 Take the function f of x equals x cubed plus x squared minus 5x plus 3. This is just a good old polynomial function. How can we graph this function right here? Well, when it comes to polynomials, polynomials domain is always pretty easy. The domain of this function will always be all real numbers negative infinity up to infinity. There's no situations where you can divide by zero, no square roots, no logarithms. Domain will be all real numbers. Intercepts can be a challenge at times, right? The y-intercept, like I said, it's not too difficult. The y-intercept, well, we just have to look at f of 0. When it comes to a polynomial function, the y-intercept is just going to be this constant right here because when you plug in 0, everything else will disappear. The y-intercept will just be that constant right there. Let's write it down for future reference. We get y equals 3. X-intercepts are going to be a little bit more challenging because we have to solve the equation, like we said before, f of x equals 0. f of x equals 0. Now, for the most part, the homework is going to give you things that factor super easy without much difficulty to them whatsoever. This example right here is a little bit of a challenge. It's not at all obvious how to factor this thing. There are some techniques and I would recommend you go to the lecture notes if you want some more examples, not more examples, but if you want to see the details how one factors this thing, please read these in the lecture notes here. But if one goes out factoring this thing, you see that it factors as x plus 3 and x minus 1 squared. So this tells us specifically that our x-intercepts are going to equal negative 3 and positive 1, all right? And our y-intercept here was 3. Now, I'll be honest, when it comes to James Stewart's list right here, I go out of order all the time. And so honestly, I like to plot things along the way. Don't necessarily wait until the end. So we know that there's x-intercepts at negative 3 and 1, so let's actually put those. I have some good lines down here. So what did we figure out here? It was, had an x-intercept at 1. So I'm gonna draw a dot right here at 1. And it's a good idea to label your picture so it's clear to everyone what you have here, 1 comma 0. We had an x-intercept at 3, didn't we? Negative 3, I mean. So we get negative 3 comma 0. And then we also had a y-intercept. It was at 3. Get a picture like this. And then so that's 0 comma 3, all right? And so we'll come back to this in just a second. So we've plotted the x-intercepts and the y-intercept. So let's come back up to symmetry. Well, with our function, we can test for symmetry. You can take f of negative x there, but really there's not gonna be any symmetry for this function. Again, I often just skip this step personally. If it doesn't become immediately obvious to me whether they're symmetry or not, I just kind of run away from symmetry. Sorry, but it's reality. If you take f of negative x, that doesn't simplify to be f of x or negative f of x. In terms of discontinuities, well, it's a polynomial so it's continuous, right? There's really no issues to do with discontinuities here. Things that we should look out for with discontinuities is gonna be dividing by zero. That often gives us a discontinuity. Jump discontinuities will come about oftentimes from a piecewise function. So not a lot to worry about there. In behavior, well, when it comes to a polynomial function, the in behavior is determined entirely by the leading coefficient, or the leading term I should say. Since you have this x cubed as the leading term, as x goes to infinity, as x goes to infinity, you're gonna get that x cubed approaches infinity as well. And as x approaches negative infinity, x cubed will approach negative infinity as well. And that's because odd powers will keep the sign. So if you go to negative infinity, odds will go to negative infinity. Even powers will always go to positive infinity. And so for polynomials, you just have to look at the leading term there. All right, let's start now with our first derivative here. Now, unfortunately, we can't see the original function. So let me write it back on the screen for us right here. So our original function was x cubed plus x squared minus 5x plus three. So this was the original function y. So as we calculate y prime, we can just use the usual power rule, in which case we give very quickly y, a three y squared plus two x, I'm sorry, why'd I say y? Three x squared plus two x minus five. And as we wanna find the critical numbers, we need to figure out what makes this thing go to zero. This is a quadratic polynomial. This one takes a little bit more to factor. Again, you can use the quadratic formula, you can use some various techniques of factoring like this factorization by group supplies right here. And I'm gonna refer you to the lecture notes if you wanna see some more details about this right here. And again, for anyone who's not sure, there is a link in the description of the video for the lecture notes where you can get the script that I'm following right now. You can download it from my website. But if you factor this thing, you'll end up with x minus one and three x plus five, which then tells us that our critical numbers are one and negative five thirds. Now, this might feel a little bit like deja vu, but it's like, wait a second, x equals one. We saw that one before, it was an x intercept, but it could be the next intercept is also a critical number, not a big deal. Now, at this moment, we could start to do the first derivative test to help us know whether these are extreme values, maxes or minns, but we're actually gonna do the second derivative test in just a moment, because we're gonna calculate the second derivative. And therefore, we'll actually come back to these things to determine whether they're max or min. Now, with respect to these points right here, we do wanna graph these critical numbers, whatever they turn