The last conic section we'll learn about is a parabola. Parabolas only have one part to their graph. Circles on the other hand have two parts to their graph. And the reason they have two parts is because we have an x squared term and we have a y squared term. Ellipses also have two parts to their graph. And again, notice that the equation for the ellipse has an x squared term and it also has a y squared term. And finally, we saw the hyperbola. The hyperbola also has two parts to its graph. And notice, in the equation, we have the x squared term and we have the y squared term. But a parabola only has one piece. The parabola can either open up or down, or the parabola can open left or right. Each of these is one graph. Of one parabola. These two equations are the general equations that describe parabolas. This one describes parabolas that open up or down, and notice that it's y equals. This equation describes parabolas that open to the left and that open to the right. And notice that it starts with x equals. Throughout this lesson, we'll learn more about each of these values and what they do to the graph. But for now, let's take a look at the most basic parabola. y equals x squared.
The most basic parabola has the equation y equal x squared. Now, we can tell that the center, or really the vertex of this parabola is at the origin, 0 comma 0. The parabola opens in the upward direction, and it has an a value of Since it's not listed, we know that the coefficient is a 1. Another interesting part about parabolas is that they're symmetric. If there's a point on the left hand side of the graph, then there's another point equally distant away from its axis of symmetry on the right hand side of the graph. For example, we have the point negative 3, 9. Then, on the other side of the graph, we have the point 3 comma 9. We also have the point negative 2, 4. And on the right side of the graph, we also have the point 2 comma 4. Two units in the horizontal direction right, and then four units vertically up. The lowest point, or the turning point for our parabola, is the vertex. It's at the origin, 0 comma 0. Now we could make an x and y table and create a table of values to get our graph. But an easier way to get a couple points on the graph is to look at the a value. We're going to zoom in on this parabola. We can find the basic shape of this parabola by looking at the a value. The points to the left and to the right of the vertex will be exactly 1 unit above the current point. So we move one unit left, and one unit up. We move one unit up since the a value is one. And then to get the symmetric point, we move in the opposite direction. One unit right, and one unit up. So, we can see that we have a point of negative 1 comma 1, and a point of 1 comma 1. Let's make another graph and see some comparisons. For the graph y equals 2x squared, we also get a parabola. This parabolas vertex is at the origin, its still 0 0. But what's different the a value. Now the a value is 2. So to get the basic shape of this graph, we'll move 1 unit 3 and 1 unit left from the vertex. Then we'll go a units up in this case 2 units. So 1 unit left and 2 units up. This gives us our first point at negative 1, 2. Then we can move 1 unit right. And 2 units up to give ourselves another at 1, 2 and here is a 3rd parabola y equals 3x squared. Notice that the turning point of the vertex where this parabola is still at 0,0, the origin. To find other points on this purple parabola we're going to use our a value. We can move 1 unit left and 3 units up. This gives us coordinate of negative 1, 3. To find another point on the graph, we'll go the opposite way. We'll go 1 unit right, and this time 3 units up. This gives us a point of 1 comma 3. The a value is always a quick way to get 1 point that is to the left of the vertex. And 1 point that is to the right of the vertex whenever parabola is in the form of y equals. Notice 2 that as the a value gets bigger our parabola stretches vertically or in other words it gets more narrow.
Let's consider a different equation, y equals negative 5x squared. We know the vertex for this parabola will be at the origin, 0,0. We can easily see that, since every we plug in an x value of zero, zero squared will be zero. And zero times negative 5 would give us a y value equal to zero. But what are some other points on the graph. What I'd like you to do is use the a value to determine two other points that are on this parabola. When you think you've got them, check them off on the graph.
The two other points on this parabola are negative 1 and negative 5 and positive 1 comma 5. Good thinking if you checked those 2. What's so different now is that our graph is going to open downward since we have a negative value in front of the x squared. To find our basic shape, we'll start at 0, 0, the vertex. Next we'll move one unit left and five units down. This gives us our first point of negative 1, negative 5. Remember, the a value will always tell us how many units to move up or down when our parabola is in the form y equals. And since our parabola is symmetric about the y-axis, we can also go one unit to the right, and five units down. This gives us our other point of 1 comma negative 5. One of the key takeaways from this exercise is that a parabola in the form y equals will always open upward, or it will open downward. If the a value is positive, like in these three, we know the parabola opens upward. But if the a value is negative, like in these two graphs, we know the parabola will open downward.
