1 Shapes of Power Functions

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1 MA Lecture 06 1 Wednesday, 1/28/09 1 Shapes of Power Functions I would like you to be familiar with the shape of the power functions, that is, the functions of the form f(x) = x n, (1) for n = 1, 2, 3,.... Let s look at the graphs of the first several. The graphs of f(x) = x and f(x) = x 2 should look familiar. One s a line, and the other is a parabola. These are shown in Figure 1 f(x) = x f(x) = x 2 Figure 1: The graphs of f(x) = x 3, f(x) = x 4, and f(x) = x 5 are shown in Figure 2 f(x) = x 3 f(x) = x 4 f(x) = x 5 Figure 2: As you look at these graphs, you should see some similarity between the functions with odd powers. They go down to on the left, and they go up to + on the right. The functions with even powers also look similar, and they go up to + on both sides. The other thing I d like you to notice is that near the x-intercept, the graph gets flatter as the power gets larger. That is, f(x) = x 5 is flatter than f(x) = x 3, and significantly flatter than f(x) = x. This is because, for small values of x like x = 0.1, the higher the power the smaller the number, (0.1) 3 =.001 and (0.1) 6 =

2 2 SHAPES OF POLYNOMIAL FUNCTIONS 2 B A D C Figure 3: 1.1 Quiz 06A 1. Look at the graph on the left of Figure 3. It has the graphs of f(x) = x 3 and g(x) = x 6 drawn together. Which one is Graph A? (a) g(x) = x 6 (b) f(x) = x 3 (c) there isn t enough information (d) can t possibly be either one (e) none of these 2. The graph on the right of Figure 3 shows the graphs of f(x) = x 3 and h(x) = x 4. Which one is Graph C? (a) h(x) = x 4 (b) there isn t enough information (c) can t possibly be either one (d) f(x) = x 3 (e) none of these 2 Shapes of Polynomial Functions A great deal of information about the shape of a polynomial function can be taken from its factorization. For example, consider the seventh-degree polynomial function f shown here in expanded and factored form. f(x) = x 7 6x 6 3x x 4 69x 3 162x x = (x + 3)(x + 1) 2 (x 2)(x 3) 3 (2) Theoretically, you can always factor a polynomial down to linear and quadratic factors. As you saw with the cubic functions, the quadratic factors have an effect on the shape of the graph, but we will focus on the polynomials that have only linear factors. Note that there are seven linear factors in the factorization ((x + 1) occurs twice and (x 3) occurs three times). This corresponds to the leading term of the expanded expression being x 7. If you multiplied the linear factors out, the x 7 comes from multiplying the x s from each factor. The constant term comes from the constants of all the linear factors multiplied together ( 2) ( 3) ( 3) ( 3) = 162 (3) The middle terms come from combining like-terms, so it isn t as obvious where they come from. We ll look at two things. The leading term x 7 will tell us about the shape of the graph at the ends, and the factors will tell us about the shape in the middle. First of all, the leading term, x 7, dominates the graph for large values of x. In particular, for large values of x, this function will behave very much like the power function x 7. In this case, since 7 is odd, the graph

3 3 HOW TO SKETCH A GRAPH FROM THE FACTORS 3 will go down on the left and up on the right. This can be seen in the large scale graph shown on the left in Figure 4. Figure 4: Second, each factor gives us a zero (or x-intercept). Multiplying by really small numbers results in small numbers, so each factor dominates in some sense near its zero. Consider the function f(x) = (x 2)(x + 3) 2. We have zeros at x = 2 and x = 3. At x = 2, for example, we can see how the factor (x 2) is the main force near x = 2 by looking at a couple of function values. First of all, if we put x = 2 into the function, we get f(2) = ((2) 2)((2) + 3) 2 = = 0 25 = 0. If we move a small distance from x = 2, say to x = 2.1, we see f(2.1) = ((2.1) 2)((2.1) + 3) 2 = = = If we move a little further, say to x = 2.2, we get f(2.2) = ((2.2) 2)((2.2) + 3) 2 = = = The squared factor goes from 25 to to 27.04, which is not that big a deal, but the function values go from 0 to to 5.408, which is much more dramatic. This is because the (x 2) factor, while only changing a tenth, will actually double from 0.1 to 0.2. A closer-in look at this function is shown on the right in Figure 4. Since (x + 1) 2 is a double factor, the graph near the zero at x = 1 will look like a parabola. Looking closer at the graph, we can see this on the left in Figure 5 (note that the software package I m using put the y-axis at an odd place). We have a triple factor, (x 3) 3, that gives us the zero x = 3. The graph around this zero is shown on the right in Figure 5. Here, the graph looks kind of like the power function x 3. Remember For a polynomial function, the highest degree term dominates end behavior, and the powers of each linear factor dominate near the corresponding zero. 3 How to sketch a graph from the factors I will expect you to be able to sketch the graph of a polynomial function taking into account the following information. I. The leading term determines the end behavior. That is, whether the graph goes to + or on the left and right. If the function factors completely down to linear factors, then the degree of the leading term and the number of factors will be the same.

