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Zeros&asymptotes Example 1 In an early version of this activity I began with a sequence of simple examples (parabolas and cubics) working gradually up to the main idea. But now I think the best strategy is to start with an example that is rich enough to display the big idea. I would work this with the class with a diagram of the graph on the screen at the front and get them to explain to me why the factored form of the polynomial really does provide the qualitative form of the graph. I arm them with the page from the workbook that allows them to make notes on the story at each root. Following that rather chaotic class discussion, I go through the rather elegant procedure of moving from right to left keeping track of the sign of y as we pass through each root. Consider the polynomial function f(x) = x 7 15x 6 + 89x 5 265x 4 + 414x 4 320 + 96x The graph f(x) is given at the right. I used technology to draw it. Without technology, that would not have been so easy to do, even armed with calculus. It turns out that f(x) can be factored as f(x) = x(x 1) 2 (x 2)(x 3)(x 4) 2 and thus the zeros of f (where f(x) = 0) are the values x = 0, 1, 2, 3 and 4. The point of this example is that this factored form gives us a lot of information about the graph that we didn t have with the expanded formula. To be precise, the factored form allows us to easily see where y is positive (graph lies above the x-axis), where y is negative (graph lies below the x-axis), and where y is zero (graph intersects the x- axis). Solution The factored form gives f(x) as a product of factors, and thus the sign of y is the product of the signs of these factors. And the sign of each factor is easy to find for any given x. Indeed, a linear term (x a) is positive for x > a and negative for x < a. Thus the only time it changes sign is at x = a. Here s what I think is the neatest way to make the argument. Start with x greater than 4. At this point, all terms are positive so y is positive. Now let x decrease and take account of what happens as we pass through each root. And of course we only have to note what happens to the factor belonging to that root. Well from the form of the equation we see that that as x passes through each root, the sign of y: does not change at x=4 changes at x=3 changes at x=2 does not changes at x=1 changes at x=0. At each root we can record the sign of y in the next interval. This is enough to give us the sign of y in each interval. Given that, we can make a pretty good start at drawing the graph f(x). peter.taylor@queensu.ca 1

Example 2 At the right we have the graph of the rational function: x 2 x + 1 Note from both the equation and the diagram that the equation has only one root: x = 2. Note that the sign of y is also clear from the diagram: for x greater than 2, y is positive for x between 1 and 2, y is negative for x less than 1, y is positive (a) Argue that this is also clear from the equation. This time y is a quotient of two terms (rather than a product) but its sign is still determined by the two signs. We use the same argument as before. For x greater than 4 all terms are positive so y is positive. Now let x decrease. The sign of y: changes at x=2 making y negative changes at x= 1 making y positive This is enough to give us the sign of y in each interval. But we have a new kind of behavior at x = 1. For x very close to 1, the denominator x + 1 is very close to zero, whereas the numerator gets close to 3. That causes y to blow up in magnitude and using what we know about the sign of y, we see that it will take a large negative value just above 1 and a large positive value just below 1. That s what causes the vertical asymptote at x = 1. (b) Here s a good question. What happens to y as x gets large, both positive and negative? From the graph, it looks like y increases on the right side and decreases on the left side, but how far? Does it approach a limit? Is it asymptotic to the x-axis? It s hard to get this information from the graph but the equation is not so transparent either as x gets large, both numerator and denominator get large making it hard to see what happens to the quotient. The students can of course use their calculators to get a good idea of the answer, but how would an algebraic argument go? There s a nice way to rewrite the quotient that will deliver the answer divide top and bottom by x: x 2 x + 1 = 1 2/x 1 + 1/x Now what happens to this as x gets large in absolute value? The numerator and the denominator both approach 1. In terms of the graph, the curve approaches a horizontal line of height 1. That s called a horizontal asymptote of the equation. peter.taylor@queensu.ca 2

