 Welcome to our review for exam 2 for math 1050 college algebra. As usual, I'll be your instructor today, Dr. Andrew Misseldine. With exam 2 now upon us, let's talk about the types of questions we're going to see in this exam. Just like exam 1, this question or this exam will have two types of questions. There'll be actually nine multiple choice questions, which are worth five points and then there will be six response questions that range from five to 12 points each and we'll talk about the specifics of them in just a moment. There are some specific, well I should say semester specific details about this test like time, place of manner and such. I refer to you to Canvas for that information as you won't find it in this video. The basic structure and the resources available to you for this test are essentially identical to what you had on exam 1, so again refer to Canvas for those. In this review video, I mostly just want to focus on what are the types of questions that you're going to see for exam 2. Now as you're preparing for exam 2, I want you to be aware that this exam is going to cover lectures 9 through 21. So for our lecture series, this covers what we're calling chapters 2 and chapters 3. Chapter 2 was about linear functions and their applications. Chapter 3 was about quadratic functions and their applications, so lectures 9 through 21 will be covered in this exam. So without further ado, let's get to some of the specific question types you should be prepared for as you prepare for exam 2. So starting with the multiple choice section, of course, we'll do question number 1. You see question 1 here, it's asking you to calculate the slope of the line that passes through 2 specific points right here. In particular, question number 1 is going to ask you a question involving the very first lecture of this unit, lecture 9. Lecture 9 was a was a introduction review of linear functions, some things that we found in that section in that lecture. We learned how to solve linear equations, we learned how to solve linear inequalities, we learned how to graph lines and come up with linear equations to graph those lines. And therefore the slope was a very important ingredient there. To answer question like number 1, you probably do need to know the slope formula m equals y2 minus y1 over x2 minus x1 that rise over 1. But you should also be familiar with the important equations of a line, right? So we have the slope intercept form of a line y equals mx plus b, right? m is the slope that we get from this formula, v is the y-intercept. You probably also want to know the point slope form y minus y1 equals m times x minus x1, where x1 comma y1 is a specific point on the line, m is the slope that's computed right here, like I mentioned before. And oftentimes we use the point slope form to then transition into the y-intercept form. So those are some important formulas we got from lecture 9, but also the ability to solve linear equations, to solve linear inequalities, that's an important thing as well. And so again, for question number 1, it will cover lecture 9, something from lecture 9 about linear functions. It could be like what you see on the screen, what you have to then just compute the slope of a line from the formula or maybe given the equation you're asked to identify the slope. You could be asked to build the equation of a line, put in slope intercept form, but like I said, you'll probably start with the point slope form. Can you solve a very simple linear equation? Can you solve a simple linear inequality? Can you find the intercepts of a line? That's an interesting question. Of course, the y-intercept is right there in the formula. If you want to find the x-intercept, you plug in y equals 0 and then solve for x. And so those are some variants you could see for question number 1. Question number 2 will be a lot more specific than say question number 1 was. Question number 2, you are going to be given an absolute value equation. It'll be an equation involving absolute value, and that's absolutely true. I know horrible pun there. Absolute value we covered at the very end of chapter 2, specifically lecture 15. So you should be able to be prepared to solve an equation involving absolute value. Really the things to review for absolute value is essentially the following observation. If absolute value is equal to some positive number like say 7, then you're going to get two solutions to that equation. You're going to get x equals plus 7 and minus 7. So you have to consider the two cases because absolute value forgets the sign. If you take the absolute value of some variable x, you don't know if it's positive or negative. So when solving it, you have to consider the two possibilities plus 7 or minus 7 in the situation. Now of course if the absolute value is equal to 0, that same strategy works there as well. You're going to get x equals plus or minus 0. But there is no difference between plus or minus 0. It's just a 0. So if absolute value does equal 0, you get a unique solution. Be prepared for that. But it is also possible that absolute value equations have no solution. If the absolute value is equal to something negative like negative 2, well absolute value is always non-negative. The absolute value of x is always greater than or equal to 0. So if the absolute value equals a negative, we would actually have to then do no solution in that situation. And that's the only time that an absolute value equation has no solution when you force an absolute value to equal 0. All right. One other variant I do need to point out on this