WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In a previous lecture, we introduced the concept of the graphs of polynomial functions.
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Now, we are going to go further and actually talk about how to develop those graphs.
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OK, in order to obtain the graph of a polynomial function, you make a table of values,
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connect these points with a curve, and then use information about the end behavior of the function.
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Earlier on, when you had to graph linear and quadratic functions, you used the technique of making a table.
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So, it is the same idea here, but with the additional concept of thinking about end behavior.
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For example, if you were to graph a function such as f(x) = 3x³ - 2x² + x,
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you would approach this by finding some x and f(x) values and graphing those out.
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Then, look at the areas of the function that tell you about end behavior.
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I am not going to go ahead and make the table on this one.
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But let's say that you were to graph it out, and it came out something like this.
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OK, using points, and also finding the zeroes, can help you graph.
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In addition to just finding various points, sometimes you can approach making a graph of a polynomial function by also finding the zeroes
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(and recall that the zeroes are where the graph crosses the x-axis, so they are the x-intercepts), and then finding some other points.
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And then, you think about what happens to the graph at the ends.
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Well, recall that, if you have an odd-degree polynomial, such as this one, the ends are going to go in different directions.
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So, one will increase; the other will decrease; and then you have to look at the leading coefficient.
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The leading coefficient: if it is positive (if it is greater than 0), as x increases, y will also increase.
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So, this is odd degree with a positive leading coefficient: x and y are both increasing, and as x gets very, very small,
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y is going to get very, very small, as well; so this is odd degree, where a of n, the leading coefficient, is greater than 0.
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If you have an odd degree where the leading coefficient is negative, then the graph is going to pretty much be the opposite,
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where you are going to have y decreasing (f(x) decreasing) as x gets very large.
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And when x gets very small, f(x) is going to increase; this is odd degree with a negative value for the leading coefficient.
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Let's look at even degrees right here.
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If you are talking about a function with an even degree, such as the largest exponent of the fourth power or the sixth power,
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so it's an even degree, both ends go in the same direction--they both face up, or they both face down.
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And the leading coefficient tells you which direction that is.
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With a leading coefficient that is positive, you are going to get a graph where both ends are facing up.
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So, as x gets very large, or as x gets very small, y is going to get very large; this is even with a leading coefficient that is positive.
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Now, if you have an even degree where the leading coefficient is negative, then both ends will face down.
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So, this could be an example of even with a negative leading coefficient.
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This is review from a previous lecture; but recall that, in addition to finding what is going on
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here in the middle of the graph, and finding your zeroes--finding various points--
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you also need to know about end behavior in order to develop a graph of a polynomial function.
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OK, I mentioned talking about finding the zeroes.
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Well, sometimes it can be difficult or time-consuming to find the exact location of a zero.
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But if you are just trying to sketch out a graph, you can use the **location principle**
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to estimate where a zero will be--where a graph is going to cross the x-axis.
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So, let's look at what this is saying: this is saying that, if you have a polynomial function f(x),
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suppose that there are numbers a and b, such that f(a) < 0 and f(b) > 0.
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Then, there is a zero, f(x), between these two points, a and b.
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So, let's make this concrete: let's say that this is my point a--that I am going to let a equal 1.
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And this is saying that if f(a) is less than 0 (I find f(a)--let's say it is somewhere down here: f(a) = -3), a is at 1, and f(a) is -3;
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at (1,-3), that is (a,f(a)) on the graph; there is a negative value to the function.
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OK, now let's say I find another value b, and let b equal 3.
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So, here I had a equal 1, and let b equal 3; f(a) = -3; let's let f(b) equal 4.
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Now, what this is telling me is that, somewhere between this point and this point,
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we went from a negative value of the function to a positive value of the function.
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That tells me that there has to be a zero somewhere...actually, it could even be over here; but somewhere between 1 and 3, this is going to cross.
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So, I don't know that it is exactly here; but I know that somewhere between 1 and 3, there is a zero,
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at an x-value somewhere between x = 1 and x = 3.
