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Contents

## Udacity Expo Game

Grant is still at the Udacity Glasses Expo. And to get people more excited about his family of glasses wiper products, he's decided to add a game to his booth. Anyone who wins the game gets a free pair of deluxe glasses wipers, the ones with the awesome curlicues on them. The game is pretty simple. The person playing uses this giant slingshot that Grant built to try to get a ball through a hoop that's attached way up high on the ceiling. Grant has another idea of a way to soup up his booth though. The game isn't quite enough for him. A gigantic sign with his brand name on it. That way no one will ever forget what his company is called. The question is, where should he hang it? He needs to make sure that it doesn't get in the way of shots from the slingshot that would otherwise have gone through the hoop.

## Linear Slingshot

He's decided that the best way to solve the slingshot sign problem is to use math. If he can come up with an equation describing the trajectory of a ball that makes it through the hoop, he'll be able to find every single point that the ball has to pass through to get there. And this will tell him exactly where not to hang his lovely new sign. If we call the horizontal position of the ball at any given time x, and we call its vertical position at the same time y, then here's a list of the x and y values of the ball at different points throughout its trajectory. If we think of the space above the ground at Grant's display booth as an x y coordinate plane, where x is horizontal and y is vertical, then these are just the points that the ball passes through on its way to the hoop. Looking at the coordinates of these points, do you think that we can find a linear equation that goes through all them? Please answer yes or no.

## Linear Slingshot

No, we can't. So what on earth do we do now? Pretty much all of the equations we've dealt with in the course so far, we've been able to put in the standard form for a linear equation. I guess we're going to have to come up with some new kind of equation.

## Slingshot Graph

Maybe we'll get some ideas about what to do about these points if we graph them. That might help us move a little bit closer to an equation. So which of these graphs on the coordinate plane right here matches the coordinates in the x y table over here? Please select the letter next to the graph that you think contains all of these points.

## Slingshot Graph

The answer is graph E, this reddish curve right here. We can see that it goes to the point 4, 0, the point 16, 18, and the point 28, 0. All of the other graphs on here go through at least one of these points, but none of the others goes through all three. And in fact, remembering that this curve is supposed to be the trajectory or the path that Grant's ball takes through the air, the curve really should look like this. The point down here, 4, 0, is where the ball leaves the slingshot. The point at the top of the curve is where it goes through the hoop. And the point down here is where it lands on the ground.

## Slingshot Equation

Now that we have a graph that fits the three points we knew were part of the ball's path, the next question is what kind of equation fits with this graph? Here are a few options for you. Which of these equations do you think might work? Just a hint. Pick a few points on this parabola, and then check and see if all three of them satisfy any of these equations.

## Slingshot Equation

The answer is y equals negative x squared over 8 plus 4x minus 14. This is the only equation that is satisfied by all three of the sets of coordinates that we saw in our x y table before. And it's the only equation of these equations that will fit any point that we pick along this path. Since every point along the curve satisfies this equation, this must be the equation that produces this curve. Now that's great, except that this equation and this graph both look really different from anything we've played with before. It looks like we've got some exploring to do.

## Find the x Intercepts

Even though this parabola looks really different from the graphs of linear equations we've been playing around with recently, we actually know a fair amount about it. Start off by telling me the x-coordinates at the point where the curve intersects the x-axis or the x-coordinates of the x-intercepts. There should be two of them.

## Find the x Intercepts

The x-intercepts of this graph are at 4, 0 and 28, 0. So the two numbers we need for the x-coordinates are 4 and 28.

## Find the y intercept

Next, what is the y-intercept of this graph? You may notice that it's not actually shown here because I cut the graph off above where it is. But you can figure this out based on the equation alone. Remember what you know about the coordinates of the y-intercept from the fact that it has to lie along the y-axis.

## Find the y intercept

The y-intercept is at 0, negative 14. We know that the x-coordinate has to be just the line x equals 0. If we substitute 0 in for x in this equation, these two terms both turn into 0, and we end up with y equals negative 14. So that's the y-coordinate.

## Find the Highest Point

How about the highest point on the graph? What are the coordinates of that?