Based on what you've seen so far, in which direction do you think this parabola would open? Would it open up, down, to the left, or to the right?
This parabola would open to the left. Great thinking if you got that one right. One of the key things to recognize is that this equation is in the form x equals. Parabolas that are in this form, either open to the left or to the right. Since the a value is negative 1, we know the parabola will open to the left. If I had a different parabola like x equals 3y squared, then we can see that, that parabola opens to the right, since the a value here is positive. Let's take this one away for a moment and just concentrate on the first parabola. Again, its vertex is at 0,0. This is the turning point for the parabola. Notice also that there's an axis of symmetry that goes through the vertex. It means that if there's a point on this side of the graph, then directly across from it is another point. For example, we know the point negative 4,2 is on the graph. So, we could also know that the point negative 4 negative 2 is also on the graph. If we zoom in on our parabola, we can also find other points by moving from the vertex. Since our parabola's in the form, x equals. This time, we'll move up and down one unit. Next, we'll move a units in the x direction to find the two points. So, a is negative 1, so I can go one unit up and one unit left. This gives me the point negative 1,1. And if I want to find another point I can move one unit down and one unit left to get this point negative 1 negative 1. Let's also see this with this graph. x equals points of three one and 3 negative 1. We start at the vertex and go one unit up and one unit down, since we're in the form x equals. And this time we're going to go three units right since the a value is positive 3. So we'll go one unit up and three units right, one unit down and three units right. This gives us the other 2 points on our parabola.
We can also shift parabolas, or move them around on the coordinate plane. Let's take a look at our original equation, y equals x squared. If we add 1 to the equation, we can see that the entire graph shifts up 1 unit. Notice before that the vertex was at 0,0. And now, our vertex is at 0, 1. By adding one right here to the end of our equation, we really increase the y values from before by 1. This is why the entire graph, the orange graph, shifts one unit up. So a point that was at negative 1, 1 is now at negative 1, 2. And a point that say was at two comma four is now at two comma five. Notice how the y value increased by one. We can also make this graph go up by two units, three units, four units, or even eight units. So, we can see that adding a constant term to the end of our parabola, will shift it up. So, if we subtract a number, it will shift it down. The parabola, y equals x squared minus 3, is our original parabola, only it's been shifted down three units. And again, notice that the orange parabola had a vertex of 0, 0. And this new parabola has a vertex of 0, negative 3. We could even change this to be a negative 5, so that way the orange parabola would shift down 5 units. And again, notice that the vertex of this last parabola is 0, negative 5. We can see that the y coordinate of the vertex comes from here, this negative 5 is right here on our vertex. The x coordinate of our vertex is 0. And it turns out it comes from solving this binomial or setting this binomial equal to 0. X minus 0 is still x, so we know that this term is really still x squared. And if I graph this equation, I can see that there exactly the same. I'm just writing this in a different form. So, now we can quickly see where this vertex comes from in our equation. The 0 is here, and the negative 5 is here. Now, we've seen shifting the parabola up and down. Well, we can also see a parabola shifted left and right. So, we know by adding or right, we're going to change this binomial. If I subtract 1 from x, it really moves the graph one unit to the right. Now the vertex is here. Notice that the sign of the x coordinate of the vertex is opposite from the one in here. For a y equals parabola, we can set this binomial equal to 0 to find the x coordinate of the vertex. So in this case, x would equal one. The y coordinate of the vertex is simply found by the number that's added or subtracted at the end of the equation.
So, based on the shifting and what you know about a vertex for a parabola. What do you think is the vertex for this parabola? When you think you've got it, type that answer here.
This parabola will have a vertex at negative 3, negative 4. 3 units left, and 4 units down. We can start from the basic graph y equals x squared. We know that that vertex is centered at the origin. If we subtract 4 from this equation we know that the entire parabola shifts down four units. Essentially all the y values from the original graph that we had before shifted down four units. Notice to that the y coordinate of the vertex is now negative 4. And we see this reflected in the equation. We have a negative 4 tacked on to the end. So we know that the y coordinate of the vertex is negative 4. And to find the x coordinate of the vertex we take our binomial and we set it equal to 0. Solving for x will get x is equal to negative 3, the x coordinate of our vertex.
Let's try another one like that. Where do you think would be the vertex for this parabola? Write that answer here.
