When you're graphing (or looking at a graph of) polynomials, it can help to already have an idea of what basic polynomial shapes look like. One of the aspects of this is "end behavior", and it's pretty easy. Look at these graphs:
As you can see, even-degree polynomials are either "up" on both ends or "down" on both ends, depending on whether the polynomial has, respectively, a positive or negative leading coefficient. On the other hand, odd-degree polynomials have ends that head off in opposite directions. If they start "down" and go "up", they're positive polynomials; if they start "up" and go "down", they're negative polynomials. All even-degree polynomials behave, on their ends, like quadratics, and all odd-degree polynomials behave, on their ends, like cubics.
Which of the following could be the graph of a polynomial whose leading term is "-3x4"?
The important things to consider are the sign and the degree of the leading term. The exponent says that this is a degree-4 polynomial, so the graph will behave roughly like a quadratic: up on both ends or down on both ends. Since the sign on the leading coefficient is negative, the graph will be down on both ends. (The actual value of the negative coefficient is irrelevant for this problem. All I need is the "minus" part of the leading coefficient.)
The only graph with both ends down is Graph B.
Describe the end behavior of f(x) = 3x7 + 5x + 1004
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The real (that is, non-complex) zeroes of a polynomial correspond to the x-intercepts of the graph of that polynomial. So you can find the number of real zeroes of a polynomial by looking at the graph, and conversely you can tell how many times the graph is going to touch or cross the x-axis by looking at the zeroes of the polynomial (or the factored form of the polynomial). A zero has a "multiplicity", which refers to the number of times that its associated factor appears in the polynomial. For instance, the quadratic (x + 3)(x - 2) has the zeroes x = -3 and x = 2, each occurring once. The eleventh-degree polynomial (x + 3)4(x - 2)7 has the same zeroes, but in this case, x = -3 has multiplicity 4 and x = 2 has multiplicity 7, because of the number of times their factors occur.
The point of multiplicities with respect to graphing is that any factors that occur an even number of times (twice, four times, six times, etc) are squares, so they don't change sign. Squares are always positive. This means that the x-intercept corresponding to an even-multiplicity zero can't cross the x-axis, because the zero can't cause the graph to change sign from positive (above the x-axis) to negative (below the x-axis), or vice versa. The practical upshot is that an even-multiplicity zero makes the graph just barely touch the x-axis, and then turns it back around the way it came. You can see this in the following graphs:
All four graphs have the same zeroes, at x = -6 and at x = 7, but the multiplicity of the zero determines whether the graph crosses at that zero or turns back the way it came.
The following graph shows an eighth-degree polynomial. List the polynomial's zeroes with their multiplicities.
I can see from the graph that there are zeroes at x = -15, x = -10, x = -5, x = 0, x = 10, and x = 15, because the graph touches or crosses the x-axis at these points. (At least, I'm assuming that the graph crosses at exactly these points, since the problem doesn't tell me. When I'm guessing from a picture, I do have to make certain assumptions.) Since the graph just touches at x = -10 and x = 10, these zeroes occur an even number of times; the other zeroes occur an odd number of times. The odd-multiplicity zeroes might occur only once, or might occur three, five, or more times each; there is no way to tell from the graph. And the even-multiplicity zeroes might occur four, six, or more times each; I can't tell by looking.
But if I add up the minimum multiplicity of each, I should end up with the degree, because otherwise this problem is asking for more information than is available for me to give. I've got the four odd-multiplicity zeroes (at x = -15, x = -5, x = 0, and x = 15) and the two even-multiplicity zeroes (at x = -10 and x = 10). Adding up their minimum multiplicities, I get 1 + 2 + 1 + 1 + 2 + 1 = 8, which is the degree of the polynomial. So the minimum multiplicities are the correct multiplicities.
x = -15 with multiplicity
x = -10 with multiplicity 2,
x = -5 with multiplicity 1,
x = 0 with multiplicity 1,
x = 10 with multiplicity 2, and
x = 15 with multiplicity 1
I was able to compute the multiplicities of the zeroes in part from the fact that the multiplicities will add up to the degree of the polynomial, or two less, or four less, etc, depending on how many complex zeroes there might be. But multiplicity problems don't usually get into complex numbers.
It isn't standard terminology, and you'll learn the proper terms when you get to calculus, but I refer to the "turnings" of a polynomial graph as its "bumps".
For instance, the following graph has three bumps, as indicated by the arrows:
Compare the numbers of bumps in the graphs below to the degrees of their polynomials:
You can see from these graphs that, for degree n, the graph will have, at most, n - 1 bumps. The bumps represent the spots where the graph turns back on itself and heads back the way it came. This change of direction often happens because of the polynomial's zeroes or factors. But extra pairs of factors don't show up in the graph as much more than just a little extra flexing or flattening in the graph.
Because pairs of factors have this habit of disappearing from the graph (or hiding as a little bit of extra flexure or flattening), the graph may have two fewer, or four fewer, or six fewer, etc, bumps than you might otherwise expect, or it may have flex points instead of some of the bumps. That is, the degree of the polynomial gives you the upper limit (the ceiling) on the number of bumps possible for the graph, and the number of bumps gives you the lower limit (the floor) on degree of the polynomial.
What is the minimum possible degree of the polynomial graphed below?
Since there are four bumps on the graph, and since the end-behavior says that this is an odd-degree polynomial, then the degree of the polynomial is 5, or 7, or 9, or... But:
The minimum possible degree is 5.
Given that a polynomial is of degree six, which of the following could be its graph?
To answer this question, I have to remember that the polynomial's degree gives me the ceiling on the number of bumps. In this case, the degree is 6, so the highest number of bumps the graph could have would be 6 - 1 = 5. (I would add if I were going from the graph to the polynomial, but here I'm going from the polynomial to the graph, so I subtract.) Also, I'll want to check the zeroes (and their multiplicities) to see if they give me any additional information.