## Find the Highest Point

The highest point on the graph is right here where Grant's ball is going to go through the hoop. Looking at our graph, it looks like that point is at sixteen, eighteen. This point that lies halfway along the parabola, which we can think of kind of cutting the parabola in half, and which in this case is the peak of the curve, is known as the vertex. We'll talk a lot more about this in a little bit.

## Parabolas and their Equations

So now we've seen several curves that look like this and several equations that have forms similar to this one. So it's important that we talk about what each of these really are. This kind of curve is called a parabola, and a parabola is a curve that's produced by equations that have certain features. So let's talk about what those features are. The equation that produces any parabola can be written in what's called the standard form for the equation of a parabola, y equals Ax squared plus Bx plus C. Now in this equation, A, B, and C are all constants, and there's one more important restriction. A, this first coefficient, is not allowed to be equal to 0. But why not? That is what I would like you to tell me in this quiz. Why do you think A can't equal 0? If that happened, would that stretch the parabola? Would it shift the parabola? Would it flip the parabola upside down? Would the graph, instead of a parabola, become a vertical line? Would it be a straight line instead? Or would the graph completely disappear? Pick the explanation down here that you think is the best choice.

## Parabolas and their Equations

The answer is that the graph would become a straight line if the coefficient a equaled zero. If we substitute zero in the place of a, then this entire first term becomes zero since zero times anything is zero. Then we would just be left with y equals Bx plus c. We can recognize this as the slope intercept form. Of the equation of the line. Where B here is serving as the slope, and C is serving as the Y coordinate of the Y intercept.

## Which Graph Parabolas I

Considering the general form that you just saw, which of these four equations graph parabolas? There might be more than one that does, so check as many as you want.

## Which Graph Parabolas I

These three equations all graph parabolas. Even though they might not quite look like it right now. With a little bit of rearranging though, all of these can be put into that general form for the equation of a parabola. Which you'll remember, is y equals a x squared plus b x plus c, where a is not equal to zero. Now this first equation, 3 x squared plus 3 x minus 7 equals zero, does not contain a y variable. The only variable it has is x. So this is just an equation that we can solve for x. But it's not something that we can plot. There's no information here about y that would let us graph on a coordinate plane. Or at least, not on the kind of coordinate plane that we've been interested in. With two variables. This equation actually already is in standard form. The y just happens to be on the right side, and the other expression on the left side. But you can see that A, equals negative 1. B, the second coefficient is 5, and C is negative 6. This one is also in standard form. We have A equals 1, B equals 0, and C equals 1. However, the only requirement we have, about general form Was that a couldn't be 0. Either of the other two constants can be and will still get a parabola, I'll show you the graphs for all of these in a second. But first let's look at this last equation briefly. The only thing that is preventing it from being in this general form is the negative sign in front of the y. If we just multiplied both sides of the equation, by negative 1 and rearrange the terms on the right hand side, we end up with an equation of a parabola in the form of. We can recognize. Y equals X squared minus 2X plus 1. Now let's pick up these graphs. This third parabola down here is produced by tehe equation y equals negative x squared plus 5X minus 6. This purple corresponds to Y equals X squared plus one. And the orangeish yellow one of y equals x squared minus 2x plus 1. Now just take a second here and think about what's different about these graphs and what's different about their equations. Just form some ideas in your head and keep them in mind for later. We'll keep revisiting a lot of different kinds of parabolas. And of course the differences in their equations that produce them.

## Which Graph Parabolas II

So here's a second set of equations. And I'd like you to answer the same question. Which ones of these graph parabolas?