This parabola's vertex is centered at 5 comma negative 2. Great work if you've found this as the vertex. We start with the basic graph y equals x squared. And we can shift all the points down by subtracting 2 from that equation. Think about the values of y and how they change. If you know that y equals x squared is, then subtracting 2 from x squared will give you a y value that is decreased by 2. Next, we shift the graph to the right by 5 units, by subtracting 5 from the x term in the binomial. When finding the x coordinate when y equals negative 2, we know that this part of the equation, our binomial squared, would have to equal 0. We know if x minus 5 squared equals 0, then we would have 0 minus 2. So y would just equal negative 2. That's the y coordinate of our vertex. So this means that we set our binomial equal to 0 to get an x coordinate of our vertex, have x equals positive 5. Notice again that the sign is different for the x coordinate of our vertex than the sign in the binomial here. Our vertex has an x coordinate of positive 5, but the binomial has the number negative 5. But the sign of the y coordinate is the same. The y coordinate the vertex is negative 2, and we see negative 2 at the end of our equation.
What we've been looking at for the last few examples is vertex form. Vertex form indicates the vertex of a parabola, h and k. For parabolas that open up and down, the h value is the opposite sign of the one in the binomial, and the k value is the one on the end. Remember we can always find the x coordinate of this vertex by setting the binomial equal to zero. And if we look at the number in front of the squared term, we can determine whether or not the parabola opens up, or if the parabola opens down. If the a value is positive or greater than 0, the parabola opens up. And if the a value is negative, or less than 0, then the parabola must open down. Horizontal parabolas are a little bit more tricky. These parabolas start with x equals and either open to the right or to the left. We find the vertex by looking at h and k. Notice here that h is added on the end or subtracted in some cases. And k is the number inside the binomial. Notice that h is now the number on the end, where k is the number in the binomial. We can find the y coordinate of the vertex by setting this binomial equal to zero. The x coordinate of the vertex, for horizontal parabolas, will just be the number, added or subtracted, on the end. And finally, if the a value was greater than 0 or positive, we know the graph opens to the right. Whereas if the a value is less than 0, or negative, then our graph opens to the left.
Let's see if you can use that knowledge to find the vertex of this parabola. Where do you think it is? And if you think you've got it, type it in here.
The vertex of this parabola is at the point 1 comma 2. Great work if you found this point. Keep in mind that this parabola is in the form x equals, so we know that it's going to open either to the left or to the right. Since the a value is a positive 1, we know that the graph opens to the right. Now this number is important for the vertex its really these 2 numbers. Since this is a horizontal parabola in the form x equals this number is the x coodinate of our vertex its positive 1. To find the y coordinate of our vertex, we simply take our binomial, and we set it equal to 0. So we'll have y minus 2 equals 0. Adding two to both sides, we get y equals positive 2. This is the y-coordinate of our vertex. And notice again how it matches the vertex form for a horizontal parabola. This is h and this is k, 1 comma 2. Our vertex.
Here's a chance for you to see what you know. What do you think is the vertex for each of these parabola's and which direction do you think each parabola opens? Enter the vertex here for this parabola and then choose whether it opens to the left to the right upward or downward. And if you have trouble with these, I highly recommend playing around in decimals. Try to create your own graphs and look for patterns. Maybe just start with some y equals graphs, and then play around with some x equals graphs, and see how they change. That should help you out if you get stuck. Good luck.
Here are the solutions. Great work if you got most of these right. For these two parabolas, we know that they're going to open up or down, since they're in the form y equals. We look at these numbers or the a value to determine whether or not they open up, or whether or not they open down. Since the a value is negative here, we know that this parabola must open down. And since the a value is positive here, we know that this parabola opens up. To find the vertex of this parabola, we simply look at the h and k value. h would be positive 4, by setting this binomial equal to zero. And k would be positive 2. The number added on the end. For this parabola, the vertex is 1,3. Again, we set the binomial equal to zero. So that way, we get an x coordinate of x equals 1. The y coordinate of the vertex is the k value, or just positive 3. These two parabolas here open left or right, since they're of the form x equals. By looking at the a value, we can see that one of them is negative, and the other one is a positive 1. This means that the negative 1 must open to the left and that the positive 1 must open to the right. Remember, negative a values for x equals parabolas make the parabola open to the left, whereas positive values of a for x equal parabolas make the parabola open to the right. And finally, to find the vertex for this parabola, we look at the value of h here and the value of k here. H is simply positive 2, our x coordinate of the vertex. And our y coordinate of the vertex is positive 1. We set this binomial equal to 0, to get y equals 1. And we find this vertex in a similar way. Here's the h value for our vertex. The x coordinate is positive 2. And here's the y value for our vertex, its negative 3. Remember we set the binomial y plus 3 equal to 0 and solve for the y coordinate of this vertex.
