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Contents

- 1 Moving away from Standard Form
- 2 Moving away from Standard Form
- 3 Choosing the Right Constant
- 4 Choosing the Right Constant
- 5 Writing in the New Form
- 6 Writing in the New Form
- 7 Where is the vertex
- 8 Where is the vertex
- 9 Completing the Square
- 10 Completing the Square
- 11 Read off the Vertex
- 12 Read off the Vertex
- 13 Multiplying the Leading Coefficient by 3
- 14 Multiplying the Leading Coefficient by 3
- 15 Replacing x and y
- 16 Replacing x and y
- 17 Distribute and Square
- 18 Distribute and Square
- 19 Giving us the vertex
- 20 Giving us the vertex
- 21 Naming the Form
- 22 Naming the Form
- 23 Moving toward Vertex Form
- 24 Moving toward Vertex Form
- 25 Factor out the Leading Coefficient
- 26 Factor out the Leading Coefficient
- 27 What should we add
- 28 What should we add
- 29 Is replacing that easy
- 30 Is replacing that easy
- 31 Adding to Both Sides
- 32 Adding to Both Sides
- 33 Final Answer in Vertex Form
- 34 Final Answer in Vertex Form
- 35 Write down the vertex
- 36 Write down the vertex
- 37 Pick the Graph
- 38 Pick the Graph
- 39 Flipping the Coefficient
- 40 Flipping the Coefficient
- 41 If a is negative
- 42 If a is negative
- 43 Identifying Coefficient Signs
- 44 Identifying Coefficient Signs
- 45 Tell me the vertex
- 46 Tell me the vertex
- 47 Wiper Price for Maximizing Profit
- 48 Wiper Price for Maximizing Profit
- 49 Key Points
- 50 Key Points
- 51 Axis of the Parabola
- 52 Axis of the Parabola
- 53 Tangent Lines
- 54 Tangent Lines
- 55 Comparing Tangent Slopes
- 56 Comparing Tangent Slopes
- 57 Differences of Squares
- 58 Differences of Squares
- 59 Factor
- 60 Factor
- 61 Completing the Square
- 62 Completing the Square
- 63 How do we find x intercepts
- 64 How do we find x intercepts
- 65 Rounding x intercepts
- 66 Rounding x intercepts

At the end of our last lesson, I showed you an example of how to start with the equation for a parabola in standard form and then work through modifying it, so that it's in the form that we can just read off the coordinates of the vertex. Now you're going to get a chance to find this. But don't worry. I'll help you through it. Let's say we're dealing with the equation, y eqiuals x squared minus the quantity x minus h squared. So the first thing that I notice about this form over here that we're aiming for is that we have y minus a constant. So one logical first step would be to get the constant that's on the right side, to be over on the left side instead. So the first thing I'd like you to do then is to make that change, getting the constant term to be on the left hand side instead. And rewrite the equation in this box.

To get the constant term to be on the left side instead of the right side, we just need to subtract 15 from both sides. This gets rid of it on the right and makes this so we have a new constant on the left. Now we're left with y minus 15 equals x squared minus 8x.

The left side of our equation looks good now. This is exactly the sort of thing that we're looking for, y minus the constant. The right side is not quite there yet though. Eventually, we want this to be x minus some constant and that whole thing squared. Well, if you think about it, we're actually well on our way. If you think about this right side in the general form of this equation, we can multiply out the quantity x minus h squared to get x squared minus 2hx plus h squared. You've seen things like this a lot of times before. So here, it looks like we have this term covered, the x squared term, and we also have the term the x to the first covered. What we're missing is this constant. What constant do we need to add on to the end of x squared minus 8x in order to be able to factor it in the form x minus h squared. So, just think of what constant belongs here in order to make this expression on the right-hand side, the perfect square of some polynomial of two terms.

We had a lot of practice with this when we were working with factoring earlier on. We're just working in the opposite direction this time, almost. When we look at the multiplied out version of x minus h squared, we notice that this constant term at the end, h squared, is just equal to the coefficient of the middle term divided by 2 and then squared. So if we do that same thing to the negative 8 here, we know that our constant at the end or h is going to be equal to negative squared is 16.