## Which Graph Parabolas II

Only these two graph parabolas. Which might surprise you. In particular, this equation, y equals x squared plus 2 xy plus 1 does not graph a parabola. Even though there is a x squared term, there's also this term 2 xy. Looking back at the general form, we don't see any terms here that have both x and y in them multiplied together, so this does not fit the bill. We'll see this graph in just a second. And, of course, you'll recognize this graph as a straight line. This is linear. Think to yourself what the slope is and what the y-coordinate or the y-intercept is. This last answer may also have been a little bit tricky for you. I know that right now this doesn't look at all, like this general form equation of a parabola. But if we distributed multiplication by what's in this first set of parenthesis towards on the second one, we get something pretty interesting. In fact, let's look at the graphs for all four of these right now. So here are the graphs for all 4 of the equations we just looked at. This funny red curve, which has a big break in the middle, is the graph that corresponds to the equation that we've never seen anything like before. Y equals x squared plus 2 xy plus 1. How funny that this isn't even in one piece. This is called a hyperbola. We're not really going to deal with them much. But it's important to recognize that having a term like xy makes an equation no longer equation of a parabola. This solitary straight line is of course y equals 5 x plus 7. The yellow curve, is the parabola y equals 3 x squared plus 2 x. And this last curve left, this big blue arc, is y equals negative 1 8th, x minus 4, times x minus a term with an x squared in it, as the only term of degree 2. And it does, indeed, graph a parabola.

## How to Draw a Parabola

Now that we know what parabolas look like, it's time to talk about how to actually draw them. I mean, what you do with your hand to make them appear. First things first, let's pick an equation to experiment with graphing with. I want to use y equals x squared minus 2x minus 3. Just like with linear equations, for quadratics we can use x y tables to come up with a bunch of points that should lie on any given line or curve. In fact for quadratics, since we're graphing these fancy curve parabolas, I find it's much easier for me if I write down a bunch of points. When you're drawing a line, you really only need to mark two points and then connect them together. So I'm going to use these points to graph this curve right now. I'll of course label my axes. Here I'm just going to use a scale of 1 for each grid mark. Wonderful. Now that I have all of these points marked, I can start sketching a line that looks like it'll connect all of them. I like to start at the vertex, at this lowest point in this case, and then moved in a curved way up to the next point to the left. Definitely takes practice to get your hand to move smoothly like you want it to, but there, that's not too bad. Moving to the right is usually a little bit harder for me. Takes a bit more sketching. So you can see I am doing little strokes to make sure I'm moving my pen in the direction I really want the line to go. Now it's a half a sort of rough sketch. I can go over it to make it smoother. And voila, we have a parabola.

## Two Equations

A few minutes ago, I showed you the graph of y equals negative 1 8th times x minus 4 times x minus 24. Now this time I've graphed a graph that looks pretty similar to that. I just changed this to a 28 instead. Anyway I'd like you to inspect this graph that we have now. Now while your'e looking at this graph, I'd like you to think back to the first parabola we dealt with in this lesson. Y equals negative x squared over 8, plus 4 x, X minus 14. Remember, this was the curve that we were using to describe how the ball would fly out of the slingshot and go through the hoop at Grant's display booth. For this quadratic equation, we found it's y-intercept, it's x-intercepts and it's maximum, as these points right here. Take a second and compare these points to this graph, and then think about how these equations might be connected. Here are a few different options for how these two equations might be connected to one another. And I'd like you to pick all of the ones that apply. Is there no connection between them at all? Do their graphs share a few points? Do their graphs go through all of the same point? Do they graph the same curve? Or are the equations equivalent to one another?

## Two Equations

The last three choices are correct, and they're all really saying the same thing. If two equations are equivalent to one another, then they have all the same solutions. And if they have all the same solutions, and if they have all the same solutions, then they graph all the same points, which means they graph exactly the same curve.

## Information from Equations

So these two equations give us two different ways of expressing the exact same curve. But considering this one with the parentheses might not look quite as pretty as this other one, why would we ever want to use the second form? Well, from it we can actually find out some information about the curve before we even graph it. If I label this first form of the equation 1 and this second form of the equation 2, then what does equation 2 automatically tell us about that we can't see right off the bat with equation 1? Thinking about the curve that it graphs, this parabola over here, can we learn very quickly about the maximum or minimum, about the x-intercepts, or about the y-intercept from equation 2?

## Information from Equations

Right here in the equation with the 4 and the 28, we can see the x-coordinates of the two x-intercepts on this graph appear subtracted from x, and the two factors that are multiplied together. Interesting. Keep this in mind as we move forward.

