use-the-information-in-the-table-to-answer-each-question-a-what-are-the-three-real-zeros-of-the-po-2

Question

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).

Heather Zimmers · Accepted Answer

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&#x27;t be quadratic because it&#x27;s not a parabolas so can&#x27;t have degree to. We know it&#x27;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&#x27;re writing an equation, and there are many correct answers, So we&#x27;re just gonna write a sample equation. And so what we&#x27;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&#x27;s start by multiplying X plus two and X minus one that we get X squared plus X minus two. And then we&#x27;re gonna multiply that at my X minus four. I&#x27;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&#x27;ll move to minus two and multiply that Bible terms and we get minus two X plus eight. Let&#x27;s combine Elect terms X cubed minus four groups. Let&#x27;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&#x27;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&#x27;t have a cubic equation, so I need to add another factor. And what other factor would I add? Well, let&#x27;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&#x27;re going to add another factor of X minus one, and I&#x27;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&#x27;s combine our like terms the next to the fourth. Let&#x27;s look at our cubic terms. We have minus four x cubed Let&#x27;s look at our square terms. We have minus three X squared. Let&#x27;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&#x27;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&#x27;re going down, coming back up in touching and then going back up something like that.

Christina Shute · Answer

But this problem, the first thing one identified is it degree, even are off based off the ground, missing or in behaviors or the same, which tells us that the degree is even then, the next thing we want to do is identify our cereals. That&#x27;s where does the graph touch or across the X axis? So we have a zero at X equals negative three. X equals negative one and then X equals positive. Three. Not only this, we need to know their multiplicity. Four. Multiplicity. We&#x27;re just looking at whether or not it&#x27;s even or odd. So if the multiplicity is even then the breath will bounce off of the X axis. But if it&#x27;s odd the graph, we&#x27;ll straight across the exactness. So with that in mind looking that identify if the multiple city and region B zero says even someone hand negative three. The graph crosses so it hasn&#x27;t multiple city Let&#x27;s pod negative one. You get nearly balance is telling us that the most blessed is eaten. I&#x27;m gonna get a positive three. It crosses said that multiple cities off. Now the next step is identifying the minimum possible to really so again we&#x27;re going back to the degree this time we&#x27;re gonna look at are turning points because they&#x27;re related to a degree. So say we have a degree off in that means we have in minus one turning points. The good news is this goes both weights. So if we have in lice one turning points, then you have a degree of it. So going to our graph we can count are turning points where the graph changes direction. We have one here to and three telling us we have a degree of at least four. Next, we want to write this any factor for So you start out with why equals and then we have each of our zeros remember standard forms and have X minus c to the power of him. So am is the multiple city see is the zero subtraction is our operation. We have y equals X plus three because the zero is negative three. And that&#x27;s an odd multiplicity. So it&#x27;s gonna leave that, as is to the power of one multiplied Why X plus one again? Because zero is negative. One of the operation is in subtraction, with an even multiplicity gonna square that last night we have X minus three with odd multiplicity leaving that as one does, quick checking that if we were the most by this out. So we have 23121 resulting in a degree of at least four. So that&#x27;s our factor. One last but not least, we need to identify our domain and our range. So our domain is the range of X that the graph covers. Looking at how the breath will continue up and out, all real numbers on the X axis be addressed in this function. Looking at the y right and which is the range itself in the positive direction will go to insanity. But in this native direction, we only go down to about negative. So you say negative nine inclusive, which is what that squared off brackets for. 11 positive instantly. And that&#x27;s all the information that we need. It

