Absolute Value Graph: Explaining |2-x| Horizontal Translation Right

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SUMMARY

The discussion clarifies that the absolute value function |2-x| translates horizontally to the right by 2 units. This is established by recognizing that |2-x| is equivalent to |x-2|, indicating that the vertex of the graph shifts to the right. The key insight is that the value of x that makes the expression inside the absolute value zero determines the location of the vertex, which in this case is at x=2. Thus, the graph of |2-x| is a standard |x| graph shifted 2 units to the right.

PREREQUISITES
  • Understanding of absolute value functions
  • Familiarity with horizontal translations of graphs
  • Basic algebraic manipulation
  • Graphing techniques for functions
NEXT STEPS
  • Study the properties of absolute value functions
  • Learn about horizontal and vertical translations of graphs
  • Explore the concept of transformations in algebraic functions
  • Practice graphing various absolute value equations
USEFUL FOR

Students learning algebra, educators teaching graph transformations, and anyone seeking to understand the behavior of absolute value functions in mathematics.

buttretler
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Can someone explain to me why |2-x| would have a horizontal translation to the right? When I've always been taught that anytime you see a [+] it will translate to the left. The graph would be a regular |x| graph but it is shift 2 spots to the right. Thanks to anyone for help
 
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Ineedhelppp said:
Can someone explain to me why |2-x| would have a horizontal translation to the right? When I've always been taught that anytime you see a [+] it will translate to the left. The graph would be a regular |x| graph but it is shift 2 spots to the right. Thanks to anyone for help

There are several ways of thinking about this. One way: $|2-x|=|x-2|$, so you can see that it must be shifted to the right. In addition, one way of thinking about shifting is to ask yourself at what value of $x$ will the stuff inside the magnitudes be zero? That value of $x$ is going to have the corner that $|x|$ normally does at the origin. The nice thing about this second way of thinking is that it works for positive or negative shifts. Does this answer your question?
 
I think the second way I understand better, it still leaves me hazey because if |x-2|=|2-x| would mean no shift is taking place. And the only way at my level to show it is to the right, is a table of values with f(0),f(1) and f(2). Either way thanks for the further inlightenment.
 

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