5. Zeros. We deduce that the graph crosses the x-axis at the points x = 0, 1, 2 and 4, and nowhere else. And that s exactly what we see in the graph.

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1 . Zeros Eample 1. At the right we have drawn the graph of the polnomial = ( 1) ( 2) ( 4). Argue that the form of the algebraic formula allows ou to see right awa where the graph is above the -ais, where it is below the -ais and where it crosses the ais. Well the idea here is that since is given as a product of factors, then to tell where it is positive, negative or zero we onl need to have that information for each factor. And since the factors are all linear, this information is eas to get. The crossings Where does the graph cross the -ais? These are the points where =, and that will happen eactl when one of the factors is zero (and not otherwise). Now the first factor is zero onl at =, and the second factor is zero onl at =1, and the third and fourth factors are zero onl at =2 and 4, respectivel What we do in class is give the students onl the formula and ask them to make a rough sketch of the graph. The have then to cope all at once with the sign of as well as where it crosses the ais. We deduce that the graph crosses the -ais at the points =, 1, 2 and 4, and nowhere else. And that s eactl what we see in the graph. The zeros The points where the graph crosses the -ais are the points where = and are called the zeros of the function. Often in mathematical modeling, the problem reduces to one of determining where a certain epression is positive or negative. Most often the zeros hold the ke to this question as these are the places where can change sign.. zeros 1

2 Above and below The graph is above the -ais where is positive and it s below the ais where is negative. Since is given as a product, its sign can be determined from the signs of the factors. Now it can be a tedious business working out the sign of each of the factors at ever point, but there s a neat wa to get around that. Here it is. = ( 1) ( 2) ( 4) Start with a value of that is above 4. Then all the terms are positive and must be positive. So for > 4, the graph is above the ais. Now let decrease. Nothing happens to an of the signs until crosses from above 4 to below it. At this event the last factor changes sign (from positive to negative) but the other factors sta the same sign, so the whole epression changes sign. Consequentl the graph drops below the ais. What happens when is just less than 4 but still greater than 3? Well the first three terms are all still positive, but the last term ( 4) becomes negative. And that makes negative and the graph is below the ais This seemingl simple observation is so powerful, that we spell it out. For just above 4 is positive: = ( 1) ( 2) ( 4) = POS POS POS POS For just below 4 is negative: = ( 1) ( 2) ( 4) = POS POS POS NEG = NEG. What happened as passed through 4 was that the last term ( 4) changed sign. And that made change sign. We can continue in this wa. As decreases from 4 to 2, there are no sign changes, but when passes through 2, the term ( 2) changes sign, (but the other terms don't) so the whole product changes sign again and the graph crosses the ais. That puts it above the ais again. And so forth at both =1 and = we get another sign change a crossing of the ais. The idea here is ver simple. When eactl one factor changes sign, the entire product changes sign.. zeros 2

3 Eample 2. At the right we have drawn the graph of the polnomial = ( 1) 2 ( 2) ( 3) ( 4) 2. Argue that the form of the algebraic formula allows ou to see right awa where the graph is above the -ais, where it is below the -ais and where it crosses the ais Again, is given as a product of terms, and the zeros of the polnomial are eactl the numbers, 1, 2, 3 and 4 that appear in the factors. - Again we give the students onl the formula and ask them to make a rough sketch of the graph. But now there's something new. The graph does not alwas cross the ais at a zero. It crosses the ais at the zeros, 2 and 3, but it does not at the zeros 1 and 4 it just hits the ais and bounces back up. What's the difference? It is this the zeros, 2 and 3 belong to terms of degree 1 in the epression, while the zeros 1 and 4 belong to terms of degree 2. Does this make sense? Let's tr to do the same "sign" analsis as in Eample 1. Start with a value of that is above 4. Then all the terms are positive and the graph is above the ais. Now let decrease. As drops below 4, the factor (-4) becomes negative, but its square does not. So the whole product does not change sign and the graph remains above the ais. This time, the detailed analsis looks like this: But having mastered the signchange ideas of Eample 1, the now have to think "twice" It is no longer true that the entire epression changes sign as passes through 4. Wh not? For just above 4 is positive: = ( 1) 2 ( 2) ( 3) ( 4) 2 = POS POS 2 POS POS POS 2 For just below 4 is still positive: = ( 1) 2 ( 2) ( 3) ( 4) 2 = POS POS 2 POS POS NEG 2 If ou like, at =4 we have two changes of sign happening simultaneousl, for no net change. So the graph stas above the ais. As continues to decrease, we get sign changes at 3 and at 2, but at =1 we have the "square" situation again, and no change of sign.. zeros 3