out to be maximum or minimum, if it's a max or min, we wanna graph them. So we are gonna have to compute y equals f of one. We already know that's a zero. We also have to compute y equals f, or y equals f of negative five-thirds. That one's not as obvious. You'll have to do a calculation there. It's gonna turn out to be 256 divided by 27. Feel free to use a decimal right here if you want to. And so before we go on, I do wanna graph those critical points on the graph right there. We already had one, x equals one, right? That was one of them. So let's do the other one. The x coordinate was negative five-thirds. Just so you know, negative five-thirds is really close to negative two. So we're gonna draw a little bit to the right of negative two. And then the 2256 over 27, that's a tiny bit bigger than nine. So we're gonna draw that just about right here. This is our critical one. And again, label all the points, negative five-thirds comma 256 over 27, like so. And so already with these dots we have right here, we have some x intercepts right here and right here. We have a point. I'm already sort of believing that this, it seems very likely we have a local maximum right here. But again, we'll go through the whole derivative test to make sure as we talk about the second derivative here. All right, so for the second derivative, we have to take the derivative now of the derivative. So y double prime is gonna equal 6x plus two. This should equal zero for which you can divide everything by two, because six and two are both even. So we get 3x plus one equals zero, 3x equals negative one, and then x equals negative one-third. And so this right here represents our PPI. That is a potential point of inflection. It could be an inflection point. We don't exactly know yet. We're actually gonna build our sign chart to help us with this information here. Now, in the meanwhile, we could start graphing this thing. We have this x equals negative one-third. It'll be useful to compute its y-coordinate. When you're looking for y-coordinate, you're always gonna put this back into the original function, the original function f of negative one-third. That one turns out to be 128 over 27. And we can graph that below. Negative one-third, it's gonna be to the left of the x-axis. And 128 over 27, that's just a little bit less than, a little bit less than five. So we get a point about right here, all right? And also feel free to label it negative one-third comma 128 over 27. All right, so we could try to connect the dots right here, but I do wanna kind of do our sign chart so we see exactly what's going on here. I'm gonna put it right here, so there's some space. So when you're building your sign chart, what I want you to do is draw a line that kind of represents the x-axis. And I want you to do a mark for each and every critical number and PPI. So we had a critical number and negative five-thirds. Draw them in ascending order. We had a PPI at negative one-third and then we had a final one at one, right? And consider what the second derivative does with these functions, at these values here. Now negative one, so the second derivative was a linear function, remember? We saw it above right here, it was 6x plus two. And because it's a linear function with a positive slope of six, it'll be increasing, increasing, increasing. For always, it'll be below the x-axis until it hits its intercept, which is negative one-third, and then it'll be positive forever afterwards, all right? And so interpreting this right here, it will, we have negative, negative, positive, positive. So if we think about our critical numbers, because it's concave down, so it's gonna be concave down at this interval right here, it's concave down at negative five-thirds, that means you have a sad face, oh, how sad that is, because it was concave down at the critical number. On the other hand, at x equals one, it's concave up because we have these positive values right here, in which case we draw a little happy face, he's so happy to see us. And so it's concave up, and it's gonna be positive right here, meaning that we have a local minimum at one, we have a local maximum at negative five-thirds. And if we think about what is the first derivative doing right here, if you have a local maximum, that means it was increasing, then decreasing, decreasing, and then increasing. And so we wanna put all this together, right? Make sure we have all the appropriate behavior about monotonicity and concavity. So this point of inflection is gonna happen at negative one-third, it's gonna be increasing, decreasing, I often like to draw the picture from the inflection point. So to the left of negative one-third, it should look like it's concave down. Negative five-thirds is a local max, whoops, kinda missed it there, whoops, a daisy. Also got a little extra inflection that I didn't need there, let me try that again. So we're concave downward, you have a local maximum. Oh, I missed it again, oh well. So you can see the behavior here that it was increasing and then decreasing because we had a maximum and then it's concave downward in this region. Then when you're past negative one-third, it should be concave up. You'll be decreasing until you hit one and then you can be increasing after that, all right? And so now we have a pretty good picture of our graph here, our polynomial. Now what I actually have on the next page is a computer-generated image of what we just created here. And you can see that, hey, I think I did pretty okay. Not the best drawing I've ever done in the world, certainly is not the worst either, but we can see that behavior we were seeing before. We see that it was concave down right here. It was concave up right here. We found that point of inflection. It was increasing, decreasing, increasing because we found this max and this min. And so we found all this information just from the formula. And so our graph is actually fairly accurate compared to what we had on the computer.