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Polynomials |
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A polynomial is the sum of terms of the form anxn, where n is a nonnegative integer and an is any real number. The degree of polynomial is the largest exponent of a term with a non-zero coefficient. The leading coefficient of a polynomial is the coefficient an where n is the degree of the polynomial. The domain of a polynomial is all reals. The graph of a polynomial can't have holes, breaks or sharp points. If you were to zoom in any portion of the graph of a polynomial it would eventually look like a straight line. The tails of polynomials of degree 2 or more get progressively steeper. Examples of polynomials:
The polynomial in (2) has degree 3 and leading coefficient 2. The polynomial in (3) has degree 0 (x0 is not usually written out). The leading coefficient is p. The polynomial in (4) has degree 1 and the leading coefficient is 5. Examples of functions that are not polynomials:
The global behavior of a polynomial shows us the general shape of the graph, as if seen from the moon. For global behavior we asking what the polynomial's (function's) graph looks like from a long distance away. On the graphing calculator this means using a window such as -100 < x < 100 and -5000 < y < 5000. The global behavior of a function tells us the behavior of the tails of its graph. The local behavior of a polynomial tells us what the polynomial (function) looks like up close, in particular close to each x-intercept. This should show all intercepts and turning points. We want to predict and then find the intercepts and general location of turning points. We also would like to know the behavior of the polynomial graph near each x-intercept if possible. The graph of a polynomials of degree 0 - i.e., y = c for some constant c - is a horizontal line. It has no turning points and its tails are flat. The graph of a polynomial of degree 1 -- i.e., y = ax + b with a > 0 or a < 0 -- is a slant line with one x-intercept, no turning points, and tails in opposite direction. The graph of a polynomial of degree 2 -- y = ax2 + bx + c with a nonzero leading coefficient a -- is a parabola that opens up if a > 0 and down if a < 0. The graph has one turning point. It can have 0, 1, or 2 x-intercepts. We have also looked at the graph of some polynomials of degree 3, those that are obtained by shifts, stretching or shrinking and reflections in the x- (or y-) axis from y = x3. The unshifted functions y = ax3 have graphs with tails in opposite directions, one x-intercept (0) and no turning points. They are increasing if a > 0 and decreasing if a < 0. What is the global behavior of an odd-degree polynomial with positive lead coefficient? Globally all odd-degree polynomials with positive lead coefficient have the general shape of y = x3 with the tail down on the left and the tail up on the right. If the lead coefficient is negative, the global behavior of the graph is similar to y = -x3 with the tail up on left and down on right because of reflection. What is the global behavior of an even-degree polynomial? If the leading coefficient is positive, the graph of an even degree polynomial will have both tails up. If the leading coefficient is negative both tails will be down because of reflection. Look at the graphs of f(x) = (x + 2)(x - 3)(x + 5), g(x) = (x + 2)(x - 3)2(x + 5) and h(x) = (x + 2)(x - 3)3(x + 5). Use a table of signs to determine general location of the graphs on intervals between x-intercepts and determine tail direction. What shape do the graphs of f, g and h have near the x-intercepts -2 and 5? What shape do these graphs have near the x-intercept 3? Why is the shape different if the corresponding factor has a exponent of 1? 2? 3? An odd-degree polynomial must have at least one x-intercept because the tails extend indefinitely in opposite directions. An even-degree polynomial does not have to have any x-intercepts (e.g., y = x4 + 1). A turning point of a graph is a point where the graph changes from increasing to decreasing or vice versa. As we see with y = x3 an odd-degree polynomial does not have to have a turning point. The polynomial f(x) = (x + 2)(x -3 )(x + 5) has two turning points and is typical of third-degree polynomials with three x-intercepts (the largest number possible). If the polynomial has even degree n, it must have at least 1 turning point (after all if both tails are in the same direction, then between them must be a point where the graph changes from decreasing to increasing or from increasing to decreasing); it can have as many as n-1 turning points. The number of x-intercepts and the number of turning points of a graph along with the type of tails can help determine the possible degree of the polynomial, particularly the smallest possible degree: count the number of turning points and add 1 for the smallest possible degree. How many turning points ("peaks" and "valleys") can a polynomial have? The largest number will occur with the largest number of x-intercepts (that is not the only time though). A very useful theorem in determining a polynomial that could have a particular graph is the Factor Theorem. Factor Theorem: Suppose f is a polynomial. Then f(a) = 0 if and only if x - a is a factor. A polynomial may have as many linear factors as the degree of the polynomial. This means that the maximum number of x-intercepts is the same as the degree of the polynomial. If we know that a polynomial has exactly the x-intercepts 2, -5, and 6, then the polymonial has the factors (x - 2)k, (x + 5)m and (x - 6)n and no other linear factors (it may have quadratic factors such as x2 + 1 which cannot be factored). The exponent for each factor is determined by the shape of the graph near the corresponding x-intercept. If that piece of the graph looks like a straight line, then the exponent is 1. If the piece of graph near the x-intercept looks like a baby parabola, then the exponent must be even (so the smallest possible exponent would be 2). If the piece of graph near the x-intercept looks like a miniature of the y = ax3 graph, then the exponent must be at least 3 and odd. EXAMPLE. Draw the graph of a fifth-degree polynomial that has x-intercepts of -3, 1, and 5, a y-intercept of 15, f(x) > 0 for x < 1, and f(x) < 0 for 1 < x < 5. Find a function that could have the graph. Steps: 1. Draw the graph, making certain it includes all of the named intercepts and no others. Make certain that the tails have appropriate behavior. In this case the tails of the graph must go in opposite direction with the one on the left going upward. The graph must stay above the x-axis for x < 1 and below the x-axis for x > 1. 2. Determine the factors (one for each of the x-intercepts) and the exponent for each one (one if the graph looks like a line at the intercept; an even integer if the polynomial looks like a parabola at the intercept, 3 or a higher odd integer if the graph looks like a baby cubic) . For this example f must stay above the x-axis for x < 1 and below the x-axis for x > 1 which means that f(x) = a(x +3)2(x - 1)(x - 5)2. Once you have chosen the factors that correspond to the x-intercepts you have pinned the graph to the x-axis at the x-intercepts. You cannot move the graph up or down. To make the graph go through a particular point such as the y-intercept, you must shrink, stretch and/or reflect it in the x-axis. The factor a in the product we used for f is our fudge factor. It will correspond to the stretching or shrinking and reflection that we need for this graph. 3. Use the y-intercept or other point given to find the value of a and/or check the tails of the polynomial. Remember: DO NOT SHIFT THE GRAPH BY ADDING A CONSTANT. In this problem we are told that the polynomial has the y-intercept of 15 so we must have f(0) = 15. Evaluating f(x) = a(x +3)2(x - 1)(x - 5)2 at x = 0 we get f(0) = a(3)2(-1)(5)2 = -225a 15 = -225a a = - 1/15 Therefore the polynomial is f(x) = - (1/15)(x + 3)2(x - 1)(x - 5)2 |
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