4 3 HOW TO SKETCH A GRAPH FROM THE FACTORS 4 Figure 5: II. III. A single factor will give you an x-intercept that crosses the x-axis like a line does. A double factor will give you an x-intercept that touches the x-axis like a parabola does. IV. A triple factor will give you an x-intercept that crosses the x-axis like a cubic (as shown on the right in Figure 5. V. For quadruple factors and more, we will draw them like III or IV depending on whether the multiplicity is even or odd. Any of these might be flipped up-side-down, because the other factors can be negative. We know what the graph of f(x) = x 7 6x 6 3x x 4 69x 3 162x x = (x + 3)(x + 1) 2 (x 2)(x 3) 3 (4) looks like, because we ve seen computer generated graphs of it. Here, I want to show you how I would sketch it without the computer. I m going to work from left to right, and I ll start with the leading term, x 7. The power is 7, which is odd, so I know the graph will go down to on the left. I could have gotten this same information by counting the seven linear factors. Next, I have four distinct zeros, x = 3, 1, 2, 3, so working from left to right, the x = 3 will be the first to consider. I m starting at and I m going to cross x = 3 like a line. This part of the graph should look like Figure 6. Figure 6: The next zero is at x = 1, and it comes from the double factor (x +1) 2. Therefore, it will touch the x-axis like a parabola, as in Figure 7. In this case, the graph is coming from above, so it has to go back up. (If it were coming from the bottom, then it would go back down). The next zero is at x = 2, and it comes from the single factor (x 2). It will cross the x-axis like a line, and since it is coming from above, it will cross the x-axis and go below, as in Figure 8.

5 3 HOW TO SKETCH A GRAPH FROM THE FACTORS 5 Figure 7: Figure 8:

6 3 HOW TO SKETCH A GRAPH FROM THE FACTORS 6 Finally, the zero at x = 3 comes from the triple factor (x 3) 3. It will cross the x-axis like a cubic, that is, it will hit the x-axis very flat. Since this is the last zero, the graph will continue upwards to +, as we knew from the beginning. This last part of the graph is shown in Figure 9. Figure 9: Right now, I m not concerned with how high the hills go or how low the valleys go. I just want to see the behavior around the zeros. 3.1 Quiz 06B 1. One of the graphs in Figures 10 and 11 is the graph of f(x) = (x + 1)(x 1) 2 (x 2). Which one is it? (a) Graph II (b) Graph I (c) Graph III (d) Graph IV (e) none of these Graph I Graph II Graph III Figure 10:

7 4 HOMEWORK 06 7 Graph IV Graph V Graph VI Figure 11: Graph VII Graph VIII Graph IX 4 Homework 06 Figure 12: The graphs of the following functions are in Figures 10, 11, and 12. Match them up. 1. f(x) = (x + 1) 2 (x 1)(x 2). 2. f(x) = (x + 1)(x 1)(x 2) f(x) = (x + 1) 2 (x 1) 2 (x 2). 4. f(x) = (x + 1)(x 1)(x 2) f(x) = (x + 1)(x 1) 3 (x 2). 6. f(x) = (x + 1)(x 1) 2 (x 2) f(x) = (x + 1) 2 (x 1)(x 2) f(x) = (x + 1) 3 (x 1)(x 2).