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And I know that because the function changed from a negative value to a positive value.
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Now, there actually could be more than one zero between those points.
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For example, I could have a graph right here that goes like this.
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And if I just graphed this point out here, I would see that the function is negative at this point, at -4: at f(-4), I have a negative value.
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And then I say, "OK, let's say over here, at f(-2), I have a positive value."
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I know that there is a zero in here; but as it turns out, there is actually more than one zero.
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I know that there is at least one zero here, but there also could be more than one.
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So, when you go ahead and make your table of values, x and f(x), if you see that you have,
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say, positive, positive, and then it goes to negative, I know that at an x-value somewhere between here and here, there is a zero.
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And then, I might have negative and negative, and then switch to positive again.
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And I know that, in here somewhere, there is a zero.
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So again, the idea is that, if you see the value of the function switch from positive to negative,
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or negative to positive, you know that the graph has to pass through the x-axis, and that there is a zero in that location.
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OK, maximum and minimum points: recall that, when we worked with quadratic functions, we talked about maximums and minimums.
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And here, today, with polynomials, we are talking about relative maximums.
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When we talked about quadratics that were downward-facing parabolas, we said, "OK, the vertex is a maximum."
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We didn't say relative; we just said maximum, because this is the highest value that the function could achieve.
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If we were dealing with a minimum value, we didn't say relative; we just said, "OK, this is the minimum,"
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because this is the smallest value that the function could achieve.
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However, when we talk about more complicated, or higher-degree, polynomials, we talk about **relative maximums and minimums**.
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And what we mean is relative to the x-values around those.
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For example, if I have a graph that looks like this, I could say that this right here--this point (we will call that a,
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and then, right here, the y-value is going to be f(a)--f(a) would be right there,
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so this point is going to be (a,f(a)))--this is a relative maximum.
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And the reason it is a relative maximum is: if I look at any value of x around in this region, the function value is lower than it is right here.
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So, it is the maximum, relative to the values of x near this point.
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A relative minimum is the same idea: if I look at this point here, this gives me a relative minimum.
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And that means that this value of the function (I will call this f(b), so this would be b, and then the y-value would be f(b),
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so this is at (b,f(b)))--this is the smallest value for the function in that region, for these x's.
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But you see here: the graph actually goes lower way over here.
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So, it is not an absolute minimum; it is just relative to that region of the graph.
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And there might be a maximum that is higher up at some area in the graph; but this is a relative maximum for that region.
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The other important point is that the graph of a polynomial function of a certain degree,
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degree n, can have at most n - 1 points that are relative maximums or relative minimums.
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So, if I have a function--let's say f(x) = 4x⁵ + 3x³ - 2x + 6,
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then here, the degree, n is 5 in this case, so the maximum number of relative maximums and relative minimums is 5 - 1, which equals 4.
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So, in a graph like this, degree 5, I can have at most 4 of these maximums and minimums.
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OK, putting together what we have discussed about finding points, the zeroes, using end behavior,
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and relative maximums and minimums, we can develop a graph of some polynomials.
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OK, looking at this first one, this is x³ - 2x; and I am just going to start this one by plotting out some points.
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OK, so let's let x equal -2; this would equal -8; and then, -2 times -2 is 4; -8 + 4 is going to give me -4.
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Here I have -1 for my x-value, and that is going to give me a -1 here, and then -1 times -2 is going to give me 2, so this will be 1.
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Now, notice something: my value for the function switched from negative to positive.
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That means that somewhere between -2 and -1, there is a zero.
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Using the location principle, I know that, somewhere between x = -2 and x = -1, this graph of this function crosses the x-axis.
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When x is 0, f(x) is 0; well, if f(x) is 0, then this point lies on the x-axis, and I actually found a zero--this is a zero.
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Now, when x is 1, that gives me 1, minus 2; that is -1.
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When x is 2, 2³ is 8, minus 2 times 2 (that is minus 4)--that is going to give me 4.