Since we're looking for something in the form X minus H squared on the right hand side, the result that we need to add on over here is 16. Now of course if I add something to one side of the equation, I also have to add it to the other side. Now as we can see from our work in the last quiz, after we add 16 to both sides We'll be one step away from writing this in the form that we were aiming tor. So I'd like you to do that step of adding 16 and then modify the right hand side of the equation so that we get our final equation in the form y minus k equals x minus h squared. Put the equation written that way in this box, with of course the appropriate numbers plugged in.

Adding 16 to both sides of our equation, gives us y plus 1 equals x squared minus 8x plus 16. All we need to do now is factor the right-hand side of the equation to make sure that it fits the form we're looking for. And sure enough, we can factor this to be x minus 4 squared. So, we have a final equation of y plus 1 equals the quantity x minus four squared. Awesome.

So here's overall what we just did. We changed the equation y equals x squared minus 8x plus 15, an equation for a parabola in standard form. To the equation y plus 1 equals x minus 4 squared. So now we should be able to really easily where the vertex of this parabola is. I'd like you to write the coordinates of this vertex in these boxes.

The vertex will be at 4, negative 1. Since we have x minus 4 and y plus 1.

Let's take a bird's eye view of the process we just went through in transforming this equation. Since we were aiming for having a perfect square at the end, on the right-hand side of the equation involving x and a constant, we knew that the initial constant we started out with couldn't be right, since right now, we can't factor x squared minus 8 x plus 15 to be in this form. That means, we need a different constant attached to the two terms with x in them. So, the first step was to get rid of the initial constant we had so that we could have a different constant, on the end of the x squared minus 8x. You've figured out the constant that belonged at the end of those two terms was 16. When we add 16 to this, we get the perfect square of a binomial. This process is called completing the square, since the constant that we add on over here ends up giving us a squared expression as our answer. Now, we don't' technically have to move the 15 away from the right-hand side to do this. But I think that it makes dealing with these two terms on their own a little bit easier, since we don't have anything else to look at on the right side. So, completing a square is a crucial part of transforming equations of parabolas from standard form, into the form where we can see the vertex. Now that we understand how to complete this square and the importance of that technique in transforming equations from one form to another, I'd like you to rewrite the equation y equals x squared plus 5x plus 6 in the form that we've been talking about. Now, if your numbers get a little bit messy in here, don't worry. That doesn't mean you're doing the problem wrong, that often happens in math. So, just go with it and then at the end, you can always check your answer by plugging in points to the equation that you start with and the equation that you end with to make sure that they're the same.

We can transform the equation, Y equals X squared plus 5x plus 6. Using completing the square, to instead read, y plus 1 over 4 equals x plus 5 over 2 squared. The first step that I do is make sure that what's on the right hand side isn't already the square of binomial. Which is, if you remember, the form that we want the right-hand side of the equation to be. You should always check this first just to make sure you don't do a ton of unnecessary work. Anyway I subtract 6 from both sides. Then I figure out what term I need to add on to x squared plus 5 x to make this the perfect square of two terms added together, and then I just factor.

Now, we have this equation in such a lovely form. I'd like you to tell me what the vertex of this parabola is.

The vertex is at negative 5 halves, negative 1 4th. Please note that x plus 5 halves is the same as x minus negative 5 halves and y plus 1 over 4 is the same as y minus negative 1 4th. So, that is why these are the coordinate of the vertex.

