Question

match the graphs with the functions.$y=(x-5)^2$$y=-5x^2$$y=-2|x|$$y=2|x|$$y=-|x+2|$$y=-x^2+5$$y=|x-2|$$y=x^2+5$$y=-(x+5)^2$$y=-|x|+2$$y=|x|+2$$y=5x^2$drag the function given above into the appropriate area below to match the graph.

Explanation:

Step1: Analyze first graph (downward parabola, vertex at (0,5))

This is a vertical reflection of $y=x^2$ shifted up 5. Function: $y=-x^2+5$

Step2: Analyze second graph (upward parabola, vertex at (0,5))

This is $y=x^2$ shifted up 5. Function: $y=x^2+5$

Step3: Analyze third graph (downward V, vertex at (0,0), steep)

This is a vertical stretch and reflection of $y=|x|$. Function: $y=-2|x|$

Step4: Analyze fourth graph (downward V, vertex at (0,0), less steep)

This is a reflection of $y=|x|$. Function: $y=-|x|$ (matches $y=-|x|+2$? No, vertex is (0,0): correction, vertex at (0,0) with slope -1: $y=-|x|$, but if vertex at (0,2) no, wait graph shows vertex at (0,0): $y=-2|x|$ is steep, this is $y=-|x|$. Wait no, fourth graph vertex at (0,0), arms to (-10,-10) and (10,-10): slope -1, so $y=-|x|$. But given functions: $y=-|x|+2$ has vertex (0,2). Correction: third graph: vertex (0,0), goes to (-5,-10),(5,-10): slope -2, so $y=-2|x|$. Fourth graph: vertex (0,0), goes to (-10,-10),(10,-10): slope -1, but no $y=-|x|$. Wait given functions: $y=-|x|+2$ has vertex (0,2), $y=-|x+2|$ has vertex (-2,0). Wait first graph: parabola vertex (0,5), opens down: $y=-x^2+5$. Second graph: parabola vertex (0,5), opens up: $y=x^2+5$. Third graph: V-shape down, vertex (0,0), steep: $y=-2|x|$. Fourth graph: V-shape down, vertex (0,0), slope -1: but given functions, $y=-|x|$ not present. Wait fourth graph: vertex (0,2)? No, graph shows vertex at (0,0). Wait given functions: $y=-|x|+2$ has vertex (0,2), so if fourth graph has vertex (0,2), then $y=-|x|+2$. Correction: recheck:

1. Top left: downward parabola, vertex (0,5): $y=-x^2+5$
2. Top right: upward parabola, vertex (0,5): $y=x^2+5$
3. Bottom left: downward V, vertex (0,0), steep (goes to (-5,-10)): $y=-2|x|$
4. Bottom right: downward V, vertex (0,0), slope -1 (goes to (-10,-10)): but no $y=-|x|$, wait given functions have $y=-|x|+2$: no, vertex (0,2). Wait bottom right graph: vertex (0,2)? If yes, $y=-|x|+2$.

Answer:

Top left graph: $y=-x^2+5$
Top right graph: $y=x^2+5$
Bottom left graph: $y=-2|x|$
Bottom right graph: $y=-|x|+2$