x-intercepts, asymptotes, and end behavior together

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1 MA 2231 Lecture 27 - Sketching Rational Function Graphs Wednesday, April 11, 2018 Objectives: Explore middle behavior around x-intercepts, and the general shapes for rational functions. x-intercepts, asymptotes, and end behavior together With a rational function in factored form, the factors in the denominator dominate the function near the corresponding asymptotes, since the other factors stay relatively stable, while the factor that is going to zero pushes the function to infinity, since infinity times something not changing very fast takes you to infinity. In the numerator, the factors also dominate as they go to zero, since zero times something that isn t changing very fast is basically zero. As a result, the behavior around x-intercepts will like slightly distorted graphs of ±x, ±x 2, ±x 3, etc. Consider the function (1) f(x) = (x + 2)2 x 3 (x 2) (x + 1) 2 (x 1) 3, and look at its graph. We see the vertical asymptotes at x = 1 and x = 1, as we would expect from the factors in the denominator. Looking at the numerator, we can see that there should be x-intercepts at x = 2, x = 0, and x = 2, and we also see that in the graph. Furthermore, notice the behavior at the x-intercepts corresponds to the power on each factor. Since the zero at x = 2 comes from the factors (x + 2) 2, the graph of f hits the x-axis like a parabola would. The zero at x = 0 comes from x 3, and the graph of f looks kind of like the saddle point of y = x 3. The factor (x 2) has a power of 1, and the x-intercept at x = 2 just goes across the x-axis, kind of like a line would. Finally, notice that the end behavior is like a line (a tilted asymptote). We could have seen that by looking at the definition of f, if it were multiplied out. Since we really only need to look at the leading terms, all we need to know is (2) f(x) = (x + 2)2 x 3 (x 2) (x + 1) 2 (x 1) 3 = x6 + x 5 = x. + So for a rational function in factored form, we can see pretty easily what the graph looks like in general. That s what I would like you to be able to do. 1

2 MA 2231 Lecture 27 - Sketching Rational Function Graphs 2 Sketching the graph of a Rational Function Sketch the end and middle behavior of the function (x + 3)(x + 1) (3) f(x) =. (x 2) 3 As you will see, if you re doing everything together, you don t have to worry about whether the other stuff is positive or negative, like we had to do last time. That s because the end behavior and behavior around asymptotes and x-intercepts can only fit together one way. Here s the general process. vertical asymptotes. We have a vertical asymptote at x = 2, and it comes from a triple factor. Let s call this a 3rd-degree asymptote. The main thing we ll take from that, is that the sign will change as we move across the asymptote, because the degree is odd. x-intercepts/zeros. We have x-intercepts (a.k.a. zeros) at x = 3 and x = 1. These both come from single factors, so let s call these 1st-degree zeros. The graph of f will simply cross the x-axis at these x-intercepts, kind of like a line. end behavior. If you multiply everything out, you get x 2 + 4x + 3 x2 (4) f(x) = x 3 6x x 8 x 3 = 1 x. Notice that all you really need are the dominant terms, which you know without much work. The other terms are irrelevant to the general end behavior. In any case, the ends behave like 1 x. Putting everything together. The main thing to remember is that f can only change signs at the zeros and vertical asymptotes, and we know the signs of f on the ends from the end behavior. Therefore, we have all the information we need to sketch the graph. Put your zeros and vertical asymptotes in. Mark where your x-intercepts and vertical asymptotes like this. Start at the ends. The end behavior is like 1 x, which means that the the x-axis is a horizontal asymptote, and we approach from below on the left, and from above on the right. Like this.

3 MA 2231 Lecture 27 - Sketching Rational Function Graphs 3 Then draw in your graph. Start on the left end. You need to cross straight across the x-axis at x = 3. Then straight across again at x = 1. Then to the asymptote. You know you have to go to, because you re below the x-axis, and there are no more zeros before the asymptote. The degree of this asymptote is odd, so you must go to + on the other side. Finally, you connect up with the end.

4 MA 2231 Lecture 27 - Sketching Rational Function Graphs 4 If you were on the wrong side of the x-axis, then you made a mistake. Example 1. OK. Now look at this function. (5) f(x) = (x + 2)(x 1)2 (x + 1) 2 (x 2). There is a 1st-degree zero at x = 2, and a 2nd-degree zero at x = 1. There is a 2nd degree asymptote at x = 1, and a 1st-degree asymptote at x = 2. The end behavior is like x3 x 3 = 1. We won t worry about any extra wiggles, but there may be some. The end and middle behavior must look like this. Example 2. One more. (6) f(x) = (x + 1)2 (x 1) (x + 2) 3 x 2 (x 2). So the ends are like 1 x 3, so we approach the x-axis from below on the left. The asymptote at x = 2 is odd, so change signs. We have a 2nd degree zero at x = 1, so we hit it like a parabola, then go up to the asymptote at x = 0, which is even. Then straight across the zero at x = 1 to the asymptote at x = 2, which is odd. Then finish off on the right end going down to the x-axis.

5 MA 2231 Lecture 27 - Sketching Rational Function Graphs 5 Consider the function (7) f(x) = Sketch the end and middle behavior. Quiz 27 x 7 (x + 2) 3 (x 2) 2. Homework 27 Sketch the end and middle behavior for the following functions. 1. f(x) = 2. f(x) = 3. f(x) = 4. f(x) = 5. f(x) = x 2 (x + 2) 2 (x 2) 3 (x + 2)2 x 2 (x 2) 3 (x + 2)6 x 3 (x 2) 2 (x + 2) x 3 (x 2) 3 x 4 (x 3) 3 (x + 2) 2 (x 2) 3 Odd answers on next page.

6 MA 2231 Lecture 27 - Sketching Rational Function Graphs 6 1) 3) 5)