Graph A: This shows one bump (so not too many), but only two zeroes, each looking like a multiplicity-1 zero. This is probably just a quadratic, but it might possibly be a sixth-degree polynomial.
Graph B: This has seven bumps, which is too many; this is a polynomial of degree at least 8.
Graph C: This has three bumps (so not too many), it's an even-degree polynomial (being "up" on both ends), and the zero in the middle is an even-multiplicity zero. Also, the bump in the middle looks flattened, so this is probably a zero of multiplicity 4 or more. With the two other zeroes looking like multiplicity-1 zeroes, this is a likely graph for a sixth-degree polynomial.
Graph D: This has six bumps, which is too many. On top of that, this is an odd-degree graph, since the ends head off in opposite directions. This can't be a sixth-degree polynomial.
Graph E: From the end-behavior, I can tell that this graph is from an even-degree polynomial. The one bump is fairly flat, so this is probably more than just a quadratic. This might be a sixth-degree polynomial.
Graph F: This is an even-degree polynomial, and it has five bumps (and a flex point). But looking at the zeroes, I've got an even-multiplicity zero, a zero that looks like multiplicity-1, a zero that looks like at least a multiplicity-3, and another even-multiplicity zero. That gives me a minimum of 2 + 1 + 3 + 2 = 8 zeroes, which is too many for a degree-six polynomial. The bumps were right, but the zeroes were wrong. This can't be a degree-six graph.
Graph G: This is another odd-degree graph.
Graph H: This is an even-degree graph, and there aren't too many bumps, seeing as there's only the one. Looking at the two zeroes, they both look like at least multiplicity-3 zeroes. So this could very well be a degree-six polynomial.
Graphs B, D, F, and G can't
possibly be graphs of degree-six polynomials.
Graphs A and E might be, and Graphs C and H probably are.
To help you keep straight when to add and when to subtract, remember your graphs of quadratics and cubics. Quadratics are degree-two polynomials and have one bump (always); cubics are degree-three polynomials and have two bumps or none (with a flex point instead). So going from your polynomial to your graph, you subtract, and going from your graph to your polynomial, you add. If you know your quadratics and cubics very well, and if you remember that you're dealing with families of polynomials and their family characteristics, you shouldn't have any trouble with this sort of problem.
Once you know the basic behavior of polynomial graphs, you can quickly sketch rough graphs, if required. This can save you the trouble of trying to plot a zillion points for a degree-seven polynomial, for instance. Once the graph starts heading off to infinity, you know that the graph is going to keep going, so you can just draw the line heading off the top or bottom of the graph; you don't need to plot a bunch of actual points.
Without plotting any points other than intercepts, draw a graph of the following polynomial:
This polynomial has already been put into factored form, which saves me the trouble of doing the solving for the zeroes. I'll just solve the factors, noting the multiplicities as I go. The zeroes will be:
x = - 5, with multiplicity 2 (so the graph will be
just touching the x-axis here)
x = -1, with multiplicity 1 (so the graph will be crossing the axis here)
x = 4, with multiplicity 3 (so the graph will be crossing the axis here, but also flexing)
x = 7, with multiplicity 1 (so the graph will be just crossing the axis here)
Also, adding the degrees of the factors, I see that this is a polynomial of degree seven (that is, an odd degree), so the ends will head off in opposite directions. Because the leading coefficient is negative, the left-hand end will be "up" (coming down from the top of the graph) and the right-hand end will go "down" (heading off the bottom of the graph). So I can start my graph by penciling in the zeroes, the behavior near the zeroes, and the ends, like this:
If I multiplied this polynomial out (and I'm not going to, so don't hold your breath), the constant terms of the factors would give me 5 × 5 × 1 × (-4) × (-4) × (-4) × (-7) = 11,200, which is rather large. This would explain the large denominator of the leading coefficient: by dividing the polynomial by a sufficiently-large number, they made this polynomial graphable. Otherwise, the graph would likely go off the picture between the zeroes. (Not all texts notice this, so don't worry about this consideration if it doesn't come up in class.) When x = 0, I get the y-value (-1/5,600)(11,200) = -2, so I can pencil this in, too:
I'm not supposed to find other plot points, so I'll just sketch in a rough guess as to what the graph looks like. I'll go further from the axis where there is more space between the zeroes, and I won't be so primitive as to assume that the y-intercept point is the minimum point, what with the midpoint between the two nearest zeroes being at x = 1.5. Granted, the flexy zero at x = 4 will push the graph a little to the left, but the bump is still probably to the right of the y-axis.
So I'll "rough in" an approximate drawing, and then draw my final answer as a heavier line, erasing my preliminary sketch-marks before I hand in my solution.
This compares favorably with the actual graph of the polynomial. Some of my details (like my max and min points) were a little off, but the general graph was pretty good.
The nice thing about this last example is that you were given a polynomial in factored form. If you are asked to graph a polynomial that is not already in factored form, your best bet is to factor it first, and then graph it the same way we graphed this last example.
Grading for this Lesson:
To get a 10: All answers are correct the first time, or within first revision.
To get a 9: You can have 1 incorrect answer after your original submission.
To get an 8: You can have 2 incorrect answers after your original submission.
To get a 7: You can have 3 incorrect answers after your original submission.
To get a 6: You can have 4 incorrect answers after your original submission.
To get a 5: Cheating - Plagiarism - purposeful or mistaken, which will lower your final grade for the course (so be very careful when posting your work!); lack of effort, disrespect, or attitude (we are here to communicate with you if you don't understand something);
Note: For this class it is necessary to post the questions over each answer. Failure to do so will result in asking for a revision. No grade will be given for incomplete work.
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