Question

which is the graph of $f(x)=|x|$ reflected across the x - axis, translated 3 units left, 4 units up, and dilated by a factor of 4?

Explanation:

Step1: Apply Reflection

Reflect \( f(x) = |x| \) across the \( x \)-axis: \( f(x) = -|x| \).

Step2: Apply Dilation

Dilate by a factor of 4: \( f(x) = -4|x| \).

Step3: Apply Horizontal Translation

Translate 3 units left: Replace \( x \) with \( x + 3 \), so \( f(x) = -4|x + 3| \).

Step4: Apply Vertical Translation

Translate 4 units up: Add 4, so \( f(x) = -4|x + 3| + 4 \).

Step5: Analyze the Graph

The vertex of the original \( |x| \) is at \( (0,0) \). After transformations, the vertex is at \( (-3, 4) \). The graph opens downward (due to the negative sign) and is narrower (due to the dilation factor 4). The first graph (top - left) has a vertex around \( (-2, 4) \)? Wait, no, let's re - check. Wait, the first graph: let's see the coordinates. Wait, the first graph's vertex: looking at the x - axis, the vertex is at \( x=-2 \)? No, maybe the user's first graph: wait, the problem has two graphs? Wait, no, the original problem's image: the first graph (top) has a vertex at \( x = -2 \), \( y = 4 \)? Wait, no, let's re - do the vertex calculation. For \( y=-4|x + 3|+4 \), vertex is at \( x=-3 \), \( y = 4 \). So the graph with vertex at \( (-3,4) \), opening downward, and steeper (since dilation factor 4). So the correct graph should have vertex at \( (-3,4) \), be a downward - opening V - shape, and narrower than the original \( |x| \) (since the coefficient of \( |x| \) is 4 in magnitude). So among the given graphs (assuming the first graph is the top one), if the first graph has vertex at \( (-3,4) \) (maybe my initial coordinate reading was wrong), then the correct graph is the first one (the top - most graph) as per the transformation steps.

Answer:

The top - most graph (the first graph shown)