So we see now how to transform a couple of different equations to this new form, which is really helpful for us in certain ways. But there are a lot of different equations for parabolas that don't look just like these two. In particular, what if our equation doesn't have a leading coefficient of 1? Leading coefficient here, refers to the coefficient in front of the x squared term, the term with the highest power. In both of these situations, that coefficient is 1, which we know makes factoring a lot simpler, and presumably, also changes how we complete the square. In fact, recall the equation that we came up with as our final equation, for calculating Grant's profit based on the price that he charges per glasses wiper set. Last lesson we had p equals negative 200 c squared, plus certainly not equal to 1. If we want to find out exactly where the vertex of this equation is, we're going to need to get this in that new form, using completing the square. Well, we'll get to solving this eventually, but for now I'd like us to start out with a equation similar to this, and that its leading coefficient is not 1, and over through that, to eventually be able to deal with the Grant situation. Here I've graphed the equations y equals x squared, that parent parabola we've seen so many times, and then a second parabola y equals three x squared, this green one right here. So judging from the relationship between these two graphs, how does multiplying the leading coefficient of a parabola change the shape of the parabola? Does it shift it vertically, shift it horizontally, make it wider, or make it more narrow?

It must make the graph more narrow, judging from the fact that this green curve is skinnier than this gray curve. And the only difference between our two equations is that here, x squared is multiplied by 3, and here, it's just multiplied by 1. Otherwise, the graph hasn't moved at all. There is no shift horizontally or vertically. It's just the width of the graph that's changed.

We just saw one example of changing the width of a graph, and what if I now want to take our new graph and move it down 2, and over to the right 5? Thinking back to last lesson, what should I replace x with in the equation, and what should I replace y with?

I need to replace x with x minus 5 and y with y plus 2. Remember, that the vertex of y equals 3x squared is at 0,0. So, if we want to move the graph down to and to the right 5, then the new vertex will be at 5, negative 2. Let's take a look at that new graph. Here, I've plugged in those transformations you talked about in the last quiz, replacing y with y plus 2 and x with x minus 5 to make the vertex be at 5, negative 2, like we just figured out. So remember, the 3 here, this coefficient in front of the leading term, squish the parabola down to be skinnier than the parent function. And then, the addition of these constant terms shifted the graph over.

Because we started out with knowing about the graph, we were able to write our equation for this new parabola right away in this form where we can see the vertex. What I'm interested in is what this would look like if we wrote it in the standard form for the equation of a parabola. So let's try to move toward that. Start by multiplying out the terms on the right-hand side of the equation. Write that new expression that belongs on the right-side in this box.

Multiplying the right-hand side out, gives us the expression 3x squared minus now that the right-hand side doesn't really tell us anything clearly about where the vertex is. And if we subtracted 2 again from both sides and got this fully in the standard form for the equation of the parabola, then we would really have no idea right away where the vertex was. Keep these two different forms, which are equivalent ways of writing the equation for this curve, in mind, as we move forward.

Thinking back to the different ways that we saw to write the equation in the last quiz, which of these four equations do you think gives us most direct access to the coordinates of the vertex of a parabola, where that parabola has a leading coefficient of a.

The answer is this one. Y minus k equals a times the quantity x minus h squared. This is almost the same as the equation that we were working with, in the last lesson. Remember that we kept talking about the form y minus k equals x minus h squared. And this was used for parabolas that had a leading coefficient of 1. So this makes perfect sense that if we have a leading coefficent of a instead of 1. We would modify the equation like this. Now, 2 of these equations here are actually not correct mathematical representations of what we're looking for. This one, y minus k equasl ax minus h. Doesn't graph a parabola at all. We can see that we only have an x to the first term, not an x to the second term. So this is a linear equation. That means it definitely can't be right. And the last one right here doesn't have the correct placement of the leading coefficient of a, as we can see from the last example we did. The second choice is mathematically equivalent to the third one, the correct one, except here, like we saw in the last quiz, we can't directly read off the coordinates of the vertex, or we wouldn't be able to if we had actual numbers plugged in. So this is the best choice.

So now we have this awesome new form for the equation of a parabola, y minus k equals a times x minus h squared. What I want to know is what we should call this form. I keep having to read it out each time I say it. And, as you know, I'm really lazy and I don't want to do that. I just want to have a name for it. So what should we call it? Should we call this the x-intercept form for the equation of a parabola? The y-intercept form? Vertex form? Minimum form? Or maximum form?

