Question

y = -2|x - 2|
show your work here
draw

Explanation:

Step1: Identify Vertex

The general form of an absolute - value function is \(y = a|x - h|+k\), where \((h,k)\) is the vertex. For the function \(y=-2|x - 2|\), we can rewrite it as \(y=-2|x - 2|+0\). So, the vertex \((h,k)=(2,0)\).

Step2: Analyze the Slope and Direction

The coefficient \(a=-2\). Since \(a < 0\), the graph opens downwards. The slope of the two linear parts of the absolute - value function: for \(x\geq2\), \(y=-2(x - 2)=-2x + 4\), the slope \(m=-2\); for \(x<2\), \(y=-2(2 - x)=2x-4\), the slope \(m = 2\) (wait, no, correction: for \(x<2\), \(y=-2(2 - x)=2x - 4\)? No, \(y=-2|x - 2|\), when \(x<2\), \(|x - 2|=2 - x\), so \(y=-2(2 - x)=2x-4\)? Wait, no, \(y=-2\times(2 - x)=- 4 + 2x\), the slope is \(2\), but since \(a=-2\), the correct slopes: the standard absolute - value function \(y = |x|\) has slopes \(1\) (for \(x\geq0\)) and \(- 1\) (for \(x<0\)). When we have \(y=a|x - h|+k\), the slopes of the two lines are \(a\) and \(-a\) (when \(a>0\), one slope \(a\), one \(-a\); when \(a < 0\), one slope \(a\), one \(-a\) but the direction is reversed). Wait, for \(y=-2|x - 2|\), when \(x\geq2\), \(y=-2(x - 2)=-2x + 4\), slope \(m=-2\); when \(x<2\), \(y=-2(2 - x)=2x-4\), slope \(m = 2\). But since \(a=-2<0\), the graph opens downward. So the vertex is at \((2,0)\), and we can find two points on each line:

• For \(x = 2\), \(y=0\) (vertex).
• For \(x=3\), \(y=-2|3 - 2|=-2\times1=-2\).
• For \(x = 1\), \(y=-2|1 - 2|=-2\times1=-2\).

To draw the graph:

1. Plot the vertex \((2,0)\).
2. For the line with \(x\geq2\) (slope \(-2\)): from \((2,0)\), when \(x\) increases by \(1\) (to \(x = 3\)), \(y\) decreases by \(2\) (to \(y=-2\)).
3. For the line with \(x<2\) (slope \(2\)): from \((2,0)\), when \(x\) decreases by \(1\) (to \(x = 1\)), \(y\) decreases by \(2\) (since slope is \(2\), but since the graph opens down, wait, no, slope for \(x<2\): \(y=-2|x - 2|\), when \(x = 0\), \(y=-2|0 - 2|=-4\); when \(x = 1\), \(y=-2|1 - 2|=-2\); when \(x = 2\), \(y = 0\); when \(x=3\), \(y=-2\); when \(x = 4\), \(y=-4\).

So the correct graph should have its vertex at \((2,0)\), open downward, with points like \((0,-4)\), \((1,-2)\), \((2,0)\), \((3,-2)\), \((4,-4)\) etc. The given graph in the picture has the vertex at \((0,0)\), which is incorrect. The correct graph should be drawn with vertex at \((2,0)\), and passing through points such as \((0,-4)\) (when \(x = 0\), \(y=-2|0 - 2|=-4\)), \((1,-2)\) (when \(x = 1\), \(y=-2|1 - 2|=-2\)), \((2,0)\), \((3,-2)\), \((4,-4)\).

Answer:

To draw the graph of \(y=-2|x - 2|\):

1. Vertex: The vertex of the absolute - value function \(y = a|x - h|+k\) is at \((h,k)\). For \(y=-2|x - 2|\), \(h = 2\) and \(k = 0\), so the vertex is \((2,0)\).
2. Direction and Slopes: Since \(a=-2<0\), the graph opens downward.

• For \(x\geq2\), the equation is \(y=-2(x - 2)=-2x + 4\). When \(x = 3\), \(y=-2(3 - 2)=-2\); when \(x = 4\), \(y=-2(4 - 2)=-4\).
• For \(x<2\), the equation is \(y=-2(2 - x)=2x - 4\). When \(x = 1\), \(y=-2(2 - 1)=-2\); when \(x = 0\), \(y=-2(2 - 0)=-4\).

1. Plotting: Plot the vertex \((2,0)\). Then plot the points \((1,-2)\), \((3,-2)\), \((0,-4)\), \((4,-4)\) and draw two straight lines (one for \(x\geq2\) with slope \(-2\) and one for \(x<2\) with slope \(2\)) passing through the vertex and these points, opening downward.