problem right here. If you have something like the absolute value of x plus 1 equals the absolute value of like x plus 3 or something like that. In that situation, when you have two absolute values equal to each other, it really doesn't fundamentally change the problem. And it's actually similar to this one right here. You would proceed to look at x plus 1 is equal to plus or minus x plus 3. Now in this example, I should mention that one of those choices will lead to no solution, but the others will lead to a solution. So you might only get one solution. Absolute value equations can have no solution, one solution, or two solutions. So be prepared for these possible variations. Question number three is going to be a question about complex numbers, in particular doing arithmetic with complex numbers. Complex numbers were reviewed in section, I should say lecture 16. That was our first section for chapter three, our introduction to quadratic equations, quadratic functions. And that's because sometimes quadratic equations have non real solutions. You might have to have a proper complex number. In this exercise right here, it's asking us to take the quotient of two complex numbers, three plus five i over one minus two i. So some things to remember there is that when you divide by a complex number, you should times the numerator and denominator by its complex conjugate, you switch the sign. Then you got the FOIL, something that's things out there. So you definitely have to be able to add, subtract, multiply, and potentially divide complex numbers. One of the beautiful things about asking you a division problem with complex numbers is that in order to do division, you have to also do multiplication. You have to also be able to add and subtract complex numbers. And so it really is the whole enchilada. So if you understand how to do a question like number three here, from this practice test, you're probably fine from the type of stuff we saw from lecture 16 there. The things to remember, of course, is when you add and subtract complex numbers, treat it like combining like terms, all right? You add the real parts, you add the imaginary parts. When it comes to multiplying complex numbers, it's basically the FOIL method because each complex number has the two parts, the real part, the imaginary part. The, of course, the important thing to remember is that when you FOIL at some point, you might get something like i squared. Well, since i is the square root of negative one, i squared is itself negative one. That also allows us to do things like i cubed, which is negative i, i to the fourth, which is positive one, and other higher powers of i as well. So be comfortable with doing these arithmetic calculations, the four arithmetic functions, addition, subtraction, multiplication, division with complex numbers. Moving on to question number four, as you can see here on the screen, question number four is going to be something about solving quadratic equations. So you see on this one right here, you're asked to solve the quadratic equation, 3x squared plus 5x plus 2 equals 0. When solving quadratic equations, I guess the first thing to mention is that these topics were covered in particularly lecture 17 and 18. That's where we talk about solving quadratic equations. So some basic principles we want is that when you have a quadratic equation, we like the right hand side to be 0. I mean, it could be the left hand side, doesn't make much of a difference, but you want to make one of the sides of the equations, we'll just say the right hand side for the sake of it, to be 0. And then you want to combine all the like terms on the left hand side. So that is to say we're looking for something of the form ax squared plus bx plus c equals 0. Once the quadratic equation is in the standard quadratic form, it's then prime for solving it. We could solve it by factoring, right? We could take three times two to get six and look for factors of six that add up or subtract maybe to be five. Won't say much more about that. We did talk about how to factor a quadratic polynomial in lecture 17 and how that when combined with the zero product property helps us solve quadratic equations. That'd be useful. Another approach is you could try to solve the quadratic equation by completing the square. That's what we talked about in lecture 18. That method is admittedly not as popular as perhaps the two other methods that we're talking about right here, but it is a legitimate method. So you should know how to complete the square. There actually will be some questions that mandate it, but we'll talk about that in just a second. You can also use the quadratic formula, right? So think of pop goes to B's on your head there, right? X equals negative B plus or minus the square root of B squared minus four AC all over two A. So you could solve this using the quadratic formula. This thing is in the quadratic form already. You could just plug in the numbers A, B, and C. Now, admittedly, when it comes to actually doing the number crunching, generally speaking, factoring is easier than the quadratic formula. But of course, you can use whichever of these three methods you feel most comfortable with. Sometimes factoring doesn't quite work because it could be that the solutions are irrational. But admittedly, if I'm looking at this question, I don't see any solutions with square roots or with the I, the symbol I, right? Complex I here, which itself is a square root, square root of a negative. And as such, that actually suggests to me that I probably could solve this one using factoring