Now, not every parabolic equation we encounter will be in vertex form. Here's another parabolic equation. Notice that we have an x squared term, but we just have a single y term. Remember, all parabolic equations either have an x squared term, or they have a y squared term. They can't have both. Based on what you know so far, which way do you think this parabola opens? Does it open to the left? To the right? Up? Or down?
This parabola will open upward. Good hinking if you chose up. Now we haven't even covered this yet, so don't worry if you didn't get it right on your first try. But what I wanted you to recall is that the number in front of the x squared term is the a value. This tells us about the parabola's direction. Since a equals two and is positive, we know the parabola will open up. And since the parabola was in the form y equals, we already knew that the parabola opened up, or it opened down. So left and right weren't really good options there.
Now, comes a better question. Can we graph this parabola? Well, the equation isn't in vertex form. So, we'll need to do something different. We know the vertex form has a perfect square in it. So, what if we completed the square for the x terms, so that way we could get a perfect square? Essentially, we want to change this equation, so it looks like this vertex form. I can group these x terms together, and then we can add an amount in order to complete the square. But, since we don't want to complete the square unless the coefficient of this x squared term is positive one, let's factor out two from both of these terms. Now, I haven't completed the square yet but I did factor out a two here and here. This left me with two times the quantity x squared minus 8x plus 33 on the end. I'm just leaving a spot here in order to complete the square soon. If we were to complete the square here, what number would we put here in order to have a perfect square? Type that number in right here.
We would need to add 16 in order to complete the square. Remember that to complete the square, we always add the amount b divided by 2 squared. The b value here is the coefficient in front of the x term or negative 8, so b divided by 2 would be negative 4, and when we square this result we get 16. So, we add 16 in order to complete our square.
Now, notice that we added 16 in order to complete the square for these two terms. But the 16 is inside of these parentheses, and there's a factor of two on the outside of the parentheses. This means that we didn't actually just add side of the equation? Enter that number here.
We've really added 32 to this equation. Good work if you found the number 32. We take 16 and we multiple it by the coefficient of the factor in the front which is 2 so we know we have 2x squared here. Negative 18 x here and positive factor that is not 1 in front of our expression. We didn't just add 16 to complete the square, we really added two times that amount or 32. If you think about distributing the positive 2 it should make a lot of sense.
Now here's the important part, our equation started off balanced, and then we changed it. We tried to complete the square, by adding an amount. And in this case we didn't just add 16, we really added 2 times that, or 32. So, we need to do something in order to get this equation back and balance. Since we added 32 to the right, we could also add 32 to the left, but I really don't want to do this since eventually I just want to get y alone or isolated on one side of the equation. So, what operation should we perform here and the same operation here in order to make our equation balanced? As a hint think about what you would need to do in order to make this equation not change.
Well if we add 32 on the left side of the equation, then we need to subtract 32 from the same side. This should make a total of 0 on the left hand side of the equation. And we'd be left with 2x squared like here, minus 16x here, plus 33 here. This is why these two equations are exactly the same. The other way that we can think about it is that if we add 32 on the right hand side of the equation, then we add 32 on the left hand side of the equation. We could just subtract the 32 to get the y alone, and we wind up with this equation. Since we want the equation to be in the form y equals, it's best to go ahead an subtract on both sides, so long as we subtract the 32, since we added 32 here.
So we took our original equation and we completed the square. We added 32 to complete the square, which mean we needed to subtract 32 from the same side of the equation, so that way our equation remained unchanged, or it remained balanced. Now remember our goal is to take this equation and to write it in vertex form. Something that we can identify. So, what I want you to do, is to take this equation, and write it in vertex form here. I've already put y equals, so I want you to fill in the rest. Once you've written the equation, in vertex form, type in the vertex here.
This is the equation in vertex form and the vertex is four comma one. Great work if you found these two. We simply rewrite this trinomial expression as a perfect square. X minus four squared equals this polynomial. We keep the factor of 2 in front. and then we just combine the like terms of positive 33 and negative 32 to get positive one. This equation is in vertex form, so here is our vertex. We have an h value of 4 and a k value of positive 1. So h comma k. Keep in mind that we find the h value by setting this binomial equal to 0. So x would equal positive 4. The k value is simply the number on the end, so it's positive 1.