## Intersecting Lines

We looked at solving linear equations graphically. We looked at the intersection of two lines. And we know that the values of the coordinates, of those points of intersection, would satisfy, either equation passing through that point. So we're considering, the x intercepts, of our lovely parabola, that we now have two equivelent equations for. What lines do we need to think about intersecting, at either one of these points? I'll give you a hint, that it's the same two equations at each point. And if you decide that this parabola is one of the lines that you should use, then it doesn't matter if you write the equation in this form, or in this form. Fill in the rest of these two equations right here, that we are interested in.

## Intersecting Lines

The x-intercepts are the points where the graph of our parabola, whose equation is y equals negative x squared over 8 plus 4x minus 14, or of course written in this other form is perfectly acceptable. So where that line meets the line y equals 0, which is just the x-axis.

## Equation of Intersection

Since we know that we're interested in the intersection of these two lines, what equation can we create out of these two different equations that we can solve to figure out where their points of intersection are?

## Equation of Intersection

Since we know what points of intersection, the y value for the two lines that are crossing through there will be the same, we can set 0 equal to negative x squared over 8 plus 4x minus 14. And that's the equation we need to solve. The solutions to this equation, values of x, will give us then the x values of the points where these two lines intersect.

## Removing Factors

So now we have, from the last quiz, the equation, negative x squared over 8 plus know both describe our parabola here, are equivalent to one another. So that means, we know that negative x squared over 8 plus 4 x minus 14 is equal to negative 1 8th times x minus 4 times x minus 28. So, we can set the right-hand side of this equation equal to zero through simple substitution. Wonderful, now we have another equation. This one, however, is actually pretty easy to solve if we think of a little trick. First things first, we're going to simplify. Can you think of a really, really easy way to simplify this? In particular, I'd really love to get rid of this factor of negative 1 8th. The first thing I'd like you to tell me is what we can do to both sides of this equation to get rid of this factor of negative 1 8th. In this little blue box, please write some sort of operator and then a number. Plus, minus, multiply or divide, and then whatever number you think we should do that with. Once you figure out how to get rid of this negative 1 8th, I'd like you to write the new version of the equation in this box right here. So, the equation I want here is just the equation that we have after we get rid of that factor of negative 1 8th.

## Removing Factors

We need to divide both sides by negative 1/8 to make this factor on the left side disappear. Over here, that will leave us with x minus 4 times x minus 28. And on the right side, we'll still have 0.

## The Mystery of a and b

The answer is yes. We do in fact know something about the value of at least one of these variables.

## What we know about a and b

So there is one number that either a or b or maybe even both should be equal to. What is that number?

## What we know about a and b

1. That is the number that we're looking for. Now, we think about either a or b or really any real number. There are three options in terms of their value. They can either be positive, they can be negative, or they could be neither positive nor negative, in other words, 0. Now if we multiply two positive things together, we get a positive number. So a and b can't both be positive or else we wouldn't get zero. Two negative numbers multiplied together also gives us a positive number. So a and b can't both be negative either. A positive number times a negative number, or, the reverse order, a negative number times a positive number, both equal negative numbers. That's also not what we're looking for. The only way to end up with a 0 as the product of two numbers is for one of those two numbers that are being multiplied together to equal 0 itself. So either a or b or both a and b must equal 0.

## Splitting Equations

So that means then, that we might have two ways to get an answer of 0 over here. Either a could be 0 or b could be 0 or in some cases both, if a equals b. Now that we know that either a or b or both need to equal 0 if we have an equation of the form a times b equals 0, that means that to make this equation true, we either need to have x minus 4 or x minus 28 equaling 0. So then what two equations can we get out of this one equation that will each give us a solution to this equation. I know this is kind of a tricky, subtle question, but think back to what we learned about the ab equals 0 equation to figure this out. We should end up with two expressions, each set equal to 0.

## Splitting Equations

We know that either this factor needs to equal 0, or this factor needs to equal 0, to make this entire expression on the left side equal to 0. So that means we can either have the equation x minus 4 equals 0, or we can have the equation x minus 28 equals 0.

## Two Solution for x

Now we have two simple equations that we can very easily solve. What do you get for x when you solve each of them?