because if the answers are rational numbers, whole numbers are fractions, that actually means we can solve it by factoring. So just a little bit of a tip there. The quadratic formula is perfectly fine, but I think you might actually find factoring to be easier on this one. So how do we solve quadratic equations? We can also solve quadratic equations again by these three methods. I'll just leave it at that. One other thing you should be prepared for on this question. You don't see it in this variant, but one possibility here coming from section 19, could we identify the vertex of a quadratic function? So let's say we have something like f of x equals 3x squared plus 5x plus 2. Could we find the coordinates of the vertex of that function, whether it's in the standard form, or maybe it's in the vertex form? One has to be cautious about that because when you're in vertex form, you might have to switch the sign of something. So it's probably useful to know the formula for the vertex h, which is the x-coordinate of the vertex. This is going to be negative b over 2a. Notice this is just the part of the quadratic formula if you forget the discriminant. And then to find k, there's a formula for it, but my suggestion is just do f of h. Just plug this formula into the function and simplify to get the y-coordinate. That's probably going to be your best bet. You probably want to know the vertex form. f of x is equal to a times x minus h plus k. Don't forget the square, so it's a quadratic function there. Where this a here is the same a as was here. This should be the leading coefficient. The h and k can be found using these formulas, or you can do it by completing the square. And so if the quadratic function is given to you in vertex form, then you don't have to compute the vertex. It's already in front of you. You do have to change the sign on the horizontal coordinate because it's in the horizontal zone. You don't have to change the sign on the k though. So those are some versions of question 4 you should be prepared for. Coming on to question 5, then question 5 is going to be our second and final question about absolute value. So remember absolute value is covered in lecture 15. I mentioned that, but I'll just say it again. This time you have to solve an inequality that involves absolute value. That does change things a little bit. It's very similar to solving equations, right? If you have the absolute value of x, it's greater than or equal to say some positive number like 3. That will then tell you that okay x is greater than or equal to 3, or x is less than or equal to negative 3. You have to consider both the positive and the negative case, but the sign switches on the negative case because we need times an inequality by negative sign. It switches the sign. So we have to consider that. You have those two possibilities. If x is greater than or equal to 0, in that situation, well, you would get all real numbers in that situation because absolute value is always greater than or equal to 0. So it doesn't matter what you choose for x. You get all real numbers. If you get something like the absolute value of x is less than or equal to 0, you get some negative in that situation. That does change things, right? In that situation, you'd only get the x-intercepts as your solutions, right? I guess I should have mentioned this before. If you have the absolute values less than 3, same basic idea. You have to consider x is less than 3, and you're going to have x is greater than negative 3. So put those things together. So the direction of the inequality does matter. My recommendation, of course, is just to just to consider your markers. You have negative 3 and 3 and use test points. Which one satisfies the original inequality? Is it the beak or is it the wings of your bird, right? Because this is going to be absolute value, something like that. That'll take care of all of them. But of course, there are situations where there are no markers, maybe. If you have to take the absolute value of x is less than 0, there's no solution to that. There's no markers. And that's because in that situation, your absolute value sits entirely above the x-axis. So you might get either everything or nothing, depending on the direction. So there are a few special cases you have to consider. But generally, you're going to be seeing something like this on question number 5. So be prepared for that. I'm going to erase a little bit on the screen here so that we can do question number 6. Question number 6 is going to be a question about discriminants, all right? So remember, the discriminant is the radicand inside of the quadratic formula, right? It's this number right here, d equals b squared minus 4ac. And so by just knowing the discriminant of a quadratic equation, we can determine are there real solutions, are there non-real solutions, how many solutions are there. So remember, when the discriminant is a positive number, you're going to get two real solutions, all right? When your discriminant is 0, you're going to get one real solution. Because when your discriminant is 0, it turns out your quadratic equation is actually a perfect square. And the one solution is the vertex. If your discriminant is less than 0, you're going to get two non-real solutions. They're complex numbers. All of them are complex numbers. Real numbers are complex numbers, right? Just like a tree is a plant, but not all plants are trees. If your discriminant is less than 0, you'll get two non-real solutions. So with this one right here, if it's asking which of the following quadratic