Christina Shute · Answer

for this problem we are obtaining of set of information based off of the graph. I have a rough sketch of it here that I&#x27;m gonna use to work through this information. First thing we need to determine is whether or not the degree is even or odd. To do this, we can look at the end behaviors so we can see it&#x27;s up on the right and down on the left, meaning they&#x27;re in opposite directions of each other, telling us that the degree is opt. Next thing would identify, then is the zeros of the function. That is where the graph touches or crosses the X axis. So working from negative to positive, we heard X equals negative three. X equals negative one and X equals positive too. Not only this, we&#x27;re gonna identify whether the multiplicity of each of visas otter Even so, the multiplicity. It&#x27;s odd if the graph crosses the X axis at that point, it&#x27;s even. It&#x27;s the graph bounces off of the X axis at that point. So going back to our zeros Negative three. You see that the graph crosses there, Put multiple city is no one in your bounces rendering the multiplicity even And that positive two It also bounces. So that&#x27;s also next up is to identify the minimum possible Different Where we do this is through the turning points of the breast. We have this relationship to say Okay, if I have a degree in that means have in minus one turning points. The thing is that also wants in the other direction. If I have in minus one turning point settings I have a degree of at least in so again looking at the ground turning points where the breath changes direction. So working positively of one to three and four, meaning we have a minimum possible degree A lot. Next one right arm equation in factor form. I was gonna help us do that. Do this is remembering that standard form here is in the X minus C raised to the power of in where m is the multiplicity, See is the zero subtraction Is the operation with that in mind? Begin. So why equals and then that sparks parentheses X plus three again cause the zero is a negative three. Operation subtraction, and it&#x27;s an odd multiple cities. We&#x27;re gonna go ahead and assume that&#x27;s one multiplied by X plus one. That&#x27;s it. Even multiple cities go ahead and assume it&#x27;s too. Lastly, X minus two race of power to again because that&#x27;s even so. Double checking our experiments with this minimum possible degree of five. So to that&#x27;s 21 was far. That&#x27;s good. That&#x27;s five is what woman. Last but not least, we need to identify our domain and range. So the removed owning the first to the X values. So these arrows of the ends of our graph tell us that Ferguson will continue on infinitely in Buffalo&#x27;s directions, meaning that the domain is gonna be X is all real numbers. So all real numbers of X have a value on this function. Then, looking at the range, you can see that it continues in the positive direction and the negative direction and everything in between. So the range is also going to be all your members

Christina Shute · Answer

that this problem, the first thing we need to do is determine if the degree of this function is even or odd. So the way we do that is look at our end behaviors we see on the left, then behaviors down and on the right. It&#x27;s also down so that if the end behaviors are the same, that integrate indicates that the degree is even now. The next step that we need to identify them is the zeros of the function. In other words, where just the graph cross or touch the X axis. So working from negative to positive, we have X equals negative three X equals negative one. And then, on the positive side, we have X equals positive two and X equals positive, for Next thing we need to do is identify whether each of these zeros has a multiplicity that&#x27;s even or odd. The way we do that for Multiplicity is we look at the graph and we know if the multiplicity is even, that means that the graph is gonna bounce off of the X axis. On the other hand, if it&#x27;s odd, it&#x27;s gonna cross. It&#x27;s looking at our zeros. Negative. Three. It crosses negative one. Also across is two crosses and four crosses. So all of these are gonna have odd multiplicity ease. Now the next step is identifying the minimum possible degree of the function. Now, the way we find this is by looking at are turning points. So get rid of this information&#x27;s We have more room. So if we have, say, a degree that&#x27;s in degree in that indicates that there in minus one, turning points. And the good news is this works the other way. If we have in minus one turning points, then we have a degree of at least in so counting are turning points where the graft changes directions we have one to and three. Therefore, we have a minimum possible degree of four. This enables us to write out the equation of the function in factored form. So standard form here is gonna be X minus C to the power of him in is the multiplicity that we talked about before See is the zero and then you want to notice that subtraction is the operation here. With that in mind, we begin so wise equals and then our 1st 0 x plus three again. That&#x27;s because negative three is the zero and the operation is attraction. And then it on Multiplicity. So for now, we&#x27;re gonna assume that it&#x27;s one next one X plus one again, because this year was negative. One odd multiplicity. So raise that foot power one X minus two and X minus four and then we can double check. Remember that we want a minimum degree of four. So all of these air raised to the power of one So we could just look at that over here. One plus one plus one plus one equals for just didn&#x27;t readable out. So that&#x27;s good. Send checks out and then last but not least, we need to remember that these end behaviors are down. Remember that that indicates that we have a leading coefficient. That&#x27;s negative. So we need a subtraction sign in front of all that, all right. And then final step is finding the domain and the range of this function. So domain refers to the threat of the X values included in the graph. So here we only see a small range. But remember that these arrows indicate the graph continues on infinitely telling us that the domain is all really numbers or, in other words, the negative infinity to positive infinity. The range, on the other hand, refers to the breath of the Y values. So in the positive direction, we only reach 25. But in the negative direction we keep going on infinitely so negative infinity. So we were right. That is, we say parentheses. Un inclusive, negative infinity all the way up to 25 inclusive. And that&#x27;s what that closed graphics for, and we are all done.

