Get ready for AP® Calculus
- Positive and negative intervals of polynomials
- Positive & negative intervals of polynomials
- Multiplicity of zeros of polynomials
- Zeros of polynomials (multiplicity)
- Zeros of polynomials (multiplicity)
- Zeros of polynomials & their graphs
- Positive & negative intervals of polynomials
Positive and negative intervals of polynomials
If we know all the zeros of a polynomial, then we can determine the intervals over which the polynomial is positive and negative. This is because the polynomial has the same sign between consecutive zeros. So all we need to do is check each interval that is between two consecutive zeros (or before the smallest zero and after the largest zero).
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- Think I've got this figured out:
Consecutive Negative Intervals
...if the zero is also a local maximum of the graph, then the graph will go up, with the highest point of that curve just touching the y-axis before going back down. When this happens, there will be two negative intervals in a row: (-)(-).
Consecutive Positive Intervals
...if the zero is also a local minimum of the graph, then the graph will drop down, with the lowest part of that curve just touching the y-axis before rising up again. When this happens, there will be two positive intervals in a row: (+)(+).
...if the zero occurs where the graph completely crosses the y-axis, then you will have alternating intervals:
- if the graph is ascending: (-)(+)
- if the graph is descending: (+)(-)
Does this seem right?(30 votes)
- yes. there is an easy way of finding it; it is called the table of signs.
here is a link to kind of explain it.
- How would you be able calculate at8:37how high or low the points would go (or turning points of the graph) mathematically?(8 votes)
- Other than figuring out each point in between, this is a big part in calculus once you get to it.
One option though is to take the point exactly in between two consecutive zeroes. This will be a local max or min, so you would know that this "arc" of the graph will be symmetric around that max/ min (extrema). I hope that helps answering your question(5 votes)
- this is SOOO hard to understand! can someone help? thanks! :D(5 votes)
- I'm pretty sure what Sal is saying is that the intervals between roots (zeroes) can be positive or negative, and you can find those roots by plugging in x values between your root values. Intervals will alternate between roots. Hope that helped.(8 votes)
- Did anyone else notice the sirens at2:07?(8 votes)
- how did he get the samples did he just randomly pick them?(2 votes)
- Kind of. He had a set of numbers to pick from. So for instance the first was less than -2, so he just needed to pick a number less than -2, so he picked -3. He could have picked any number less than -2 though.
Some numbers are nicer than others when picking, For instance whole numbers are usually nicer than non whole rational numbers, and rational numbers are better than irrational number. 0 in particularly is a good choice when available, which Sal used when he could.
It kinda takes some practice to pick the easiest choices, but sticking to whole numbers or at least rational numbers should make it doable.(4 votes)
- Can someone explain "consecutive zeros"(2 votes)
- A given polynomial will have a set of zeroes, which are inputs that make the polynomial equal 0. If we mark all of the real zeroes of a polynomial on the number line, then two zeroes are consecutive if there are no other zeroes between them.
For example, the polynomial x³-x has zeroes at -1, 0, and 1. -1 and 0 are consecutive, and 0 and 1 are consecutive, but -1 and 1 are not, since 0 is between them.(2 votes)
- Do the signs always have to alternate over intervals? For example, can we have a function that has zeros at a and b, then for the interval x < a it is positive, for a < x < b it is positive and continues to be positive for x > b? (i.e. an upward-opening parabola but forms somewhat of a semicircle where the minimum point typically is at)(2 votes)
- To have a continually positive or negative quadratic would be impossible. When you have a zero, the polynomial must cross the x-axis.
Looking at the interval when x < a and a < x < b as positive — which is possible — the polynomial must go down to hit b; thus, making the interval negative.