I think this should be called vertex form since, as we've seen, we can just read off the coordinates of the vertex, which is at h,k. That's why this form is so useful, and also because we can see from the coefficient of a about how wide the graph is going to be. Now, there may be some variation depending on what source you look at for what we should call this equation, but for the purposes of this class, we're going to call this the vertex form for the equation of a parabola. You may see other names elsewhere and there's not one that's absolutely right or absolutely wrong, but for the sake of speaking the same language, know that this is what I mean when I talk about vertex form.

Now that we know what vertex form looks like for equations of problems that have leading coefficients other than 1, let's see if we can get this equation in this form. We have y equals 3x squared plus 6x minus 10. Now what was that tool we used before to get equations in vertex form? It was called completing the square. So that's what we need to do here, complete the square. The process is going to be slightly different, since we have this 3 present now, instead of a 1 in front of the x squared. But, let's give it a try. Well, the first thing we did before, when we were starting to complete the square. Was recognize that this constant over here is probably not the constant we're going to need if we want a perfect square term on the right hand side of the equation. So we want to get the constant off the right hand side and onto the left hand side. So if we do that, what does our equation look like now?

You just need to add 10 to both sides and this gives us y plus 10 equals 3x squared plus 6x.

So far the process of completing this square looks pretty similar to what we did before when the leading coefficient was 1. But here's where things are going to get a little bit tricky, or at least a little bit different from what we've seen before. The next step that I'd like you to do is to factor out that leading coefficient that we have on the right hand side. So pull out this 3 from these two terms and then write what we need to multiply that by in this box. Think about why we want to do this based on the final form of the equation that we want to end up with.

If you factor out the 3 from 3x squared plus 6x, then on the right hand side of our equation we get 3 times the quantity x squared plus 2x.

Now if we compare this to the vertex form that we just came up with, we can see that we're already filling in things in most of the slots that we need. The left hand side is good to go. We have our a set with this 3, and we almost have this x minus h squared figured out. This isn't quite a perfect square yet though, like x minus h squared is. So what do we need to add on to the expression x squared plus 2x? Which we have inside these parenthesis, in order to have a polynomial that we can factor to be in the form, x minus h squared, where h is unconstant. Remember that if h is negative here, we end up with x plus some constant. So don't worry if you're wondering how to get x minus a number here. Just think about what we need to complete the square.

The number we need to add here is 1. Likely to be 4, we can get this by dividing the coefficient of the x to the first term by 2, and then squaring it. 2 over 2 squared is just 1. So that's what we add on

We just decided that we need to add a 1 into this expression, in order to complete the square. So we're going to want to have x squared plus 2 x plus 1 inside these parentheses. My question is can we go ahead and replace x squared plus 2 x with x squared plus 2 x plus 1, without changing anything else in our equation? Please answer yes or no.

The answer is no, we can't do that. If we're adding a 1 inside the parentheses, then we're adding something, some quantity other than 0 to the right side of the equation. And that means we have to do it for the left side as well, otherwise you'll completely change what graph we have.

So, we know that we need to add something to both sides, but what number is it? If we want to have the right side end up being 3 times X squared plus 2X plus 1, what number do we need to add to both sides of the equation?

The answer is 3. This is actually pretty hard I think, especially the first time you think about it. Since we know we want to have this quantity on the right hand side, we need to think about what we're actually adding when we entered this 1 inside the parenthesis. If we look at the entire expression, including the coefficient of 3 outside of the parenthesis, we actually need to add a 3 times 1 in order to get this. So that's going to give us the equation y plus 13 equals 3 times the quantity x squared plus 2x plus 1. Awesome.

We are one step away from having our equation in vertex form. Look back at what vertex form is, and write that final equation in this box.

All we need to do is factor the right hand side of the equation. You can leave the left hand side exactly the same. Keep that leading coefficient and recognize that x squared plus 2x plus 1 is just equal to x plus 1 squared. So if our final equation is y plus 13 equals 3 times the quantity x plus 1 squared.