Christina Shute · Answer

for this problem. The first thing we need to know is whether the degree of this function is even or odd. The way we identify that is looking at the end of behaviors on the left and the right. They&#x27;re both up. So they&#x27;re the same on left and right, indicating that we haven&#x27;t even degree they were opposite from each other. Then that would mean it&#x27;s odd degree now. The next step is identifying the zeros of the function. In other words, where does the graph touch or cross the X axis? So, working from the negative to the positive direction we have X equals negative three X equals negative, too. X equals positive one and X equals positive, too. Not only this, we&#x27;re going to identify if the multiplicity of each of the zeros is even or odd. The way we salt for that is we say OK, long simplicity. Looking at the graph. If the graft crosses the X axis at that zero, that indicates an odd multiplicity. But if the graph merely bounces off of the X axis at that point, that indicates an even multiplicity. So looking at our graph here, negative three crosses and positive one crosses So those are both gonna have odd multiplicity ease. But then we look at negative two and positive two and those both merely bounce off the X axis, indicating that they have even multiple cities. All right, now that we have that we can solve for the minimum possible degree of this function. Now the way we do that is we look at our turning points. So the way turning points are related is we say OK, given we have a degree of in that indicates we have in minus one turning points but we don&#x27;t have the degree. However, we do have turning points and it works the other way around If we have in minus one turning points that we have a degree of at least in so we could look at are turning points. We got up. We have one here to three for five turning points. Therefore, we have a minimum possible degree of six. That then enables us to write out the function in factored form. Remember, standard form here is gonna be X minus c. All of that raised the power of in. So it is that multiplicity that we were talking about C is the zero. And then subtraction is the standard operation. So he said it equal to why and we start with our first years. We have parentheses, X plus three again cause negative three is the zero and subtractions. The operation that&#x27;s an odd multiplicity is we&#x27;re gonna assume that&#x27;s raised part one multiplied by X Plus two. And that&#x27;s an even multiple cities. We&#x27;re gonna assume that&#x27;s two for now and then X minus one. That&#x27;s odds us to part one and then X minus two. That&#x27;s even zero squared up. Now, these multiple cities, we assumed the lowest possible number. So you don&#x27;t we&#x27;ll check that against this minimum possible degree of six said one plus two plus one plus two that equal six. That&#x27;s good. So our assumptions work here, and then the last thing we need is do you don&#x27;t main and the range of this function so domain refers to the breath of the X values in the graph. So basically saying how far in the negative and a positive direction does the graph go now, these arrows here indicate that the graph continues on infinitely, indicating that the X values range from negative infinity all the way to positive money. Another way we can write that is me. SE X includes all real numbers. That&#x27;s the same. Is writing from negative. Infinity took positive infinity. Now the Y values are a little different. So, looking in the negative direction, we don&#x27;t extend always to negative infinity. We stopped and negative 54 and then in the positive direction we do go away to infinity. It&#x27;s the way we want to write. That is, we start in the negative direction and we use a squared off bracket to mean negative 54 is included negative 54 comma all the way to positive infinity. And then we use a parentheses because you can&#x27;t really include infinity and that Quin quince problem.

Heather Zimmers · Answer

so, for it&#x27;s problem on the one thirty seven. Ah, see? No, it&#x27;s a You see, the zeros should be connected to one and four and at X equals one. It was from Netflix connected. So it must touch the or accept it must touch the rent from Bath in that Ah, so that means that one is at least a repeated the route. And ah, this one, it&#x27;s what Us Exco. So either negative infinity or possibly it is positive. So it must be It must have even tea growing and that these possible degree has drew that narrative to foreign repeatedly through that one so that these passport the group should be on these four leading coefficient should be positive because it goes who invented the values costed right and possible equation ofthe f end sketch it. So here we&#x27;re going to use have some fruit that&#x27;s cross team two and one and the Strip created Ruth and has truth for so a list of dates

Problem

(a) Find a quadratic function $f$ (with integer c…

Problem 137 Easy Difficulty

Answer

a) The graph touches the $x$ -axis
b) The graph touches the $x$ -axis
c) the degree is always even and $\geq 4$
d) Positive

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

Precalculus with Limits

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

Ron Larson, David C. FalvoPrecalculus with Limits

Catherine R.

Anna Marie V.

Kayleah T.

Michael J.

a) The graph touches the $x$ -axis
b) The graph touches the $x$ -axis
c) the degree is always even and $\geq 4$
d) Positive

Ron Larson, David C. Falvo

Precalculus with Limits