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Oh, and notice also, here, another zero.
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There is a zero between these values (between the corresponding x-values); somewhere between 1 and 2, there is a zero.
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I have a zero between -2 and -1; I know I have a zero right here; and then, I have another zero between 1 and 2.
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So, I am going to go ahead and plot these out; when x is -2, the function is -4.
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When x is -1, the function is 1; when x is 0, the value of the function is 0;
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when x is 1, we have a -1 right here; and when x is 2, the function value is 2.
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I can see, between here and here: this had to cross the x-axis in between here and here.
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Now, I connect these with a smooth line; I found my points; I know that there is a zero
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somewhere in here; that there is one here; and that there is one here.
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Now, I am going to think about end behavior; and the degree equals 3, and that is odd.
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The leading coefficient, a of n, is 1, so that is positive.
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And what that tells me is: it confirms that my graph is correct, because if it is an odd degree, the two ends are going to face in different directions.
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And since it is a positive leading coefficient, that means that, as x gets very large, y gets very large.
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As x gets very small, y gets very small.
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The other thing I know is that the number of relative maximums and relative minimums is going to be one less than the degree of the polynomial.
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And since the degree is 3, I can have, at most, 2 relative maximums and relative minimums.
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And I see right here: I do have a relative maximum--this is the largest function value for the x-values in this region.
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And right here, I have a relative minimum; so I found both relative maximums and minimums.
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So, this is a pretty good graph of this polynomial, just based on using knowledge that I had and finding a few points on the graph.
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In the second example, again, we have a polynomial of degree 3; but notice that it has a negative leading coefficient.
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That is going to give the graph a different shape.
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In the previous example, I just went ahead and graphed some points; I am going to take a slightly different approach here.
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And I am going to find the zeroes; last time, I found one of the zeroes by luck at point (0,0),
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and I also used the location principle to find the approximate location of two more zeroes.
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Here, I am just going to go ahead and find them directly.
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So, what I am going to do is look at the corresponding equation, -x³ + x² + 6x.
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And I am going to set that equal to 0 to find values of x for which the function's value is 0; that will give me the zeroes.
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So, I am going to approach this by factoring; and I see that I have a common factor of x in all three terms.
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And I am actually going to factor out -x, so what I have left behind is simpler to factor.
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If I pull a -x here, I am going to have an x² left.
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If I pull a -x from x², that will leave me -x, because -x times -x is positive x².
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Here, I am also going to pull out a negative x, leaving behind -6, and checking that -x and -6 is + 6x.
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OK, now I can look here and see that I can factor this further.
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So, I know that this is going to have the general form (x + something) (x - something).
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And I know that, because I have a negative sign here; so a positive and a negative is going to give me a negative.
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Factors of 6 are 1 and 6, and 2 and 3; and I need those factors when one is positive and one is negative.
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I am going to make one positive and one negative; I want them to add up to the middle term of -1.
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Well, these two are too far apart; now, if I take -3 + 2, that is going to equal -1.
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So, I know that this is the right set of factors, and that I want 3 to be negative and 2 to be positive.
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If I use FOIL to check this, I get x² - 3x + 2x (is going to give me -x) + 2 times -3 (is going to give me -6).
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So, this is factored out as far as it can go.
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And then, recall the zero product property that says that, if, say, a times b equals 0, then either a equals 0 or b equals 0, or they both equal 0.
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So, any of these factors could equal 0, and that would give me a total value of 0.
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And don't forget about this -x; that is part of it; x + 2 could equal 0, or x - 3 could equal 0, to make this equation true.
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Dividing both sides by -1 just gives me x = 0; here, subtracting 2 from both sides gives me x = -2.
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And adding 3 to both sides, I get x = 3.
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These are the zeroes; these are the points at which the graph crosses the x-intercept.
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So, x equals 0; x equals -2; and x equals 3.
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OK, now, to flesh out this graph some more, I am going to find some other points.