Now that we have our equation in vertex form, where is the vertex of this parabola?

It's at negative 1, negative 13.

Which of the graphs on this coordinate plane over here on the right represents the equation we've been talking about? Y plus 13 equals 3 times the quantity X plus 1 squared. Please note the scale on either axis. They're not the same. I just wanted to make these curves a little bit less skinny than they would normally be so you could tell them apart.

The answer is graph C, this light green one right here. We found out in the last quiz that the vertex of this equation is at negative 1, negative 13. So if you look for the curve on this graph that has that vertex, we have two actually that meet right here, C and D. If we plug in other points to this equation though, we can see that C is the one that fits.

So we just found the graph of y plus 13 equals 3 times x plus 1 squared. But what if I want to change the sign of a leading coefficient? Let's make it negative 3 instead of positive 3. Which graph matches this new equation?

Graph D matches this new equation. Note that the vertex hasn't changed, since the values of h and k haven't changed. Well if you plug in values now, since D is the only other curve besides C that has this vertex, that must be the answer. You could, of course, also plug in other points to verify this.

What do you think that we can say in general about our problem in vertex form that has a negative leading coefficient? What does this say about the shape of the graph, or rather the direction that the graph is pointing in.

If a is negative in the equation of this form, then the parabola opens downward. We saw before that the equation for the green curve was y plus 13 equals positive 3 times, times x plus 1 squared. And we can see that this is opening upward. The only thing that changed when we switched this to a negative was that the parabola now opens downward which is graph D. Except for playing the opposite direction Graph D is exactly the same as graph C. It has the same shape and it even has the same vertex.

Considering your new found understanding of the sine/g of the leading coeffecient for a parabola, which of the parabolas over here has a positive leading coefficient, and which ones have negative leading coefficients?

A, C, and E have positive coefficients since they point upward, and B, D, and F have negatively coefficients since they open downward. Your're welcome for giving your nice little pattern in your answers.

Now that we understand what the constants in the vertex form actually stand for and what role they play in creating the graph of an equation, let's continue practicing completing the square, so you can get more and more equations into vertex form and then graph them easier. So tell me where the vertex of the parabola Y equals -6X squared plus 20x minus 15 is. To figure this out, please use complete in a square to put this in Vertex form and then tell me the Vertex.

We know that we're going to need to complete the square to this equation for a parabola that's in standard form right now to be in vertex form eventually. So the first step to do that is to first check out whether or not the expression we have on the right side of our equation, the expression involving x, is in fact a perfect square right now. To my knowledge there's nothing that we can multiply by itself, or at least nothing pretty, to get this. So we're going to add 15 to both sides so that we can pick our own constant complete the square with On the right side. So that gives us y plus 15 equals negative 6 x squared plus 20. And since we have a leading coefficient over here that is not equal to 1 I need to factor out that leading coefficient so that the right side of our equation eventually has the form of a coefficient times some binomial where x with the coefficient of 1 is the first term. So I simply factor out the negative 6 from both terms on the right side. Then we need to figure out what constant term to add. This expression inside the parentheses, to make it a perfect square. Since I know that when I have a binomial squared, the coefficient of the middle term is just twice the number that we have as the second term in the binomial. We can find the constant that we're adding to the end by dividing The coefficient of the x to the first term by 2, then I need to square it since I have C squared as the coefficient on the end over here in our multiplied out polynomial. That gives me 25 over 9. Subsets belong inside the parentheses here, but it's not actually the number that we;re adding to both sides of the equation. That number is going to be negative 6 times 25 over 9. It looks like I forgot my negative signs here. Lets multiply 25 over 9 by negative 6 gives me negative 50 over 3. So that's the number that we need to add to both sides. Once I subtract 50 over into vertex form. Our final equation in vertex form then is y minus five thirds equals negative 6. Times x minus 5 thirds squared. Where is our vertex then? Well, we can read it off. It's at the point 5 thirds, 5 thirds.