Example 3 (a) Make a reasonable sketch of the graph of the equation x(x 2) (x + 1)(x 1) 2 Pay attention to the asymptotic behavior, occasions when either x or y become large in magnitude. Solution. Note that the equation has roots at x = 0 and x = 2. The equation allows us to find the sign of y at any point. There are four terms in the y-expression and even though only two of these are in the numerator with the remaining two in the denominator, the sign of y is still the product of the signs of the four terms. The linear terms change sign only at their roots while the quadratic term (x 1) 2 never changes sign and is zero only at x = 1. Thus we argue as follows. For x greater than 2 all four terms are positive so y is positive. Now as x decreases, the sign of y: changes at x=2 does not change at x=1 changes at x=0 changes at x= 1 This is enough to give us the sign of y in each interval. The graph is given at the right. Note that the two terms in the denominator have roots at x = 1 and x = 1. This indicates that there will be vertical asymptotes at these two x-values. However at x = 1, there is no change of sign, and the graph goes down the asymptote on both sides. Note that we are not asking for the more detailed sketch we would expect from a calculus student. We are looking for the students to get the sign right and the behavior of the graph at the roots. Using the vertical scale they can try to get the height of the graph about right at different points. (b) What happens to y as x gets large, both positive and negative? Is there a horizontal asymptote? A good way to answer this question is to rewrite the quotient by dividing top and bottom by. One way to do this gives us: x x (x 2 x ) (x + 1) ( x 1 x ) 2 = (1 2/x) (x + 1)(1 1/x) 2 ~ 1 (x + 1)(1) 2 As x gets large (positive or negative) the expression (x+1) in the denominator gets large sending the expression close to zero. In terms of the graph, the x-axis is a horizontal asymptote. peter.taylor@queensu.ca 3

Example 4. Here we will work with the parameterized family of graphs x(x a) (x + 1)(x 1) (0 a 2) (a) The endpoint cases a = 0 and a = 2 are sketched below. But before I give these to the students, I ask them to go to the board and come up with rough graphs for these two cases. a = 0: (x + 1)(x 1) a = 2: x(x 2) (x + 1)(x 1) Here s the interesting question: if we allow a to increase gradually from 0 to 2, we will see the shape of the graph change continuously, starting at a = 0 as the graph on the left and winding up at a = 2 as the graph on the right. What does this journey look like? The puzzling behavior here concerns the asymptote at x = 1. At a = 0 it goes down the asymptote on the left side and up the asymptote on the right side. But at a = 2 this behavior is reversed: up on the left side and down on the right side. How can that happen gradually? There must be a point where it happens suddenly but what can the graph look like at such a point? And how can there be sudden changes when the parameter changes continuously? Your job is to understand completely what this continuous transformation looks like. Certainly you could pull Desmos out of the cloud and input the general equation with a slider for a and see exactly how it happens. But before you do that, see if you can figure it out. Draw rough graphs of some key configurations along the way. Provide some discussion to help the reader understand exactly how a curve plunging down to infinitely can suddenly change to plunging up. Or does it? (b) A good place to start the investigation is to check out the halfway point at a = 1. Indeed the mathematical form of the equation suggests that there ought to be a significant shift at that point (do you see why?). Okay, when a = 1: x(x 1) (x + 1)(x 1) = x (x + 1) That s a pretty simple equation and since the (x 1) term in the denominator has been canceled out, the asymptote at x = 1 has disappeared. I ask the students to draw the graph. a = 1: x (x + 1) peter.taylor@queensu.ca 4

(c) Okay. For a < 1, the asymptote at x = 1 goes up on the left and goes down on the right. And for a > 1, it s the reverse the asymptote at x = 1 goes down on the left and goes up on the right. And at the transition, a = 1, the asymptote disappears. The question we are interested in is exactly how does this happen? What does the transition look like? I ask the students to draw what they think the graph might look like just before and after a = 1, for example, at a = 0.99 and a = 1.01. Well here they are: a = 0.99: x(x 0.99) (x + 1)(x 1) a = 1.01: x(x 1.01) (x + 1)(x 1) It s fun using Desmos to see the journey in real time. The two branches seem to lean towards one another in what only can be described as an expectant kiss. At the moment of the kiss, the asymptote disappears! and right away returns as the curves go off in the opposite directions. https://www.desmos.com/calculator/yyxu4cvxsy peter.taylor@queensu.ca 5