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I found three points--I found my zeroes; now I am just going to find some other points on the graph in this region.
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When x is -1, if you work this out, it comes out to f(x) is -4.
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When x is 1, this will give you -1 + 1 + 6, so that would give you 6.
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When x is 2, working this out, it will give you f(x) is 8.
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All right, so when x is -1, f(x) is 4; when x is 1, f(x) is 6 (about up here).
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And notice here that this tells me that there is a zero in here--that between x is -1 and x is 1, there is a zero right in here.
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And I defined that zero between x is -1 and x is 1; there is a zero that lies in here.
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Let's see, when x is 2, f(x) is way up here; it is going to be way up here somewhere--it is at 8, so we will just put it right there.
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All right--oh, and when x is...actually, I wrote that incorrectly; when x is -1, this is going to be -4.
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And that is why, when you get the switch from negative to positive, there is a zero right in here (in the switch from here to here).
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So, I have some points graphed, and I have my zeroes; so I am going to go ahead and connect these points.
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I am also going to use end behavior; and since this is an odd degree, one end is going to go up, and the other is going to go down.
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And since it is a leading coefficient that is negative, then I am going to have--as x becomes very large, y would be very small;
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and as x becomes very small, over here, y becomes very large.
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So, end behavior helped me figure out these portions of the graph.
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Also, I know that, because my degree is 3, the greatest number of relative maximums and minimums I can have is 2.
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2 relative maximums and minimums at most--that is the greatest number I can have.
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And I do have a relative minimum here and a relative maximum there.
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So, graphing is based on finding the zeroes, graphing a few points, and then thinking about end behavior.
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And notice, also, that here, when my graph switched from a value of -4 (a negative value for the function)
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to a positive value, up here at 6, I knew that there had to be a zero in between those; and I did find that zero.
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OK, I am going to approach this, also, by first finding the zeroes.
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Let's work with the corresponding equation and set this equal to 0.
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Now, looking at this, there are no common factors to pull out; and I have four terms, so I am going to factor by grouping.
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Grouping means that we are going to put x³ - x² together, and add that to -3x + 3.
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Now, looking at this, I have a common factor of x²; that leaves behind an x and a -1.
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Here, I have a common factor of -3; I pull that out--that leaves an x behind here, and a -1, because -3 and -1 would give me the 3 back.
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Now, I see that I have a common binomial factor of x - 1; I am going to pull that out in front.
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When that is pulled out, that leaves behind x² - 3.
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OK, using the zero product property, I know that x - 1 = 0 or x² - 3 = 0.
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If either of these expressions is 0, then the equation will hold true; this left side will be equal to 0.
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So, this is easy: just add 1 to both sides--that gives me x = 1; and then, over here,
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I am going to add 3 to both sides, and that is going to give me x² = 3.
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I am going to take the square root of both sides, and that is going to give me x = ±√3.
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OK, so I have zeroes at x = 1, x = √3, and x = -√3.
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Well, the square root of 3 can be estimated at about 1.7; so let's rewrite this as 1.7 and -1.7, so that we can graph it.
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OK, I have a zero here at 1; I have a zero here at 1.7 (which is about there) and at -1.7 (which is right about there).
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All right, again, I need to just find a few more points to get a complete graph.
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I am going to let x equal -2; and if I plug that into here, I will get out a value of a function that is -3.
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I am going to let x equal -1, and if you do the calculation on that and substitute -1 in here, you are going to find that y is 4.
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Plotting that out: when x is -2, f(x), or y, is -3, right here.
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When x is -1, f(x) is 4; and you see the location principle at work--that, since we switched from negative to positive,
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that means that there is a zero somewhere in here, somewhere between x is -2 and x is -1.
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And we already know that there is a zero between those two, and it is at -1.7.
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OK, finding a few more points: let's let x equal 0; f(x) is going to be 3.
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When x is 2, plugging that in here, we get a value for the function of 1.
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So, when x is 0, f(x) is 3; when x is 2, f(x) is 1.