## Two Solutions for x

Solving for x, we get x equals 24 for this equation, and x equals 28 for this equation. Interestingly enough, these are the x-coordinates of the two x-intercepts of our graph. So remember what we actually did? We started off by picking out 0 as the y value for which we wanted to find corresponding x values. On this parabola. Plugging in that value for y, or in other words substituting in 0 in the place of y in the equation, we solve for x. To get the corresponding x coordinates at the various points on the graph where the y coordinate is 0. So the reason that we have two answers here is because there are two points on the graph, 4,0 and 28,0 that are the same Y value of zero.

## Intersecting the x Axis I

In the last few quizzes, we figured out that if could get a quadratic equation in this form right here, with two factors involving an x to the first power multiplied together. Then we could pretty easily find out where its x intercepts are. Or where that parabola cuts the x axis. Awesome. Just to cement that idea, let's do one more practice example involving these concepts. Let's say then, that we have the equation y equals x minus 4 times x plus 3. If I want to find out, where this curve, intersects with the x axis, or in other words, where its x intercepts are, what equation do we need to solve? Please write your answer in this box. Also, at this point, we don't need to multiply out anything, or simplify at all. This is really just a one step problem at this point.

## Intersecting the x Axis I

Since the x-axis is the line y equals 0, you just need to substitute 0 in as the value of y, which gives us the equation 0 equals the quantity x minus 4 times the quantity x plus 3.

## Satisfying Values of x

Now we're going to want to solve our new equation, which you'll notice I switched the order of the sides, because I just like looking at it better this way. We now have the quantity x minus 4 times the quantity x plus 3 equaling 0. Now the question is, which values of x satisfy this equation? I'm not going to tell you how many values there are, but I'd like you to write as many as you find all in this box. And if you get more then one answer, I'd like you to separate each of your answers by commas.

## Satisfying Values of x

We know that either this factor, x minus 4, or this factor, x plus 3, have to equal 0 in order to make this entire expression equal to 0. That means the answers for either one of these equations will satisfy this one. So our two numbers that x can equal then to satisfy this equation are 4 and negative 3. Let's look at a graph of this real quick. To help you visualize this a little bit better, here's the graph of the parabola corresponding to our original equation, which you'll recall was y equals x minus 4 times x plus 3. It's really easy to mistake this equation that we're solving with the equation for the parabola. But please remember that this is the equation we get when we plug in a certain y value. This does not relate x to y, and therefore it can't graph all these different points. Looking at the full problem, we can however notice the value of what we found. Here's the point, negative 3, 0, one of our x-intercepts, which we found right here, and here's the other one, 4, 0. Please remember that x-intercepts are actually points, because what we're solving for here are the x-coordinates of those points.

## Intersecting the x Axis II

Now it's your turn to try one totally on your own. Here's a quadratic equation. Y equals the quantity 2x plus 1 times the quantity 3x minus 2. And I'd like you to tell me where the parabola that corresponds to this equation intersects the x-axis. Please make sure that whatever answers you give, whether there's one or two or three or however many, should be written as points. So I don't just want x values. I want the actual points of intersection.

## Intersecting the x Axis II

The first step here is to recognize that what we're looking for is the value of x-coordinates where y equals 0 along this curve. So we need to set y equal to find values for x that will make this true. So we can either have 2x plus 1 equals 0, or we can have 3x minus 2 equals 0. If either one of these equals 0, then this entire side of the equation will equal 0, and we'll be good to go. Now we just need to solve for x. Solving for x gives us a value of x equals negative these in, we will make this equation true. But remember, I asked you for points. And in particular, I asked you for what the x-intercepts of this equation are. We can just read them off now, using these as our x-coordinates and remembering that our y-coordinate has to be 0 for an x-intercept. Voila. Two lovely points.

## Factoring

In one of our earlier quizzes when you were deciding which curves were parabolas and which ones weren't, we saw the equation y equals 3x squared plus equation we need to start by setting y equal to 0, and then the next step after that will be to try to get the expression on the right hand side, with the x's in it, into a form where we have one factor multiplied by another factor. Where both factors have an x to the 1st power in them. And in fact, you already know how to do this. All this involves is factoring. Remember, that means looking for a common factor between terms, and putting it outside a set of parenthesis so that it will And multiply in. So, what is a way that we could rewrite this equation by pulling out a common factor that both of these terms have. Please fill in one of those factors here, and another one here.

