's check the third graph: when \( x = 2 \), the marked point has \( y=-3 \) (since \( y =-\frac{3}{2}\times2=-3 \)). Also, the slope: for \( x>0 \), the slope is \(-\frac{3}{2}\), which means for a run of 2 (from \( x = 0 \) to \( x = 2 \)), the rise is \( - 3 \) (from \( y = 0 \) to \( y=-3 \)). For \( x<0 \), the function is \( y=-\frac{3}{2}(-x)=\frac{3}{2}x \), so the slope is \( \frac{3}{2} \), which means for a run of \( - 2 \) (from \( x = 0 \) to \( x=-2 \)), the rise is \( - 3 \) (wait, no: when \( x=-2 \), \( y =-\frac{3}{2}|-2|=-3 \), and the slope for \( x < 0 \) is \( \frac{3}{2} \) (since \( y=\frac{3}{2}x \) for \( x < 0 \)), so from \( x = 0 \) ( \( y = 0 \)) to \( x=-2 \) ( \( y=-3 \))? Wait, no, \( y=\frac{3}{2}x \) when \( x < 0 \), so when \( x=-2 \), \( y=\frac{3}{2}\times(-2)=-3 \), which is correct.
Wait, let's check the three graphs:
- The first graph: The vertex is at \( (0,0) \), opens downward. Let's check \( x = 2 \): the \( y \) - value of the dot. If we calculate \( y=-\frac{3}{2}\times2=-3 \). In the first graph, the dot at \( x = 2 \) seems to be at \( y=-3 \)? Wait, maybe I made a mistake earlier. Wait, the third graph: when \( x = 2 \), the dot is at \( y=-3 \), and the slope: from \( (0,0) \) to \( (2,-3) \), the slope is \( \frac{-3 - 0}{2-0}=-\frac{3}{2} \), which matches \( y =-\frac{3}{2}x \) for \( x\geq0 \). From \( (0,0) \) to \( (-2,-3) \), the slope is \( \frac{-3-0}{-2 - 0}=\frac{3}{2} \), which matches \( y=\frac{3}{2}x \) (since for \( x < 0 \), \( y =-\frac{3}{2}|x|=\frac{3}{2}x \))? Wait, no, \( y =-\frac{3}{2}|x|=\begin{cases}-\frac{3}{2}x, & x\geq0\\\frac{3}{2}x, & x < 0\end{cases} \). So for \( x=-2 \), \( y=\frac{3}{2}\times(-2)=-3 \), and the slope between \( (0,0) \) and \( (-2,-3) \) is \( \frac{-3-0}{-2 - 0}=\frac{3}{2} \), which is correct. And the graph opens downward (since \( a=-\frac{3}{2}<0 \)).

Wait, maybe the third graph is the correct one. Let's verify with \( x = 1 \): \( y=-\frac{3}{2}\times1=-1.5 \), but the marked points are at \( x = 2 \). So when \( x = 2 \), \( y=-3 \), which is what the third graph has (the dot at \( (2,-3) \)). The second graph has a dot at \( (2,-2) \), which is incorrect because \( y=-\frac{3}{2}\times2=-3 eq - 2 \). The first graph: let's see the dot at \( x = 2 \), if it's at \( y=-3 \)? Wait, maybe the first graph's grid is different. Wait, the key is: the function \( y =-\frac{3}{2}|x| \) has a vertex at \( (0,0) \), opens downward, and for \( x = 2 \), \( y=-3 \). So the graph where at \( x = 2 \), \( y=-3 \) (the third graph) is correct.

Answer:

The third graph (the one with the blue dot at \( (2, - 3) \) and opening downward with vertex at the origin)