How do you graph $y = | x | - 2$ $y=abs(x)-2$ ?

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How do you graph $y = | x | - 2$ $y=abs(x)-2$ ?

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Answer 1 · 2017-07-20T12:40:01

The graph looks like this:

graph{y=absx-2 [-10.125, 9.875, -5, 5]}

Explanation:

The absolute value function $y = | x |$ $y=absx$ looks like this:

graph{y=absx [-9.875, 10.125, -3.68, 6.32]}

This is a function you should probably know well for the future. Anyway, in order to plot $y = | x | - 2$ $y = absx - 2$ , we need to shift the $y$ $y$ values down by $2$ $2$ , since we are subtracting two from what $y$ $y$ is.

I would start graphing this graph by plotting the vertex (the point where the graph has a tight corner). On the original graph it is $(0, 0)$ $(0,0)$ , so on the new graph it will be $(0, - 2)$ $(0,-2)$ .

graph{x^2+(y+2)^2<=.04 [-9.875, 10.125, -3.68, 6.32]}

Now you need to draw the two rays coming from this point. To do this, graph one point on either side of $(0, 2)$ $(0,2)$ and connect it to $(0, 2)$ $(0,2)$ . Let's use $x = - 2$ $x=-2$ and $x = 2$ $x=2$ .

$y = | - 2 | - 2 = 2 - 2 = 0$ $y = abs(-2)-2 = 2 - 2 = 0$

$y = | 2 | - 2 = 2 - 2 = 0$ $y= abs2-2 = 2-2 = 0$

So now we know the points $(- 2, 0)$ $(-2,0)$ and $(2, 0)$ $(2,0)$ are on our graph.

graph{(x^2+(y+2)^2-0.04)((x-2)^2+y^2-0.04)((x+2)^2+y^2-0.04) = 0 [-10.125, 9.875, -5, 5]}

Now all that you have to do is connect these points.

graph{y=absx-2 [-10.125, 9.875, -5, 5]}

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How do you graph $y = | x | - 2$ $y=abs(x)-2$ ?

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Explanation:

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