Of all the work we've done to get equations in vertex form, I think we're very well equipped at this point to help Grant deal with his profit better. Remember the equation we came up with before P equals -200 C squared plus 18,400 C minus for the price that he sets for each wiper set also in dollars. So considering this curve, what is the maximum profit that Grant can make? And what wiper set price does he need to pick in order to gain that profit.

What we're looking for in this equation is the maximum value that p can have and the corresponding c value or the c value at which that p value occurs. Well remember what this curve looks like overall. It was a problem that opens downward. That means that the maximum point here is just the. The vertex. So that's exactly what we want to find. The C value and P value that we're looking for are just the coordinates of our vertex. Let's get this in vertex form, then. So we need to go through out typical process of completing the square to get this forumla in vertex form. Note that it actually skipped one step. That I included in the past. When I subtract this constant from both sides of the equation, I know that I'm doing that in order to get the right side to equal our perfect square. In the past I've written out what this entire polynomial is multiplied out, and then factored it as a step after that, but we don't really need to do that. The reason is that I already know the factored form of the polynomial that we're creating by adding this before I even add on the constant term. The number that I'm adding to c, this negative 46 in this case, is just the coefficient of the c to the first term divided by 2. Since that's the thing that we're squaring in order to come up with what we're adding the end to make this a perfect square. So going through all these steps gives us our final equation of p minus 290,000 equals negative 200 times the quantity c minus 46 squared. From here we can read straight off that our vertex must be at 46, that we were looking for. Grant will earn a maximum possible profit of $290,000 per month if he makes his wiper sets cost $46 each.

Now, we have all the tools we need to draw a parabola from an equation. Regardless of what form it's in, when we first look at it. We can find the vertex, and we can find the x and y intercepts. This will definitely give us enough points to connect into a smooth curve. So let's put all of this together. If we have the equation, y equals x squared plus 5x plus 4. What are the y-intercepts of the graph, its x-intercepts, and its vertex?

We can find the y-intercept of this curve by plugging in x equals 0. That gives us a y-coordinate of 4, and we already know the x-coordinate is 0, so the y-intercept is 0,4. X-intercepts are where y equals 0. Now, we need to factor to figure out what x values makes this equation true. The two equations that yield solutions that could each make this equation true are x plus 4 equals 0 and x plus 1 equals 0. So now, I just have to solve each of these. Our x-intercepts are then at negative 4,0 and negative 1,0. Lastly, let's find the vertex by putting this in vertex form. Our final equation in vertex form is y plus 9 4th equals x plus 5 halves squared. That means, that our vertex is at negative 5 halves, negative 9 4th. If you think about the fact that we talked about what parabolas are for the first time just a couple of lessons ago, it's pretty impressive how much you can do to manipulate them and find out about their graphs already. This is really awesome. Good job.

So here, I've drawn the graph for you of the equation we were just working with, y equals x squared plus 5 x plus 4. Now, of course, this is in standard form and we put it in vertex form earlier, so I could have written it that way as well. We have all our important points that we've picked out, labeled. But there's one more thing about the graph that we can talk about and this is actually super important. Notice that this and all other parabolas are symmetrical. There's a vertical line running right through the vertex that divides this parabola exactly in half. What I want you to know is what the equation of this vertical line is.

Since this is a vertical line, we know that its equation is going to be x equals some constant, since every point along this line has the same x-value. Since this line goes through the vertex, which has an x value of negative 5 halves, the equation this line[INAUDIBLE] must be x equals negative 5 halves. Since this line goes through the vertex, which has an x-value of negative 5 halves, the equation in this line must be x equals negative 5 halves. I like to keep this idea of the symmetry of parabolas in mind as we move forward. You probably already noticed on your own anyway, just from looking at all of our graphs. But think about why a parabola would end up with this kind of shape, because of the way its equation is written. In particular, think about the fact that the expression involving x is of degree 2, and how squaring something might create a symmetric pattern just like this.

