8. Bilateral symmetry


 Cecil Powers
 2 years ago
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1 . Bilateral smmetr Our purpose here is to investigate the notion of bilateral smmetr both geometricall and algebraicall. Actuall there's another absolutel huge idea that makes an appearance here and that's the question of choosing the right variable so that our algebraic epressions tell us what we want to know. In this case, what we want to "see" (algebraicall!) is the smmetr of the graph. Eample 1. At the right is the graph of the parabola =. We observe a smmetr about the ais. If we held a mirror along the ais, the parabola would reflect into itself. That sort of "mirror" smmetr is known as bilateral smmetr. It seems prett clear from the diagram, but that's not a proof that it holds. Find a convincing argument that the graph is bilaterall smmetric. Solution. What do we have to show? that the height of the graph at an point is the same as the height at the corresponding point on the other side of the mirror. What does this mean algebraicall? It's best to have a specific point: take the point at =3. The graph at that point has height = 3 = 7. Now what's the corresponding point on the other side? it's the point at = 3. What's the height at that point? It's the same as before. = ( 3) = 9 = 7 Wh did this work? we can see that it was because 3 and 3 have the same square. Well what worked for =3 will work for an value of, essentiall because the square of a number and the square of its negative are alwas the same. Let's just do it: Height at =a: = a Height at = a: = ( a) = a. The graph has the same height at a and a for an a. That tells us that the graph is bilaterall smmetric about the ais Our students all 'know" that the parabola is bilaterall smmetric, but can the produce a convincing argument that this is so? bilateral smmetr 1
2 Eample. At the right is the graph of the polnomial f() = + 1. Again we see a bilateral smmetr but this time about the line =. And again this seems clear from the diagram, but we want an algebraic verification. Solution. We have to show that the height of the graph at an point is the same as the height at the corresponding point on the other side of the line =. For eample take the point at =. That's units to the right of the line =. The corresponding point on the other side is two units to the left of the line = which is at =. The graph at these two points has heights so the heights are the same. f() = () () + 1 = f() = () () + 1 = Will this work for an pair of corresponding points? How do we find a general argument that this is so? Well the thing about and is that the are the same distance from, so that s the aspect of a point we want to keep track of its distance to the right or left of. That is, we want to write an as +t so that t will tell us where is with respect to A powerful wa to think about this t is as an alternative coordinate. For eample, = will have tcoordinate, and = will have tcoordinate, and = will have tcoordinate. The mission of the tcoordinate of a point is to tell us how far to the right of = the point is located. And a negative tcoordinate means the point is actuall to the left of =. t The point shown has an coordinate of 9 and a tcoordinate of 5.. bilateral smmetr
3 Changing coordinates Now let s see what happens when we write the polnomial using the tcoordinate instead of. The equation relating t to is Then =+t f() = f(+t) = (+t) (+t) + 1 = 1 + t + t 3 t + 1 = t +. Look at that the epression depends onl on t (and not on t) just like Eample 1. And that means that it will have the same value at an pair of points whose tcoordinates are negatives of each other like = and = or an other such pair. For eample, consider the pair =9 and = 1. The are both 5 units from, and in fact the have tcoordinates +5 and 5. And the above epression tells us that the parabola will have the same height 5+ = 7 at both of them. We have here a proof of the bilateral smmetr about =. What s the point here? It s that for our present purposes ( seeing the smmetr) was simpl not the best coordinate to use. The "coordinate of choice" was t. B the wa. Let s take our tepression and change it back to. To find t in terms of, we take the transformation =+t and solve for t: t= and plug this into the epression f() = t + = ( ) + and we have the epression back again, but in a different form. In fact what we have is the completed square form of the parabola. Now this is interesting. This business of changing to the t coordinate is reall completing the square in disguise. Or perhaps, a good wa to think about completing the square is as a coordinate transformation. When we write f() = ( ) + we are focusing attention on as an entit of interest (a new coordinate?), and in fact, with a tin bit of practice we can simpl look at that epression and sa: oh es, that tells me that the graph is smmetric about =. f() = ( ) + With a tin bit of practice we can simpl look at that epression and sa: oh es, that tells me that the graph is smmetric about =.. bilateral smmetr 3
4 Eample 3. At the right is the graph of the quartic polnomial f() = ( 1) ( 3) ( 5). It appears to be bilaterall smmetric about the vertical line =3, and indeed the smmetric pattern of the roots 1, 3 (twice) and 5 might also suggest this smmetr. Change variable and verif that this is so. 5 3 Solution. We want to keep track of the distance of an point to the right or to the left of 3, so we let =3+t and then replace in the epression: f() = (3+t 1) (3+t 3) (3+t 5) = (t+) (t) (t ) = t (t ) Look at that. The epression depends onl on t so it will have the same value at two points whose t coordinates are negatives of one another. Now if we have two points and their tcoordinates are negatives of one another, that means that the will be to the right and to the left of =3 b the same amount. So the polnomial has the same value at two points which are the same amount to the right and to the left of =3, and that tells us that the graph is smmetric about the vertical line = What this eample demonstrates is that the completing the square idea is reall much more general than as a tool for analzing parabolas. In its "coordinate transform" guise it can work for an function.. bilateral smmetr
5 Problems 1. B making a suitable change of variable, write the following quadratic polnomials in a form which displas their bilateral smmetr. Illustrate with a sketch of the graph. (a) f() = + 1. (b) f() =. (c) f() = (d) f() =. (e) f() = ( + )( ).. B making a suitable change of variable, write the following functions in a form which displas their bilateral smmetr. Illustrate with a sketch of the graph which ou can obtain with the use of graphing technolog. (a) f() = (+1) ( 1) ( 3). (b) f() = (+1) ( 3). (c) f() = (+1) ( 1) ( ) (d) f() = ( 1) ( + 3)( 5) (e) f() = ( 1)( )( 3) ( ) 3. Find all solutions of the equation (+1)(+)(+3)(+) = 1.. The quartic polnomial f() = has a vertical ais of smmetr. Your job is to find it and then change variable so that the smmetr is revealed. Now we don't want ou to use technolog to draw the graph first. What we want ou to do is tr an "unknown" transformation = k+t where k is a parameter to be determined. Look at the epression for in terms of t and decide what k should be. 5. At the right is the graph = f(). Make a rough sketch of the graph = f( ) for. Pa attention to properties of bilateral smmetr bilateral smmetr 5
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