Example 5. Here we work with the parameterized family of graphs (x 2)2 (x c) 1 c 1 (a) The endpoint cases c = 1 and c = 1 are sketched below. I will shortly give the students these two graphs but before that I ask them to produce rough sketches a good exercise for NPVS (nonpermanent vertical surfaces!). In the right-hand graph the behaviour on the interval [0, 3] has been magnified to better display the roots x = 1 and 2. (x 2)2 (x + 1) (x 2)2 (x 1) Here s the question: if we allow c to increase gradually from 1 to 1, we will see the shape of the graph change continuously. What does this journey look like? Again, the puzzling behavior concerns the asymptote at x = 0, the y-axis. For c = 1 it goes up the asymptote on both sides and for c = 1 it goes down the asymptote on both sides. How can that transition happen gradually? It would seem that there must be a point where there s a sudden shift but what can the graph look like at such a point? Again, before they are permitted to enjoy the luxuries of Desmos, they have to try to figure it out. (b) A good place to start the investigation is to check out the journey s halfway point at c = 0. Indeed the mathematical form of the equation suggests that there ought to be a significant shift at that point, as will get an x-cancelation. Okay, when c = 0: (x 2)2 (x 0) (x 2)2 = x I ask the students to draw the graph. They discover, using their sign analysis, that the vertical asymptote has to go both up and down down on the left side of the y-axis and up on the right side. peter.taylor@queensu.ca 6

Okay. Let s consolidate. We are considering the family: (x 2)2 (x c) 1 c 1 The story of the sign of y is quite simple it changes once and only once at x = c: for x > c for x < c y is positive y is negative That tells us, for every value of c, whether the asymptote goes up or down the y-axis. When c is positive, it has to go down the axis and when c is negative, it has to go up the axis. And finally, as we have just seen, when c is zero, it goes up on the right side and down on the left side. (c) Now the question is, what does the graph look like for c very close to zero. How does it prepare for the big up-down shift? I put the students in small groups at the (white or black) board and ask them to draw the graph for c = ±0.1. They don t of course have calculus, but they can still do a pretty good job. Take, for example, c = 0.1. They know that the graph looks like a parabola opening up at x = 2. Then, as x decreases (moving left) it crosses the x-axis from above at x = c = 0.1. So it must have a maximum somewhere between 0.1 and 2. And then as it gets closer to x = 0, it moves rapidly down the y-axis. That gives them a reasonable picture of the graph (which is plotted at the right). This is also a good problem for a grade 12 calculus class as they can show that as c approaches zero, that peak gets higher and higher only to fail at some point and come crashing down. It won t succeed at getting all the way up until c hits zero. The story is much the same on the other side of the y- axis. The curve tries harder and harder to descend but only succeeds when c = 0. Here s an animation from c = 1 to c = 1: https://www.desmos.com/calculator/henlu59sec and here s one from c = 0.1 to c = 0.1: https://www.desmos.com/calculator/1cglkdnsmc c > 0 y>0 y>0 c c y<0 y<0 c < 0 graph crosses the x-axis at x = c peter.taylor@queensu.ca 7

Exercises. 1. At the right we have the graph of the polynomial (x + 1)(x 3) Note that both the equation and the diagram give us the roots of the equation: x = 1 and x = 3 Note that it is clear from the diagram that: for x greater than 3, y is positive for x between 1 and 3, y is negative for x less than 1, y is positive Show that this is also clear from the form of the equation. Solution. The variable y is a product of two terms and its sign is determined by the signs of those two terms. In fact there s a powerful way to make the argument. Start with x greater than 3 and see what happens to the sign of each factor as x gradually decreases: for x greater than 3, both terms are positive so that y is positive as x decreases the signs both stay positive until x passes through 3 this causes the term (x 3) to change sign but (x+1) stays positive thus y changes sign and becomes negative as x continues to decrease nothing changes until x passes through 1 this causes the term (x+1) to change sign but (x 3) stays negative since only one term changes sign, y changes sign and becomes positive and y stays that way for the rest of the journey. peter.taylor@queensu.ca 8