Let's continue along this vein of symmetry of parabolas. Now when we were talking about linear equations, a super vital part of everything that we talked was the slope of a line. And we can find the slope of a line by taking any two points along it and taking the ratio of the difference between their y coordinates to the difference between their x coordinate. We haven't however considered this slope of a curve yet of a graph of anything that isn't just a straight line. This may seem a little bit funny to think about but I think its pretty important. There is a single sub that applies to every single point along this curve. But we can talk about slope any local sense. Its pretty easy to see what the slope of a single part of the curve is by picking a point on the line, lets say this one and drawing a line that just touches the curve at that point and has the same slope as the curve does right there. So for every point on a curve we can say that there is a line tangent to that point, that has the same slope as the curve at that point and barely touches the curve right there. So it's actually really hard to draw an exact tangent line, but this is a pretty good approximation. We pick two points that are infinitesimally far away from this point or in other words two points that are as close as we can possibly pick, to this point that we're interested in. We can draw the line connecting those two points and then extend that line, tend up with the tangent line. Now you have to wait for calculus to get a really deep and rigorous understanding of tangent lines, but they're still really useful for us for understanding the shapes of parabolas and of other curves as well. So now I've drawn 3 different lines that are tangent to 3 different points on this parabola. I've labelled them A, B and C and for each of those, I'd like you to tell me what the sign of the slope of the line is. Is it positive, negative, or 0? Check the proper box for each line.

The slope of line A is negative. The slope of line B is 0, since it's horizontal. And the slope of line C is positive. This is pretty interesting. I think this gives us a different way of quantifying how the shape of our curve is symmetric. As we move from one side of our parabola across the vertex to the other side, the sign of the slopes of our tangent lines as we move from point to point to point, switches direction. This is going to be true of any parabola. Although, you can imagine that if this parabola flipped upside down, you might have filled in different answers for these three lines.

Here are two more tangents lines. Actually two was one of the three that we had in the last quiz. And these two lines are tangent to two points on the curve that have the same y value but different x values. My question is how the slopes of lines one and, and two compare to one another. Which of these do you think is correct? Do you think that m1 and m2 are equal, that m1 equals negative m2, that m1 equals negative 1 over m2 or that m1 equals m2 over 2? Remember that m1 is the slope of line 1 and m2 is the slope of line 2.

m 1 is just equal to negative m 2. These slopes are the same magnitude, but opposite signs. This makes sense, intuitively, in terms of how our graph is shaped. The slope right here, is as steep as the slope right here, just pointing the other direction. This is going to be true of any two points, that are directly opposite one another, across that line of symmetry we drew before. This is a way then of quantifying that symmetry of the graph. You can imagine also that if we flipped one side of the graph over that line of symmetry to the other side it would map directly onto it.

Let's talk a little bit more about the technique of completing the square. Now we've used this a lot to figure out how to get equations for parabolas in vertex form but we can also use it to factor expressions and to solve other equations. It's the technique that could be used in a ton of different circumstances. First things first, let's see how we can use this to factor. Now earlier on in this lesson, we saw that we could write this in a modified version of the equation in vertex form as y equals the quantity x minus 4 squared minus 1. Notice that the first term here is a squared term and the second term, apart from the minus sign, is also a squareed term. Because we have one squared term that's being subtracted from another squared term, we can call this an instance of difference of squares. In genereal, if we have two squared terms, a squared and b squared and we write a squared minus b squared then remember that we can factor this in an interesting way. This is just equal to a plus b times a minus b. Match up our difference of squares here, with this general form for the difference of squares. Can you tell me what a and b are, in the case that we have right here on the right side of our equation?

A is equal to x minus 4, the first thing that we're squaring. And b is equal to

Now what I'd like you to do is factor our expression X minus 4 squared minus 1 in this way, like we can with all differences of squares. Remembering what A equaled and what B equaled in the last quiz, write this as A minus B times A plus B.

a plus b is x minus 4 plus 1 and a minus b is x minus 4 minus 1. Simplifying this gives us the quantity x minus 3 times the quantity x minus 5. Please note that this is exactly what we got when we factored. So really when compared to factoring completing the square was a lot of work in this case. But there are plenty of cases where we have expressions that we can't factor in which completing the Completing the square will actually help us. It'll make our lives easier. Here it wasn't really necessary because we could of factored the initial expression.