## Factoring

Notice that 3x squared and 2x both have a factor of x in them. If we divide with 2. So factoring this, 3x squared plus 2x, gives us x times the quantity 3x plus 2.

## More x Intercepts

Now that we have our equation in this form, factored and everything, I'd like you to tell me at what values of x does this parabola cross the x-axis. I'll write an x equals here outside of the box to remind you that we just want values of x or x-coordinates of the points where this happens. Now, I'd like you to write all the different values you get in this box, separated by commas.

## More x Intercepts

In order to satisfy this equation, we need to either have x equals 0 or need to have 3x plus 2 equals 0. X has conveniently already been solved for here, and this equation gives us a solution x equals negative 2/3. So x can either take on a value of 0 or it can take on a value of negative 2/3 to make this equation right here true. The x-coordinates of the points where this parabola crosses the x-axis are 0 and negative 2/3.

## Degree of a Quadratic Equation

We saw earlier that we can rate the standard form for the equation of a parabola as y equals Ax squared plus Bx plus C. So the equation that produces any parabola should be able to be written in this form. And recently what we've been doing is taking equations that are written in this standard form and replacing y over here with 0. So in general that gives us something like this. Now we know that this is no longer an equation for a parabola. This is not going to give us any points to graph since we only have x as our single variable. This value of 0 isn't varying, so it's not a variable. It's just a constant. This kind of equation, or an equation that you can write in this form, is called a quadratic equation. Now a quadratic equation has one of the same restrictions that the standard form for the equation of parabola did. This coefficient A here is not allowed to equal 0. What I would like you to tell me now is what the degree of a quadratic equation is. Think back to what you learned about the term degree way back the very beginning of the course.

## Degree of a Quadratic Equation

The answer is 2. If we look at this equation and we require that a can't be 0, then we'll always have a term with an x to the 2nd power in it. And this will always be the term with the highest degree. So that means that quadratic equations have this degree as well.

## Quadratic Equation vs Equation of a Parabola

So you've now seen these two equations that look pretty similar to one another several times. We have y equals Ax squared plus Bx plus C. And we have 0 equals Ax squared plus Bx plus C. What I want to do with this quiz is make sure that we understand the difference between the two of these equations. When do we use each one, and what purpose do they each serve? So I've given you some different options over here, four different options. And each one of these describes either this equation or this equation. So for each of these four things, please pick the equation that it describes. Which one of these graphs a parabola? Which one finds the x-coordinates of the x-intercepts of a parabola? Which one relates y to x? And which one can be solved? Please pick the equation for each that best fits this description.

## Quadratic Equation vs Equation of a Parabola

Remember that this first equation, y equals Ax squared plus bx plus C, is the equation of a problem written in standard form. Because this contains two variables, x and y, we can use it to graph points on a two dimensional. Plane. For any value of x that we might want to consider, we can come up with a corresponding value for y. And then, of course, plot that point on our plane. We can't do that with this equation over here. Because there is no variable on the left side. The only variable in the quadratic equation is x. So clearly, the equation of parabola relates y to x. Whereas, the quadratic equation does not. However, the quadratic equation is very powerful and useful for us in it's own ways. It gives us a very straightforward way for finding the x-coordinates of the x-intercepts of the parabola that it corresponds to. And of course, we could find that equation, the equation of the parabola, by replacing the zero with y again. The other thing that's very useful about a quadratic equation is that it can actually be solved. Because we have one variable x in this equation and there's only one equation, we can isolate x and find some number of values that it needs to equal in order to make this equation true. So the equation of a parabola can yield a graph, A quadratic equation can yield solutions. For values of x, that tell us a lot about these graphs of parabolas. I know this is a really subtle distinction, so if this seems a little bit confusing at all, that makes perfect sense. I myself messed this up several times. Several crucial times, before I really got the difference, so just keep reminding yourself of the distinction, and when we use each one.