equations has one real solution, at least one real solution, that means the discriminant is either positive or zero. That would count as well. So we mostly just need to throw out which of these have negative discriminant. You don't have to solve the quadratic equation. You just have to determine how many solutions there are. And so the discriminant is a very nice shortcut in that regard. So we spent four lectures on systems of equations. In particular, lectures 11 through 14 in our lecture series are about systems of equations. So there's going to be quite a few questions about systems of linear equations on this test. In terms of the method used to solve them, you really want to be familiar with all of them. Like question number seven, the way it's phrased is it's with augmented matrices. You do have to have fluency with these augmented matrices in order to or answer this version of question seven. But there are alternate versions of question number seven, which have you solving systems through elimination, through substitution. So you should be familiar with all these of these techniques on this one and asked which of the following augmented matrices are consistent. So we should know this vocabulary here. Consistent means that the system has a solution. It could have one solution or it could have infinitely many solutions. Inconsistent means that it has no solution. So really, we want to figure out which of these things have no solution and throw it out. We could also ask things like when are the equations in a linear system? When are they dependent? That would suggest that we have infinitely many solutions. When are the equations independent? Or that would mean that we have a unique solution in that situation. So we should be familiar with those cases. Again, but we also should know the basics of how does elimination work. Like if I gave you a linear system and was like, okay, what elimination would sort of be the best choice right now? Could you answer that? Or if you were trying to solve a system by substitution, could you answer the question, what would be the best substitution to do right now? That is kind of like do the next step of a problem. That's really what question seven is about. It's not about solving a system of linear equations. It's basically like we're in the middle of solving a linear system. What's the next step using these, one of these techniques? That's what you should be prepared to do. And so again, these topics were covered in lectures 11 through 14. We had four lectures about solving systems of equations. Question number eight is going to be an application of linear functions. So a linear model. And this is the type of stuff we did, of course, in lecture 10. There were lots of different linear models that we learned about in that lecture. There are two that are extremely important. One of them is manifest here on the screen right now. That's simple interest. You should know the formula for simple interest and be able to do some simple interest calculations. It's just a linear model, right? The amount is equal to the principle times one plus RT, where R is your interest rate. Typically, that will be an interest rate for years. So like APR is annual percentage rate there. T is a measurement of time. That measurement of time needs to reflect the measurement of time that's in the rate. So for talking about APR, annual percentage rate, then T is a measurement of years. So like on this one, we have four years. P is the principle. That's the amount of the original loan. And then A is the amount that one pays back. The principle plus the interest. So that's why there's one plus RT. One times P gives you back the principle. P times RT gives you the interest on the principle that one gives back. So be able to calculate and solve some problems using simple interest here. Another important application you should be aware of. Very simple but still very powerful. The idea of direct variation, right? So we have two variables that are directly related to each other. So like say Y equals KX. That is, some constant K, your constant of variation there is the K. Some K times X gives you Y. And so be able to do some very basic variation problems involving direct variation, X and Y. Or they could have other names. Now that's not to say that simple interest in direct variation are the only linear models you should know. But these ones are ones you should memorize, the formulas are put them on your notes as you prepare for the test of course. But we have some other story problems we did in lecture 10 about linear growth. And so if we just have some information about, oh, this quantity is increasing or decreasing over time, like maybe the population or some commodity is growing over time, or perhaps an infectious disease is decreasing over time, things like that. It'll tell you that the spread or decay of the quantity is growing or decaying linearly. And so use the linear equation in that situation. Question nine brings us to the end of the multiple choice section of this test, in which case we're asked to solve a quadratic inequality. Solving these is very similar to how we solve absolute inequalities. My, my suggestion is actually to solve the underlying quadratic equation. So x minus one times x plus four is equal to zero. You can solve that by factoring, which this one's will be factored for you. So that's pretty great. You can complete the square, you can, you could use the quadratic formula. I can promise you that nearly all of them, you could probably solve by factoring just for the sake of convenience. You have