Let's look at different equation then. One like this one. Y equals X squared plus 6x plus three. Now we can't factor this. Know that the only factors of 3 are 3 and 1, and they don't add to equal 6. So we're going to need to come up with a different way, a different method, of figuring out where this curve touches the x axis. I'd like you to write this Right side of the equation in completed square form. So, in other words, complete the square on the right side. However, this time, don't move the 3 over to the left-hand side. Keep it on the right side and just play with the first two terms here to complete the square. Once you've completed the square with these two, tack the 3 on to the end again and simplify. Fill the new expression you have on the right-hand side into this box.

Let's start by considering just these first two terms, x squared plus 6x, and we want to complete the square with these 2 terms as the first 2 terms of that squared polynomial. Then the constant we need to add at the end is 9, since 6 over 2 is 3 and 3 squared is 9. However, if we want to add 9 inside here, without actually adding anything total to either side of the quation, then we need to talso subtract 9 from somewhere on the right side. So that's exactly what I did. I wrote plus 9 inside the parenthesis, and then a minus 9 next to the plus 3 outside. Remember that because all of these terms are just added together, these parenthesis here are really just for notation. They're just to help me visualize that these terms belong together. This doesn't, in any way, change the way these terms are combined. The factored form of x squared plus 6x plus 9 is x plus 3 squared, and negative 9 plus 3 is just minus 6. So the completed square form of y equals x squared plus 6x plus 3 is y equals the quantity x plus 3 squared minus 6.

Now, you might be wondering why on Earth this is helpful at all, well, remember, what we're trying to do here. Since we can't factor this expression, we need a different way to find the x-intercepts. Here, we can tell that we're getting close to being able to do that. At some point, we want to set y equal to zero, and now we only have x written in one spot in our equation, rather than in two. We know how to take square roots, so I feel like we're getting pretty close. At this point, we need to think about the difference of squares again. Remember that if we have a squared minus b squared, we can factor this as a minus b times a plus b. Here, we have one term that's clearly squared, so maybe we can use this technique to help us out. Now, 6 doesn't look like a squared number, but that doesn't really matter. We can stil,l of course, take the square root of 6 and get a number. It may not be an integer but it is a real number. What I would like you do now is keeping all that in mind, write out the two factors in this form that we have coming from this expression right here. This is going to look pretty different from anything we worked with before and you may not think it looks very pretty but just go with it, this is going to be super, super useful in everything we have coming up.

If we're thinking of x plus 3 squared minus 6, as being in the form a squared minus b squared, then in our case a is equal to x plus 3, and b is just equal to the square root of 6, since the square root of 6 squared is just 6. That means that to factor this we need to take this minus this times this plus this. So, this is our final answer, y equals the quantity x plus 3 minus the square root of 6 times the quantity x plus 3 plus the square root of 6.

Now that we've figured out how to factor, this equation, can you tell me what the x-intercepts of this graph are? Please round each of your numbers to two decimal places if you need to.

In order to find out where this curve hits the x-axis, we need to set y equal to We could either have the first factor equaling 0, which gives us the equation x plus 3 minus the root of 6 equals 0. Or we could have the second factor equals equals negative 3 plus the root of 6, and x equals negative 3 minus the root of are just the x-coordinates of our two x-intercepts, and of course the y-coordinate of each is 0. So this is actually really incredible what we've done. We figure out how to factor over not just the integers but over the real numbers. In other words, we can use completing the square and understanding difference of squares to find the x-intercepts of a ton of different equations for parabolas, not just ones that we can factor in the traditional way. Oh, and I almost forgot. The rounded version of these two answers are negative .55, 0 and negative 5.44, 0. We're going to keep working with the idea of finding x-intercepts of any given parabola in the next lesson. Awesome job on a lot of pretty complicated material.