uh. This question says: determine the x intercepts and the y intercept of the graphs of each polynomial function then sketch the graph okay. So we need to find the intercepts and sketch so the let’s focus on the y intercept. First, the y is really easy: it’s, the constant term okay. So the y intercept is down here at negative 18. Okay, that’s a point that is on the graph for sure it has to pass through there because it’s the constant. Why do i know that? Because when you let x, equal zero, that’s, 0 plus 0 plus 0 minus 18., that’s 0 minus 18 that’s, your intercept got it so that’s easy now the x intercepts. This is where we have to factor okay. So this is where the this is, where the process begins, where you have to start to choose factors or possible factors that are remember. The a value has to be a factor of negative eighteen, so x, minus one x, plus one x, minus two x, plus two 3, 6, 9 18, those ones a good idea to start with the smaller numbers. First, okay, usually because if you can pick one of these like get one of these uh factored out right away, then you’re going to have a quadratic right. This bumps down to a quadratic, probably you’re, going to be able to factor that by sight. So we just need to get one of these factored out of here. First.

So if we remember our factor, theorem um let’s check this one right here, so that’s p of one right. So if i try x equals one in here, if the remainder is zero or if the answer when i evaluate is zero, then x, minus one is a factor: let’s just try it out. We’Ve got negative one plus 2 plus 9 minus 18.. Hmm that doesn’t. Look that doesn’t look very good. This 18 is pretty big and those other ones. Don’T add up to positive 18.. So it doesn’t look like that. One’S gon na work, uh. What about let’s? Try a two let’s try a two: we need some bigger numbers in here. Right where these x’s are, we need some bigger numbers, because we got a battle with this 18 here, so let’s try let’s, try a two put a two in there um that’s, going to be negative. Eight right, two cubed is eight um plus two times uh. Four right is eight and then plus nine times. Two is eighteen. Oh, this looks really good. Those cancel off and these cancel off awesome. Okay. So that means, if i put a positive 2 in there, that means this should be a factor right here: okay, that’s, our winner and from there. What we’re going to do is we’re going to probably do some synthetic division, i would say, probably be the easiest. So let’s divide this polynomial by x, minus 2.. So i put the a value positive 2.

There i put the coefficients along the top row, making sure i’ve got the coefficients for each descending um term, here so 3, 2, 1, 0., so negative, 1, 2, 9 and negative 18.. This number that i end with here should be a zero, so i’m, going to anticipate that if i’ve done everything else correctly, so synthetic division we drop down. This number we multiply, we add, do it again multiply add multiply, add awesome. Okay, so that’s synthetic division went over that pretty quick but we’ve been over that number of times. So, look up that video ask some more questions about that if you need to that’s synthetic division, so that means that i have this x minus 2 as a factor don’t forget about that, but i also have now. This was a cubed, so this is squared or negative x, squared and that’s, plus 0 x, plus 9.. Okay. Those are my factors now all right, so let’s rewrite that x, minus 2 negative x, squared plus 9.. Now, you might say: okay, i don’t know if i can factor this. That looks like a difference of squares but it’s, not quite a difference of squares. Well, technically, it is because you could switch around 9 and uh 9 and x squared and then it would be 9 minus x squared, which is totally a difference of squares. So you could do it that way, if you don’t like doing it this way. What also you could do is factor in negative out, and that would change these signs.

You see this, and now you have the difference of squares that looks just like you’re used to either way is, is going to work either way is going to work so let’s. Try this x minus 2 with a negative out front negative 1 right, and this is going to be x, plus 3 times x, minus 3.. Okay. Now, when we multiply all those together, we should get our polynomial our original polynomial. I won’t do that right now, but um i’ll just stop and check for from check my algebra here, okay, so what this gives us now is this gives us. You know this is the f of x, and so i can look at these factors now and i can start to piece together my graph so over here on the right. I know that i have a y intercept of negative 18.. I also know that i have a y intercept of two, so i got a point there. I have a y intercept of here’s positive 3 as well 3 and i have a intercept of negative 3.. Okay. I know i have those i’m gon na just zoom in this graph a little bit uh, because we’re gon na need a little bit more space. I think okay, so i’ve got the x intercepts now. Okay x, intercepts so x equals negative three two and three. Those are the x intercepts or more specifically, you’d write them as actual points which that’s more proper.

Those are your x intercepts and now i need to sketch. So what else? What else do i need for sketching? I need i need to find out the end. Behavior. Okay, i also need to uh figure out the zero behavior, so let’s check the end behavior this is degree three, so that means it either goes something like this from three to one or it comes down like this right from two to four, because this is negative. It’S definitely the two to four one: okay, it comes down left to right because that’s negative, so my end behavior is over here somewhere, okay and then over here that’s. What i usually do just kind of the ends of the arrows you can adjust those later. If you want, and of course now what we want to do, is we want to consider what happens here remember this is what happens if, if we have let’s just say if we have an intercept here, if the multiplicity remember that’s what we’re looking for we’re looking For actually the exponents on all the factors, if the exponent is a one it’s, simply a cross straight line cross. If the exponent is a three it’s tangent two and crosses so it flattens out a little bit there. Okay it’s this. Looking! If uh, we have a two, then it kind of touches and bounces back up and if it’s a four or six or even but larger uh it’s, it comes tangent to a little bit earlier and then stays kind of more a flatter bottom.

Okay, that’s the difference there, all right so with our multiplicities here i have 1 1 1. awesome. So what that means is that this should simply cross now we’re going to have to come down and hit this negative 18.. What we don’t know, what we don’t know is sort of where it comes down to how far it comes down, but we know that it does come down at least to 18., and i would suggest that, because the midpoint of these two intercepts is over here, that That’S, probably where your your minimum is going to be so it’s going to be something like you know, this okay it’s probably going to be a somewhat symmetrical. So i would guess that so this got to come through this negative 18. and then it’s going to come up here and this isn’t to scale or very neat, of course, like this and then down and then of course that follows our end: behavior there, okay. So your graph kind of looks something like this. You don’t know how high this goes. You don’t know how low this goes, but you do know this intercept this one and this one and this intercept as well. Okay, so as best you can – and i guess this gets – gets pretty messy there on my tablet, but this is sort of what you’re you’re looking at okay. So i think that covers everything for that question. Was there anything that i missed or anything that uh you’d like to ask me about further about this question any of this process? You’Re good? Now, okay, thanks for the question, anybody else have any questions about this.

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