Now, let's see if you can do one on your own. What do you think is the vertex form of this parabola? I want you to write it here, and I've already listed the y equals for you. Once you have the equation in vertex form, also tell me the vertex.
This is the equation in vertex form, and the vertex is at 2, 1. Good thinking if you found these two. Now, this one was pretty tough, especially since we had a negative 3 here. Let's work through it together. Our goal is to get a perfect square so that way we can put this equation in vertex form. So we're going to try and complete the square by adding some amount to our polynomial. I can't quite complete the square yet, since I have a factor of negative 3 in front of the x squared. We want this coefficient to be a one. This means that we can factor a negative 3 from these two terms, leaving us with negative 3 times x squared minus 4x. We can quickly check to make sure that if we distribute, we would get back with what we started with. We'd have a negative 3x squared here and negative 3 times negative 4x which is positive 12x. Now we're ready to complete the square. The b term is negative 4, so b divided by 2 equals negative 2. And b divided by 2 squared equals 4. We simply square this number. So, we need to add 4 in order to complete the square for this trinomial. But we didn't just add 4 here, we really added negative 12, 4 times negative 3. Again, think about the, shooting this negative 3 to each of the three terms, and we would wind up with this expression. Now we can't just go adding numbers to equations and changing them at random. Instead we need to keep things balanced. If we subtract 12 from an equation then that means we need to add 12 to the equation as well. This makes the equation balance. It makes it so that way it's not changed. We know this since that negative 12 and positive 12 sum to 0. So we're just changing what this equation looks like, but it's really the same exact equation. Now that we're here we're ready to write our perfect square. The perfect square for this trinomial is x negative 2. If we square this amount, we'll get this expression. Our negative 3 stays in front as the coefficient, and then we have negative 11 plus 12 on the end. If we combine the like terms on the end, we get a positive 1 here. This gives us our equation in vertex form. And finally, for the vertex, we have an h value of positive 2. And a k value of positive 1. Setting the binomial equal to 0 gives us the x coordinate of our vertex. And the k value, or the y coordinate of our vertex, is just positive 1.
Now that we've seen how to graph and how to find the vertex, I want you to try graphing this parabola on your own. Now, one part of this is finding the vertex. The more important part is your creation of a graph. You can check the graph that you create against the one that I'll have in the solution video. When you think you've got a good graph, type in your vertex here and then we'll check in the solution.
This parabola has a vertex at 2, 1. Great work if you found this. We start by completing the square. We want to add a certain amount in order to make this a perfect square. We'll need to add four in order to do that. And since we add four here, let's also subtract four. This keeps our equation balanced. Remember that to complete the square here, we always add b over 2 squared. The b value is negative 4, so b over 2 is negative 2, and that amount squared is positive to write this expression as a perfect square. The positive 5 and negative 4 will just stay on the end. Here's our perfect square. And here's our equation, simplified combining these two numbers. So the vertex is at positive 2 for the x coordinate, and positive 1 for the y coordinate. Our vertex is here, and we get it by setting this binomial equal to 0. And the k value, or the y coordinate of the vertex is k. It's right here, positive 1. Now that we have our vertex, we can actually plot two other points. The one right next to the vertex. Our a value is positive 1 so this means we move one unit left and right at the vertex and then one unit up vertically. That will give us two additional points. And if we zoom in on our parabola, we can see this more clearly. We move left one and right one from the vertex. And then we move a units in the vertical direction, in this case one unit up. So left one and up one and right one and up one. This gives us our points at 1, 2 and 3, 2. Once we have these two points we can draw in the basic shape of our parabola. That's how we get this graph.
How about this parabola? What do you think would be the vertex for it? And try and create its graph. Remember, identifying the vertex is only one part of this problem. The graph is just as important, and it's up to you to make sure you get it correct. Take your time on this one, and be sure to check over your work. This one's a little bit harder. When you think you're ready to compare your graph, type in your vertex here and check out the solution video.
