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9. Point Symmetry

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9. Point Symmetry 9. Point symmetry Our purpose here is to investigate the notion of symmetry about a point both geometrically and algebrai- cally. Again, the key is to choose the right variable so that the algebraic expression reveals the symmetry of the graph. At the right is the graph of the cubic polynomial 60 f(x) = x3 – 9x2 + 22x + 3. 50 What sort of symmetry do we see in the graph? We will first 40 describe this geometrically as precisely as we are able. We 30 will then try to find an algebraic form of the equation which 20 displays the symmetry. 10 0 Geometry––describing the symmetry -2-1012345678 The curve certainly has a symmetry. It has a sort of a "centre" -10 roughly at x=3, and what it does on one side of the centre, it -20 does the opposite on the other side. -30 Let's try to be more precise. Does the notion of reflection help? 40 Let's see. If we take the right "half" of the curve and reflect it in a vertical axis through the centre we don't get the left half, but we get something that can be reflected into the left half 30 through a horizontal axis. 20 So the right half becomes the left half with a sequence of two reflections, one vertical and the other horizontal. 10 We can also do something with rotations. If we rotate the curve half a revolution about the centre it coincides with itself. 0 -2-1012345678 That's an interesting general principle. A sequence of two reflections, one vertical -10 and one horizontal, are always equivalent Tracing paper can be used here. to a rotation through 180 degrees. Trace a copy of the curve and then rotate the paper or flip it over. 9. point symmetry 1 Algebra ––the coordinate change The problem is to find a change of variable which reveals the symmetry. It’s a bit hard to know how to begin. Let's start with a nu- 40 merical check. Take a couple of points on either side of x=3, say x=2 and x=4, and compare the two heights: 30 f(2) = (2)3 – 9(2)2 + 22(2) + 3 = 19. f(4) = (4)3 – 9(4)2 + 22(4) + 3 = 11. 20 Does this fit with our symmetry hypothesis? Well rotations are not so easy to describe algebraically, but if we think in 10 terms of reflection through the centre point, we want the first point to be as much below the centre as the second point is 0 above. -10123456 Okay. What's the height of the centre? It's -10 f(3) = (3)3 – 9(3)2 + 22(3) + 3 = 15. The first point is 4 units above 15 and the second is 4 units below, so we have what we wanted! Now try to do this in general. We are perhaps now reminded that to compare heights on both sides of x=3, we want to be working with a coordinate which keeps track of the distance to the right of x=3. So we write x = 3+t. and then substitute this everywhere for x: y = x3 – 9x2 + 22x + 3. = (3+t)3 – 9(3+t)2 + 22(3+t) +3 We use the binomial expansions: (a+b)2 = a2 + 2ab + b2 . = (t3+9t2+27t+27) – 9(t2+6t+9) + 22(3+t) + 3 (a+b)3 = a3 + 3a2b + 3ab2 + b3 . = t3 – 5t + 15. Well there it is. Before we look closely at it, let’s be sure we haven't lost sight of what this expression tells us. It gives us the height of the curve at any point, not in terms of x but in terms of t, where the "t-coordinate" of a point is its distance to the right of the vertical line x=3. We might also check out the new constant term––it's the height at t=0 which we've already calculated to be 15, so that does check out. 9. point symmetry 2 Algebra––verifying the symmetry Now we need to see why this algebraic form establishes the symmetry. And for that we need to see that if we take two points whose t-coordinates are opposite in sign, one point should be as much below the centre (which has height 15) as the other point is above. Let's try it. Suppose the two points have t-coordinates +k and –k. Then the heights of the curve at these two points are: t=k y = k3 – 5k + 15. t=–k y = (–k)3 – 5(–k) + 15 = –k3 + 5k + 15. Have we got what we want? Yes we do. For emphasis, write It’s not quite so clear how the the two expressions as first pair of equations give us t=k y = 15 + (k3 – 5k). what we want, but the second, 3 rather elegant, rewriting says it t=–k y = 15 – (k – 5k). all. Very nice. The two expressions differ only in the middle + or –. That tells us that if one of them is above 15, the other is below 15 by the same amount. And that's our symmetry condition. Why did it work? That's an important question, because we did all this for one particular curve and we are curious to We come to the question of whether know how general it might be. What about other cubic poly- every cubic polynomial has a centre nomials? Do they all have a centre of symmetry? And will of symmetry. And we can reduce the same method work to establish that? that to the question of whether there is always a change of vari- The answer is yes. The above argument worked because the able that will get rid of the square pair of y-expressions had only t3 and t terms and both of these term. And the answer to that is yes. have the property that if we replace t by –t, we change the sign. It was crucial that there was no t2 term! If there had been a t2-term the argument wouldn't have worked. Can we do this with any cubic polynomial?––find a change of variable that eliminates the t2 term? The answer is yes. [Prob- lem 3.] 9. point symmetry 3 Problems A monic polynomial is one in which 1. Below we give a number of cubic polynomials and beside each we the coefficient of the leading term give the x-coordinate of its centre. Verify that the polynomial is in- (highest power of x) is 1. In this deed symmetric about this point by making a change of variable section we restrict attention to which removes the squared term. monic polynomials, just for ease of (a) f(x) = x3 – 15x2 + 12x x = 5 algebraic handling. The results we get are really quite general as any (b) f(x) = x3 + 9x2 – 10x – 10 x = –3 polynomial is a (positive or nega- tive) scalar multiple of a monic (c) f(x) = x3 – 6x2 + 20x + 2 x = 2 polynomial. (d) f(x) = x3 + 12x2 – 10 x = –4 (e) f(x) = x3 – 3x2 + 3x + 3 x = 1 2. By finding a suitable change of variable, show that the graphs of the following functions are symmetric about a point. (a) f(x) = (x–1)2 (x–3) (x–5)2. (b) f(x) = x3 (x–4)3. (c) f(x) = (x+2) (x+1) x (x–1) (x–2). 3. Show that the general cubic polynomial f(x) = x3 + bx2 + cx + d has a centre of symmetry and find its x-coordinate (in terms of one or more of b, c or d). [This essentially requires you to find a change of variable x = k+t which removes the squared term. The problem is what to take for k. Maybe you have found a pattern in #1 above which will allow you to guess what k should be. If not, just implement the change of variable, carrying k along as an unknown, and see what it has to be to make the t2 term vanish.] The shape of a cubic curve 4. At the right, are drawn graphs of the three cubic polynomials ABC A: y = x3 +4x B: y = x3 C: y = x3 – 4x It turns out that every monic cubic polynomial f(x) = x3 + bx2 + cx + d has the same shape as one of these three. Can you find conditions on one or more of b, c and d which will allow you to tell in which class the general monic cubic lies? Use this to classify the polynomials given in #1. A "proof" is not needed, but do give an account of your thought processes. 9. point symmetry 4.
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