4 Eample 3. At the right we have drawn the graph of the polnomial = ( 1) 2 ( 3) 3 Argue that the form of the algebraic formula allows ou to see right awa where the graph is above the -ais, where it is below the -ais and where it crosses the ais. The interesting feature here is the cubic term ( 3) 3. For above 3, everthing is positive. Then as drops below 3, the factor ( 3) becomes negative and so does its cube. So the whole product does change sign and the graph crosses the ais. But then at the zero =1, the product does not change sign and the graph stas below the ais. And finall there is a cross at =. In this eample, we have three kinds of zeros of multiplicit, 1, 2 and 3. We get a cross for of multiplicit 1 and 3, but not for multiplicit Note the difference between the crossings at =3 and =. At =3 the graph is ver flat when it crosses the ais, even flatter than it is at =1 (when it fails to cross). It turns out that the higher the power of the term, the flatter is the graph at the zero. That's a subject for calculus to take up. What's the general situation? At a zero of odd multiplicit, the graph crosses the ais, at a zero of even multiplicit, the graph does not. Odd and even multiplicit It would seem that there are two kinds of zeros those at which the graph crosses the ais, and those at which it hits the ais but stas on the same side. And the issue here concerns the algebraic distinction between these two behaviours. What these eamples suggest is that it's determined b the inde of the factor, which is called the "multiplicit" of the zero. At zeros of odd multiplicit, like at = and =3 in the above eample, we have a change of sign. But at even multiplicit, like at =1, we have no change of sign. But there's also an interesting difference between multiplicit 1 and 3. At a zero of multiplicit 1 (like =) we get a real "cross." But at multiplicit 3 (like =3) the graph flattens out and seems unable to decide whether to cross or not.. zeros 4

5 Problems 1. Below we give the equation of a number of polnomials in factored form. In each case make a rough sketch of the graph and eplain the basis of our decisions. We don't want ou to "sub-in" values for and plot points. [That's what computers are good at doing so leave that to them.] Rather we want ou to identif the zeros and observe whether the graph crosses or does not cross the ais at each of these. Without plotting points, ou won't reall know how "high" the graph gets between the zeros, but we don't reall care about that at this stage. It's the form we are after. (a) = ( 1) ( 2) ( 3) ( ) (b) = (+1) ( 1) ( 3) ( 4) 2 (c) = ( 1) 2 ( 2) 2 (d) = ( 1) ( 2) 2 ( 3) ( 4) 3 (e) = 2 ( 2 1) ( 4) 2 (f) = ( 2 ) ( 2 1) (+1) (g) = ( 1) ( 2) 2 ( 3) 3 ( 4) 4 2. In each case find a polnomial of degree 4 that has the given properties, and write it in the form where the leading coefficient a is either 1 or 1. f() = a 4 + b 3 + c 2 + d + e (a) f() = f(1) = f(2) = f( 1) = (b) f() = f(1) = f( 1) = and f() for all. (c) f() = f(2) = and f() > for all or 2. (d) f() = and f() for all and the graph of f is bilaterall smmetric about the line = At the right is the graph of a polnomial of degree 7. Find its equation. Check the vertical scale of our equation b plugging in a couple of suitable values of sa, =. and = zeros