An excellent example of this:
f(x) = -(x+2)²(x+1)
Which when graphed on desmos visually shows the explanation.(1 vote)
- this doesn't make the practice section make sense. How can I know what are consecutive zeros and what aren't? If anyone helps me out, explain it like I'm 5 please lol(2 votes)
- Are polynomials always continuous? I suppose the axiom that the sign does not change between consecutive zeroes doesn't hold for, say, rational functions.(1 vote)
- How do we know the interval is less than -2?(1 vote)
- Since there is no root after x = -2, the graph will continue with the same behavior that it has for everything below -2. If there was a root, the graph may bend back up to try and meet it, which would change the shape of the graph and we wouldn't be able to have x < -2 as one interval.(1 vote)
- [Instructor] Let's say that we have the polynomial p of x. And when expressed in factored form, it is x plus two times two x minus three times x minus four. What we're going to do in this video is use our knowledge of the roots of this polynomial to think about intervals where this polynomial would be positive or negative. And the key realization is, is that the sign of a polynomial stays the same between consecutive zeros. Let me just draw an arbitrary graph of a polynomial here to make you appreciate why that is true. So x-axis, y-axis. And if I were to draw some arbitrary polynomial like that, you can see that between consecutive zeros the sign is the same. Between this zero and this zero, the polynomial is positive. Between this zero and this zero, the polynomial is negative. And that's almost intuitively true. Because if the sign did not stay the same, that means you would have to cross the x-axis. So you would have a zero, but we're saying between consecutive zeros. So between this zero and this zero, it is positive again. Then after that zero, it stays negative. Once again, the only way it wouldn't stay negative is if there were another zero. So now let's go back to this example here. And let me delete this because this is not the graph of p of x, which I have just written down. Let's first think about its zeros. So the zeros are the x-values that would either make x plus two equal zero, two x minus three equal zero, or x minus four equal zero. So first, we can think about, well, what x-values would make x plus two, x plus two equal to zero? Well, that, of course, would be x equals negative two. What x-values would make two x minus three equal zero? Two x minus three equal to zero, add three to both sides, you get two x equals three. Divide both sides by two, you get x equals 3/2. And then last but not least, what x-values would make x minus four equal to zero? (distant siren wailing) Add four to both sides, you get x is equal to four. And so if we were to plot this, it would look something like this. So this x equals negative two, (distant siren wailing) x equals negative one. This is zero. This is one, two, three, and four. And let me draw the y-axis here. So the y-axis would look something like this, x and y. We have a zero at x equals negative two, so our graph will intersect the x-axis there. We have a zero at x is equal to 3/2, which is 1 1/2, which is right over there. And we have a zero at x equals hour, which is right over there. And so we have several candidate intervals. And actually, let me write this down in a table. So the intervals over which and this is really, be between consecutive zeros, intervals to consider. So I'll draw a little table here. So you have x is less than negative two. That's one interval. X is less than, actually, let me color-code this. So if I were to say the interval for x is less than negative two, so that's this yellow that I draw at the extreme left there. We could have an interval where x is between negative two and 3/2, so negative two is less than x, is less than positive 3/2. That would be this interval right over here. You have the interval, I'm trying to use all my colors, between 3/2 and four, this interval here. So that would be 3/2 is less than x, is less than four. And then last but not least, you have the interval where x is greater than four, that interval right over there. So x is greater, greater than four. Now, there's a couple of ways of thinking about whether, over that interval, our function is positive or negative. One method is to just evaluate our p, our function at a point in the interval. And if it's positive, well, that's means that that whole interval is positive. It's negative, that means that that whole interval is negative. And once again, it's intuitive because, if for whatever reason, it were to switch, we would have another zero. I know I keep saying that. But another way to think about it is, over that interval, what is the behavior of x plus two, two x minus three, and x minus four? Think about whether they're positive or negative, and use our knowledge of multiplying positives and negatives together to figure out whether we're dealing with a positive or negative. So let's do it, we could do it both ways. So let's think of this as our sample x, sample x-value. And then let's see what we can intuit about or deduce about whether, over that interval, we are positive or negative. So for x is less than negative two, maybe an easy one or an obvious one to use, it could be any value where x is less than negative two, but let's try x is equal to negative three. So you could try to evaluate p of negative three. You could just evaluate that. Actually, let's just do that. So that's going to be equal to negative one times, two times negative three is negative six, minus three is negative nine, negative nine times negative three minus four, that is negative seven. So if you were to multiply all of this out, this would give you negative 63, which is clearly negative. So over this interval right over here, our polynomial is going to be negative. So then we can move on to the next one. An interesting thing is we didn't even have to figure out the 63 part. We can just see that there's a negative times a negative times a negative, which is going to be a negative. And so let's just do that going forward. Let's just think about whether each of these are going to be positive or negative and what would happen when you multiply those positive and negatives together. Now, in this second interval between negative two and 3/2, what is going to happen? Well, we could do a sample point. Let's say x is equal to zero. That might be pretty straightforward. Well, when x is equal to zero, we're going to be dealing with a positive times a negative times a negative, a positive times a negative times a negative. And the reason why I did that is in my head. I said, okay, that's going to be a positive two times a negative three times a negative four. So a positive times a negative times a negative, so I could write it this way. It's going to be a positive times a negative times a negative. Well, a negative times a negative is a positive, and a positive times a positive is a positive. So we are positive over that interval. And if you were to evaluate p of zero, you will get a positive value. Now what about this next interval? What about this next interval here between 3/2 and four? We could try x is equal to two. At when x is equal to two, we are going to get a positive times a positive times a, two minus four is negative, times a negative. So this is going to be negative over that interval. And then last but not least, when x is greater than four, we could try x is equal to let's say five. We are going to have a positive, positive times a positive times a positive. So we are going to have a positive. And as I mentioned, you could also do it without the sample points. You could say, okay, when x is greater than four, you could say, okay, for any x greater than four, if you add two to it, that for sure is going to be positive. For any x greater than four, if you multiply it by two and subtract three, well, that's still going to be positive 'cause two times something greater than four is definitely greater than three. And for any x greater than four, if you subtract four from it, you're still going to have a positive value. So that's another way to think about it, even if you don't use a sample point. But there you have it, we figured out the intervals over which the function is negative or positive. And we don't know exactly what the function looks like, but generally speaking it's negative over this first interval. So it might look something like this. It's positive over that next interval. And then it's negative over that third interval. And then it's positive over that last interval. So we'd have a general shape like this. We don't know, without trying out more points, exactly how high or low it would actually go.