2. At the right we have the graph of the polynomial (x + 1)(x 3) 2 Note that both the equation and the diagram give us the roots of the equation: x = 1 and x = 3 Note that it is clear from the diagram that: for x greater than 3, y is positive for x between 1 and 3, y is positive for x less than 1, y is negative (a) Show that this is also clear from the form of the equation. Solution. Start with x greater than 3 and see what happens to the sign of each factor as x gradually decreases: for x greater than 3, both terms are positive so that y is positive as x decreases the signs both stay positive until the second term becomes zero at x = 3 As x passes through 3 the term (x 3) 2 does not change sign Thus y does not changes sign and stays positive as x continues to decrease nothing changes until x passes through 1 this causes the term (x+1) to change sign and thus y changes sign to negative and stays that way for the rest of the journey. (b) Here s another observation about this example. The piece of the graph around x = 3 looks like a parabola and indeed that s expected from the fact that the (x 3) terms appears as a square. But what parabola is it close to? Well when x is close to 3, the other term (x+1) is close to 4 so that the expression for y will be close to 4(x 3) 2 And at the right I have drawn the graph of this parabola along with the cubic curve. The match is good near x = 3. peter.taylor@queensu.ca 9

3. Show algebraically that the slant asymptote for the graph is the line x 3. (x 2)2 (x + 1) Of course to do this exercise the students need to know what it means for a curve to be asymptotic to a line and I ask the class to use their intuition to formulate a definition. Here s what they hopefully (after perhaps some coaching) arrive at: Definition. The graph f(x) is asymptotic to the line mx + b if the distance between the graph and the line approaches zero as x becomes large. That is: f(x) (mx + b) approaches zero as x gets large (large positive or large negative or often both). Solution. We need to show that the expression (x 2) 2 (x + 1) (x 3) approaches zero for large x. This is a good algebraic calculation. We first get a common denominator. = (x 2)2 (x + 1) x 3 + 3 = (x2 4x + 4)(x + 1) x 3 + 3 = (x3 3 + 4) x 3 + 3 = 4 This clearly approaches zero as x get large in magnitude. peter.taylor@queensu.ca 10

4. For any given value c find the slant asymptote of the equation (x 2)2 (x c) Solution. There are a couple of ways to go about this. Perhaps the obvious one is to take mx + b to be the unknown asymptote and find the values of m and b that satisfy the asymptote condition. I do that one first. This condition requires that (x 2) 2 (x c) (mx + b) approach zero for large x. Take a common denominator: = (x2 4x + 4)(x c) mx 3 b Expand: = (x3 (4 + c) + (4c + 4)x 4c) mx 3 b Now collect the like terms in x. = (1 m)x3 (4 + c + b) + (4c + 4)x 4c = (1 m)x (4 + c + b) + 4(c + 1) 4c x Now how do we choose the parameters m and b so that this approaches zero for large x? We clearly need to zero out those first two terms. Thus we need m = 1 b = (4 + c) A more direct way is to simplify the equation of the graph: (x 2)2 (x c) = (x2 4x + 4)(x c) = (x3 (4 + c) + (4c + 4)x 4c) = x (4 + c) + 4(c + 1) 4c x Now what do we have to subtract from this to make what s left approach 0 for large x? The answer is clearly x (4 + c). Thus the asymptote is x (4 + c). peter.taylor@queensu.ca 11

5. Make a reasonable sketch of the graph of the equation x(x 2) 2 (x + 1)(x 1) 2 Pay attention to the asymptotic behavior, occasions when either x or y become large in magnitude. peter.taylor@queensu.ca 12