This parabola has a vertex at 2,3 right here. Nice work if you got this correct. Just like before, we'll take our equation and we'll try to complete the square this time for the y terms. The b value here is negative 6, so b over we know we need to add 9 in order to make a perfect square here, but I can't just add 9 to an equation since I'll change it. So, if I add 9 here I should also subtract 9 that will make it balanced, I really added 0. We rewrite this part as a perfect square. And now we can just combine these two numbers on the end. This gives us an equation in vertex form. And notice that it's x equals, so we know the parabola will open to the left, or to the right. Since there's a positive 1 in front of our square term, we know that a is 1, so our parabola opens to the right. For a horizontal parabola, the x coordinate of the vertex is here. It's h. The y coordinate of the vertex is k, and it's here. It's positive 3. We can find the y value of the vertex by setting y minus 3 equal to vertex lets figure out how to graph our parabola and know that it opens this way. The a value is always in front of the squared term. So, a here is still positive 1 so lets zoom in our vertex and figure out how to graph those other two points. Here is our vertex at 2, 3 and now we go a units in the x direction. So, we'll go 1 unit up and one unit down from our vertex. Then we'll go 1 unit right and 1 unit right. This gives us our two other points and 3,4 and 3,2.
Here's your third parabola. Where do you think this vertex is? And then try and create its graph. Now, you want to pay close attention when you go to complete the square, since there's a negative 2 here. Also, think about the orientation of your parabola. Which way should it open? When you think you've got the graph and the vertex into the vertex here. And then check your graph in the solution video.
This parabola will open down and it has a vertex at negative 1,3. Excellent work if you got this vertex correct. In order to find this vertex, we need to complete the square for these x terms. This x squared term has a coefficient of negative 2, so I'm going to factor that out from these two terms first in order to complete the square. So we'll have 2 two, times x squared, plus 2x, plus some unknown number. Now, this number is the number we'll use to complete the square. And the positive 1 will just stay outside on the end. The b value is positive 2. So, we know b divided 2 equals 1. And b divided by 2 squared equals square for these terms. But we really didn't just add one here, we really added of distribution. We have negative 2x squared, we have negative 4x and we have negative 2. Because we subtracted two from this equation, we really need to add a positive 2 on the end of our equation here. Now, we're ready to rewrite this as a perfect square. That perfect square is x plus 1. We'll still have negative gives us our equation in vertex form, so we can easily identify the vertex. We set the binomial equal to 0, to get an x value of negative 1. And the k, or the y coordinate of the vertex, is just 3. This is how we get the vertex of negative 1,3. It's right here. Now, to get the other points, let's use this a value of negative 2. I can zoom in one my vertex, and then we can see we move one to the left and one to the right, and two units down to get these other two points on our parabola. Since this parabola is y equals, we know we need to move negative 2 down from the vertex once we move 1 left and one right. This gives us our points of negative 2, 1 and 0,1. Notice how the y coordinate here and here are two less than the y coordinate of the vertex here. This is why the a value is so important when it comes to graphing parabolas, and makes our graphing much more simple.
Here's the last parabola for this lesson. Try graphing it, and when your ready enter the vertex here.
This parabola has a vertex at 1, 2. Excellent work if you figured it out. We want to start by factoring a negative 3 from these first two terms so we'll have negative 3 times x squared minus 2x. We'll leave a space in order to complete the square and then we'll have negative 1 on the end. The b value is negative 2 here. So, we need to add b over 2 squared in order to complete the square. B divided by 2 equals negative 1, and this number squared is positive just didn't add 1 here, we really subtracted 3. We'll have negative 3x squared here, positive 6x here. And negative 3 here. So, in order to make our equation balanced, we'll have to do the opposite of negative 3. We'll have to add 3 to the same side of the equation. This is so that we really add 0. We don't want to change our equation, we really just want to leave it balanced. We can rewrite this as a perfect square, writing x minus 1 squared to replace this trinomial. The factor of negative 3 stays in front, and then we combine negative 1 and positive 3 to get positive 2. This is our equation in vertex form. And we can see that the vertex is h, k. Positive 1 and positive 2. Keep in mind the y value were equal to 2, then that means that this entire expression would have to equal zero. Which means the binomial x minus 1 must equal 0. So, we set it equal to 0, which means x would have to equal 1. That's the x coordinate of our vertex. So, we have this for our vertex, and then we can get two other points by moving one left and one right of the vertex, and then three units down. We can see that more clearly here. One unit left and one unit right, and then three units down. That's our a value. So, we have the points 0, negative 1 on our parabola, and we have the point 2, negative 1. And, notice that these y coordinates are three units less than our vertex.