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Use the information in the table to answer each question.(a) What are the three real zeros of the polynomial function $f ?$(b) What can be said about the behavior of the graph of $f$ at $x=1 ?$(c) What is the least possible degree of $f ?$ Explain. Can the degree of $f$ ever be odd ? Explain.(d) Is the leading coefficient of $f$ positive or negative? Explain.(e) Write an equation for $f .$ (There are many correct answers.)(f) Sketch a graph of the equation you wrote in part (e).

a) The graph touches the $x$ -axisb) The graph touches the $x$ -axisc) the degree is always even and $\geq 4$d) Positive

01:54

Frank L.

Algebra

Chapter 2

Polynomial and Rational Functions

Section 5

Zeros of Polynomial Functions

Quadratic Functions

Complex Numbers

Polynomials

Rational Functions

Missouri State University

Campbell University

Harvey Mudd College

Idaho State University

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here. We have the information from the table, and we can use that information to make a sketch of the function and that will help us to answer all the questions. So, first of all, when the function value changes from positive to negative or negative deposited, that means it must pass through zero. So that means we have a zero at X equals negative, too. We have a zero at X equals one, and we have a zero at four. And that answers our first question. And at the same time, it helps us starter graph. So according to the table, the function has positive. Why values until we get to negative two. And then it has negative y values until you to one. And then it has negative values again after one. So we should go back and maybe make that turn out a little more smoothly, and then it has positive. Why values after four. So maybe it looks something like this. Okay, so, for part B, what can be said about the behavior of the graph at X equals one? Well, looks like we had a maximum there because it kind of came right back up and then went right back down Part C. What is the least possible degree of F? Well, it looks like a Kordic graph, since it has that W shape. So the degree would be four, but it could be higher than four. It couldn't be quadratic because it's not a parabolas so can't have degree to. We know it's even because both ends go up. So it has to be uneven degree, uh, four or higher. So the degree do not be odd. It has to be even or else if it was odd one and would be up in one and would be down. So we know for party that the leading coefficient is positive because both ends go up. Okay, Now we're writing an equation, and there are many correct answers, So we're just gonna write a sample equation. And so what we're going to do is take all of our zeros X equals negative two x equals one and X equals four and find their linear factors. X plus two next minus one. Next minus four. And multiply those together to give us the polynomial, So x plus two times X minus one times X minus four Let's start by multiplying X plus two and X minus one that we get X squared plus X minus two. And then we're gonna multiply that at my X minus four. I'll start with X squared and multiply that by both terms and we get X cubed minus four X squared. And then I moved in plus X and multiply that by both terms. Now we get X squared minus four x, and then I'll move to minus two and multiply that Bible terms and we get minus two X plus eight. Let's combine Elect terms X cubed minus four groups. Let's combine these the minus four in the plus one give us minus three X squared Combined. These we get minus six X and we have our plus eight. Okay, so there's an equation for after. So what? We might want to modify and sketching the graph. Compared to what we have above is that we would want to have a why intercepted eight. So it still has zeros and negative too. And at one and four lives. Okay. Do you notice what I am missing? Do you notice that we have a court a graph degree four. That was mentioned earlier. Yet I haven't have a cubic equation, so I need to add another factor. And what other factor would I add? Well, let's look at the graphical behavior again. Remember how it just went up and touched X equals one and went back down to wait a maximum there. That means we have a double room at one. We're going to add another factor of X minus one, and I'll just tack it on at the end here. How about right here and multiply by that? Okay, so all of the terms are, uh, excuse gets multiplied by both of the terms. So we get X to the fourth minus excuse, and then the negative three x squared gets multiplied by both of the terms. So we get minus three x cubed plus three x squared. And then the negative six x gets multiplied by both terms. So we get minus six x squared plus six x and then the eight gets multiplied by both terms and we get plus eight x minus eight. So let's combine our like terms the next to the fourth. Let's look at our cubic terms. We have minus four x cubed Let's look at our square terms. We have minus three X squared. Let's look at our linear terms. We have plus 14 X and then we have minus eight. So that should be our function. And that has a Y intercept of negative eight. So that makes a lot more sense with our graph because we knew our graph had to be negative. They're on the value of X equals zero. So all we might want to do to modify this graph is make the Y intercept negative. Ate everything else. I think I would kind of leave alone. So we're going down, coming back up in touching and then going back up something like that.

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