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So now, we have enough points to get some sense of what is happening.
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And I am going to start here and connect these.
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Now, using end behavior, I have an odd degree (this is degree 3), which means the two ends are going to go in opposite directions.
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And I have a positive leading coefficient, meaning that, over here on the right, as x gets very large, y is going to get large.
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Over here on the left, as x gets very small, f(x), or y, is going to get very small.
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I am also going to look and see that my degree is 3; so I have, at most, 2 relative maximums and relative minimums.
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And I have a relative maximum here, and I have a relative minimum here; so I have them both.
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So again, I am handling this by finding the zeroes to get a few points on the x-axis,
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finding some more points to find the shape of the rest of this area of the graph,
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and using end behavior to predict how the graph will look at large and small values.
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So, let's take this and set it equal to 0.
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All right, as far as factoring this, this is a trinomial; and we are used to working with these when they are quadratic equations, and the first term is x².
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But it is really not that much different with x⁴, because x² times x² is x⁴.
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So, I am looking here, and I see that I have a positive term here, and a negative here.
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And the only way to get that is if I am multiplying a negative times a negative; that gives me a positive.
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And yet, when I sum up the outer and inner terms after multiplying, I will get a negative term here.
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Factors of 8 are 1 and 8, and 2 and 4; and I need them to add up to 6.
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So, if I added -2 and -4, I am going to get -6; so I know that these are the factors that I want to use.
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OK, I can't do anything more with this, so I am just going to leave this as it is.
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However, this is the difference of two squares, so I can factor that out a bit more to this.
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OK, so using the zero product property: x² - 2 = 0 or x - 2 = 0 or x + 2 = 0.
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So, starting with these easier ones: x = 2, x = -2; I found two zeroes.
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Now, I am going to look over here: x² = 2--taking the square root of both sides gives me x = ±√2.
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All right, so I have four zeroes; I have zeroes at x = 2, -2, and then √2 is approximately 1.4...
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so let's rewrite that as 1.4 and -1.4 to help me with the graph.
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So, plotting those out: I have a zero here, at -2, and then another one at 2.
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I have a zero here at 1.4 and -1.4.
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I need a few more points to complete my graph; so right up here, let's find x and f(x).
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When x is -1, if you work this out, you get 1; you get -6, because this x² would just be 1, plus 8; so it is going to give you 9 - 6, or 3.
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When x is 0, that cancels; that cancels; that becomes 0; that leaves me with 8 right here.
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And then finally, when x is 1, this will be 1, minus 6, plus 8; so again, I will have 3.
00:31:22.400 --> 00:31:45.000
All right, so when x is -1, f(x) is 3; when x is 0, f(x) is way up here, about here, at 8 (8 would be right about here); and then, when x is 1, f(x) is 3.
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I am going to connect these points; and I am also going to use my knowledge of end behavior.
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And since this is an even degree, both ends are either going to point up or down.
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And since this is a positive leading coefficient, both ends will actually be up.
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So, when x is very small, y is going to be large; and then, I know that after this point, I am going to go up,
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because when x is very large, y is going to be very large.
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After this point, I am going to continue on up, because when x is small, f(x) is going to be very large,
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because it is an even degree with a positive leading coefficient.
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Now, I am checking that I have degree 4; since this is degree 4, the greatest number of relative maximums and minimums I am going to have are 3.
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And I have a relative minimum there; I have a relative maximum here; and I have a relative minimum here.
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So, I found all three relative maximums and minimums.
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So again, finding the zeroes, I found 1, 2, 3, 4 zeroes.
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Then, I plotted a few other points to give me the shape of the graph.
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I used my knowledge of end behavior to figure out what is going on out here and here.
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And then, I just verified that I had the generally correct shape by seeing that I would expect, at most, three relative maximums and minimums.
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And I found that I had all three.
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That concludes this session of Educator.com on analyzing the graphs of polynomials.
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And I will see you again next lesson!