to solve the equation, this gives you your markers. So you have your two solutions. And then we think of what's the basic shape of the graph, let's say just for simplicity, a concave upward like this. And so then like absolute value, are we looking for the beak? Are we looking for the wings of the bird? And that depends on which direction inequality is going. Are we trying to be above the x-axis, below the x-axis, or something like that? Are we allowed to be on the x-axis? So try to solve these inequalities using this idea. Lecture 21, the very last lecture of this unit is where we talked about quadratic inequalities, the very last unit, less section I should say in chapter three. So be prepared to answer that one. Question number 10 is the first question in the free response section. It's only worth five points. So each of the free response questions is, excuse me, each of the multiple choice questions is worth five points. This one really is on par with a multiple choice question. It's just by the nature of the question. It doesn't really make sense to ask it as a multiple choice question. So on question number 10, you'll get another question from lecture nine about linear functions, linear equations. You can be asked to solve a linear equation. So this, you could argue, is a hard linear equation. There's really no such thing as a hard linear equation, frankly speaking, but it's very, there's more involved, let's say that. It's more involved than what you might see in question number one. Solve this one. It could be that there's no solution. Maybe the x's cancel out. Maybe every number is a solution, all real numbers because it's an identity, but most likely you should get a unique solution when you solve this linear equation. So maybe you have to distribute some things. Maybe you have to combine some like terms because there's x's on both sides of the equations. It has all of the possible ingredients you could see here. But another possibility is you're asked to graph a linear function. So maybe you're given a function like Michael's mx plus b. How do you graph that thing? Again, it's not too difficult of a question, which is why it's only worth five points. But these graphing questions don't really make sense to be multiple choice. I actually want you to graph it. To graph a linear function, I would definitely recommend you should label two points and then do the best you can to connect the dots. If you have a straight edge, that's great, but don't worry too much about it. If your line is too wiggly because you're so nervous, I suppose I could live with that. Just maybe label your points so that it's very clear exactly what the points that you're trying to draw were. Moving on to the next page, where the heat will turn up a little bit on these free response questions. Question number 11 is worth eight points. This will be a story problem involving systems of equations for the most part, right? You actually might just give you the system of equation. You're asked to solve it. It'll be a two by two system. There's a three by three system that's coming up. But most likely on question number 11, you'll be given a story problem of some kind. This will most likely be a mixture problem. And by most likely, if it's a story problem, it will definitely be a mixture problem. Those were the types of questions. The story problems I should say we saw with when it came to these systems of equations. Again, I promise you this question 11 will be a two by two. There'll be two equations, two unknowns. For a story problem, you can decide what they're called. For this one, where Alexi has some number of coins here, quarters and dimes. You might call your variables Q and D, something like that. So you have one equation involving Q and D. You have a second equation involving Q and D and you have to solve it. If you don't have to set up this system of equations, if it's given to you, I can promise you it'll be a little bit on the harder side to justify why you get the same amount of points without having to set it up. Though these mixture problems in terms of solving them will be pretty easy. But it will be a two by two problem. That's the case. In terms of solving systems of equations on question 11, you can use whatever method you want. Substitution elimination, Gaussian elimination. Heck, if you want to use Kramer's rule and you feel pretty good about it, you can do that. I personally think that the matrix approach is pretty good. But for a two by two system substitution or elimination also works out pretty good. Heck, even Kramer's rule works pretty well in that situation. It's really in the three by three systems where the matrix method really starts to shine above the other methods. So I'll make that recommendation when we get there. Of course, there's lots of lectures that cover systems of equations. It's lectures 11 through 14. But if you're specifically looking for some of these mixture problems, you want some extra practice on those, those were covered in lecture 12. So be prepared for those. Question number 12. This one's worth also eight points. This is a question which you might actually be mandated to solve the equation by completing the square or some other method. Maybe you're told to do it by factoring. Maybe you're told to do it by the quadratic formula. Question number 12 will ask us to solve a quadratic equation. So some of the principles I said earlier because you have to solve the quadratic equation in the multiple choice section as well. Those same principles apply right here, but you do have to show your work. And