GRAPH SYMMETRIES. c Gabriel Nagy

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1 c Gabriel Nag GRAPH SYMMETRIES A. Smmetr for General Equations Let us assume we are given a general equation in and, of the form: Left Epression in, = Right Epression in,. Out of this equation, we build three Test Equations Left Epression in, = Right Epression in,. Left Epression in, = Right Epression in,. Left Epression in, = Right Epression in,. (When building equation (A) we replace with on both sides. When building equation (B) we replace with on both sides. When building equation (C) we replace with and with on both sides.) With these three new equations in mind, we address the following Question: Which of the three Test Equations is equivalent to the the original equation ( )? Depending on a positive answer to the above question, the graph of ( ) will ehibit a certain smmetr, as summarized in the table below. Equation Smmetr Feature of Equivalent to ( ) Graph of ( ) (A) -ais smmetr (B) -ais smmetr (C) origin smmetr Tip: When building up the Test Equations, use grouping smbols. That is, when replacing for instance with, put [ ] in place of. EXAMPLE 1: Consider the equation ( ) (A) (B) (C) = ( 1 ) The Test Equations are (note the use of grouping smbols): = [ ] = (A 1 ) [ ] = = (B 1 ) [ ] = [ ] = (C 1 ) 1

2 (The equations following the smbol are equivalent to the ones before it.) Since (A 1 ) is the onl equation that is equivalent to the original one, it follows that the graph of ( 1 ) has onl -ais smmetr. The -ais smmetr is manifested as follows: whenever a point is on the graph, its reflection with respect to the -ais is again a point on the graph. The graph of ( 1 ) is shown in Figure 1 below. Figure 1 Figure 2 EXAMPLE 2: Consider the equation The Test Equations are: 2 = + 1. ( 2 ) 2 = [ ] = + 1 (A 2 ) [ ] 2 = = + 1 (B 2 ) [ ] 2 = [ ] = + 1. (C 2 ) Since (B 2 ) is the onl equation that is equivalent to the original one, it follows that the graph of ( 2 ) has onl -ais smmetr. The -ais smmetr is manifested as follows: whenever a point is on the graph, its reflection with respect to the -ais is again a point on the graph. The graph of ( 2 ) is shown in Figure 2 above. EXAMPLE 3: Consider the equation = 3. ( 3 ) 2

3 The Test Equations are: = [ ] 3 = 3 (A 3 ) [ ] = 3 = 3 (B 3 ) [ ] = [ ] 3 = 3. (C 3 ) Since (C 3 ) is the onl equation that is equivalent to the original one, it follows that the graph of ( 3 ) has onl origin smmetr. The origin smmetr is manifested as follows: whenever a point is on the graph, its reflection with respect to the origin is again a point on the graph. The graph of ( 3 ) is shown in Figure 3 below. Figure 3 Figure 4 EXAMPLE 4: Consider the equation The Test Equations are: = 1. ( 4 ) 2 + [ ] 2 = = 1 (A 4 ) [ ] = = 1 (B 4 ) [ ] 2 + [ ] 2 = = 1. (C 4 ) In this case all three Test Equations are equivalent to the original one. Therefore, the graph of ( 4 ) has -ais smmetr, -ais smmetr, and origin smmetr (simultaneousl!). Note that the graph of ( 4 ), shown in Figure 4 above is just the circle of radius 1, centered at the origin. 3

4 B. Smmetr for Function Graphs Recall that the graph of a function f() is simpl the graph of the equation = f(). ( f ) The Test Equations associated with ( f ) are (equivalent to) the following: = f( ) (A f ) = f() (B f ) = f( ). (C f ) In connection with the Question raised in the previous section, we introduce the two named features for f(): even and odd. These two features will be responsible for -ais and origin smmetries, thus characterizing when ( f ) is equivalent to (A f ) or (C f ). The table below eplains this terminolog. (The condition when (B f ) is equivalent to ( f ), which would provide -ais smmetr, is quite simple, so it is not given a special name. See the table for its meaning.) Named Feature Condition Smmetr of of f() on f() Graph of f() even f( ) = f() -ais smmetr constant 0 f() = 0 -ais smmetr odd f( ) = f() origin smmetr Tip: When checking if a function f() is even or odd, and the graph is not available, we must compute f( ). (We do this b replacing with [ ] in the formula.) After that, we compare f( ) with f(), or with f(). It should be noted that, unless f() is constant 0, it cannot be even and odd simultaneousl. EXAMPLE 1 (continued): Consider the function f() = Replacing with [ ] we get f( ) = [ ] = Since we clearl have f( ) = f(), it follows that f() is even. (The graph of f() is the same as the one in Figure 1.) 4

5 EXAMPLE 3 (continued): Consider the function f() = 3. Replacing with [ ] we get f( ) = [ ] 3 = 3. Since we clearl have f( ) = f(), it follows that f() is odd. (The graph of f() is the same as the one in Figure 3.) 5