generally speaking, this might be a little bit harder to do. So if you're going to solve it by factoring, finding the magic pair might not be so obvious. Maybe you have to use the quadratic formula or completing the square because maybe factoring is impossible. But the important thing to be aware of here is that if it tells you to do it by a certain method, you have to do it by that method for following the instructions, right? If I were to solve this by using the quadratic formula, even if I do everything correct, you're going to get no points because the instructions do say complete the square. And any of the methods we've learned is possible. So you should be prepared to do any of them. You should know the quadratic formula. You should know how to do it by completing the square. You should know how to do it by factoring. Of course, if factoring is possible. And so solving quadratic equations as a reminder, this was covered in sections 17 and 18. 18 mostly talked about fact, excuse me, 17 mostly talked about factoring. 18 was about completing the square and the quadratic formula. But like I said, you do have to be prepared with both, not just both. I should say both lecturers, all three techniques you should be aware of. Question number 13. This one has the highest point total for the entire exam. In this case, you will be given a three by three system of linear equations. You won't have to set it up. It will just be given to you. And you have to solve it and show all of the steps. Now, which method do you want to use? Do you want to do it by substitution elimination, some hybrid of substitution elimination? Do you want to use matrices and use row reductions to simplify it into echelon form? That is up to you. Unlike the previous question, which mandated a method, on this one, number 13, you can use any method you want. And as such, you might want to reference all of lectures 11 through 14. Now, admittedly, lecture 11 only focuses on two by two systems. Lecture 12 started showing us how to solve three by three systems. And then lectures 13 and 14 did it with matrices. So that's a pretty good approach. But still, I'll include 11 in there for reference. There's a lot of steps to do here. Do try to fit it in here. If come the actual test, you can't fit it all in here. Just see something like C attached. Actually, say it's on a scratch paper, but actually make a marker it continues on if you can't fit it all in there. You probably can though. So don't worry about it too much. And like I said, there is no mandated method. Elimination substitution, matrices or some combinations are available. You can even try Kramer's rule if you feel comfortable with it. But I should warn you, no student has ever gotten full credit on this question trying to use Kramer's rule for a three by three system. It doesn't work out super great. So I don't recommend it. It can be done. But many of the 1050 students has failed in that attempt. So I don't recommend it. Question number 14. This is going to be a story problem involving quadratic functions. So this was the main topic of lecture 20. That is quadratic modeling quadratic story problems. This will be a situation very likely you might be given the quadratic equation. Because you're not physics students necessarily. This is in a physics class. I don't need you to set up a projectile motion type problem. But I do need you to analyze it. If it was given to you, important things to know here is finding the vertex. Because if we're launching a rocket, like in this example, we want to find its maximum height. That's information we glean from the vertex of the parabola. But we might also need to find the x intercepts or the y intercepts as well. So the important thing is given a quadratic function. Can you find its vertex? Can you find its intercepts? That's stuff you really need to know how to do. And then the story problem will ask us to apply it. All right. Now while this one actually gives you the formula, there are some examples where you will be asked to set it up. So in particular, a question that we've seen the area of a rectangle that has come up a couple times where there's some type of stipulation about the perimeter. So like we've seen situations where like the perimeter is some constraint or maybe because there's a river here and you only have three sides. Or we've done situations where we're like a chicken coop and we have to separate the roosters from the hens or maybe there's also something else has to be separated. So we've done some problems like this where we have some constraint about the perimeter. And the perimeter might be more than just twice the length, twice the width. Like in this situation, you'd get two times the length, but four times the width. That's your perimeter. Anyways, in that situation, the area is still going to be length times width. But you have to substitute out one of the variables using this perimeter constraint. You solve for like say maybe L and then you plug it in there. We've done some equations like that. What would be then be the maximum area we could form for such a rectangle? Well, once you substitute the perimeter in for like say length, because there's a linear relationship between width and length, when you substitute it in there, you're going to get a quadratic relationship on the width and then you can maximize that by looking at the vertex. So that's a problem you should be able to set up and solve. Another problem that we've done was about revenue. The revenue problem is basically the following. Revenue equals P times X, where P is the price of some commodity that's being sold and X is the number of those commodities that got sold. So price times the number of sales gives you revenue. It's a financial problem. And so if the price function, so price or demand, you can call it the same way, right? If we treat price as some function of X, right? Because if we increase the price, you'll sell fewer commodities. If you lower the price, you'll sell more commodities. And so price and sales are typically going to be inversely proportional to each other. One goes up, the other goes down. We can often describe that with a linear relationship. If we have that linear relationship, we can substitute that in for P. And therefore, the revenue then becomes a quadratic relationship on the number of things you want to sell. And thus maximizing revenue would be finding the vertex of that parabola. These are two examples where you're asked to set up and solve it. And in particular, to solve the quadratic, that is, to set up the quadratic, it's just the variable time, some linear function. But you might have to set up that linear function, right? Some information about the perimeter is given. So you come up with this linear relationship of perimeter, or some information about the demand is given. Therefore, you set up the linear demand. So setting up the quadratic is pretty easy, but setting up the linear relationship is probably the hardest part about this. But as this test also includes linear models, that's a very appropriate question. So be prepared to do those ones. Those are some of the harder questions for number 14. Be prepared how to do them. And so that then brings us to the last question on this exam, question number 15. This is a graphing question, in which case you are asked to graph a quadratic function. That quadratic function will be given to you in the standard form, ax squared plus bx plus c. One of the instructions will tell you that this quadratic form needs to be put into the vertex form. That is, we have to rewrite this as f of x equals a times x minus h squared plus k. That is necessary. If you do not include the vertex form, you do not get full credit. You basically lost half the points. You're going to have to draw the parabola, yes, but that's only about half the points, five out of 10. The other five come from getting this vertex form. You do have to show your work. Probably the easiest way to get the vertex form is just to complete the square, to switch it from one form to the other. But another possibility, of course, is you can use the formula h equals negative b over 2a. You can compute that. That then plugs in for h. Once you have h, you can compute f of h using the original formula. That's going to give you a k for which you can plug in there. Then the leading coefficient doesn't change, so 3 and 3. You can actually calculate the vertex formula without completing the square, so to speak. These formulas came from completing the square, but there does need to be some work to show you how you transition from the standard form to the vertex form. That's necessary for full credit. Once you have the vertex form, you then have the transformations. This a value, this tells you if there's some type of vertical stretch. You have a vertical stretch by a factor of a. This would tell you if you reflected or not. If a is a negative number, it means you reflected across the x-axis, your parabola would be concave down now versus concave up. Those are some things there. The h value, this is going to give you a horizontal shift. So a shift to the right or left, depending on what the sign of h is. The k value tells you whether there's a vertical shift to shift up or down. So this vertex form is not just some annoying thing. It actually gives you the transformations. And then from there, we can graph our parabola. So we end up something like, to graph it, I would start off with the, I would start off with the vertex. You have to label the vertex of your graph that's mandatory. So you have to find it. You have to also label the x and y intercepts. The y intercepts generally pretty easy because you plug in x equals 0. It's going to be, in this case, you get like a negative 2. So do something like that. In which case, then from there, you can use like the axis of symmetry to try to graph the rest of it, something like that. You can use the x intercepts. You can also use the vertical stretch. Like, if you stretch it by a factor of 3, the idea is you're going to go down 3 over 1, something like that. So there's a lot of things you can do to graph this parabola. But what you do need to have is you do need to have the correct parabola, and you must label where's the vertex with its coordinates, where are the x and y intercepts with their coordinates. The x intercepts could be irrational, mind you. You can either approximate them. Like, if your x intercept turned out to be like 1 plus the square root of 2, you can approximate it using your calculator. If you don't have a calculator, just write it as exact. Like, oh, the x intercept is 1 plus square root of 2. And do your best to estimate how that is. You can use other information like y intercepts or the stretching factor to get that. So you don't necessarily have to have the calculator to do this one, not at all necessary. And so that brings us to the end of exam 2, which remember covers chapters 2 and 3 about linear and quadratic functions. If you have some other questions that weren't answered in this video, please reach out to me and I'd be glad to help you. Best of luck in your studies and I'll see you next time.