WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take a look at the graphs of quadratics.
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We will first start off by looking at the basic shape of a quadratic and how it forms the shape known as the parabola.
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We will pick up some new vocabulary such as the axis of symmetry and the vertex.
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We will look at a few quadratics that are not written in standard form.
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This will lead to a new technique that we have to use in order to apply out their graph.
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We have seen this technique once before and this is known as completing the square,
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but we will also use what is known as the vertex formula to do the same thing.
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When you are looking at the shape of a quadratic it forms a parabola.
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One way you can see this is it take the nice basic shape of x² and plot out lots of different points on a graph.
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Maybe choose some values like – 2, - 1, 0, 1 and 2, just to see what you get.
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All of these guys are being squared.
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-² would give you 4, 1² would be 1, 0² would be 1 and 1² will be 1.
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2² will be 4, and it will keep going in both directions.
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Let us plot those out.
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I’m not sure if you can see that it forms a nice parabola.
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All other quadratics will have this parabola shaped to them.
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Here are some features that you want to know about these parabolas.
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One, they are only symmetric to align known as the axis of symmetry and runs right down the middle of that parabola.
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Here are two examples that I have the axis of symmetry marked out.
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Whatever that axis of symmetry intersects with the actual graph that point is known as the vertex
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and you can see that on my little diagrams right here and right here.
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Let us call those the vertex.
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The reason why we want to know about the vertex is it will help us create a nice shortcut method for graphing these parabolas.
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That way we would not necessarily have to go back to a large table of values every single time we want to graph them.
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Think of this kind of in the context of when we are graphing out lines, when we are applying a whole bunch of points for those lines,
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we simply had to find a y intercept and a slope.
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With parabolas we do not have a notion of slope, but knowing where the vertex is can often help us plot out the rest of it.
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We can use techniques known as transformations in order to figure out where the graph is.
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This assumes that your quadratic is written in the following form of something like y= a × x - h² + k.
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I did not put in the function notation here but we could also imagine that being a y, either way will work out just fine.
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If your quadratic is written in this very special form, there are lots of different things that it will tell you.
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The (a) value out front will tell you whether it is been reflected over the x axis or if it had any type of vertical stretches.
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If (a) is negative then your parabola will be facing down.
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If it is positive your parabola will be facing up.
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It is how we can also determine how wide or narrow to make that parabola.
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If the absolute value of (a) is say greater than 1, then it will be a narrow parabola.
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If the absolute value of a is less than 1, then you know it will be fairly wide.
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It gives you a little bit about its shape.
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The h and k are probably two of the most important things you can get rid of here and they help you determine where that vertex is.
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h and k together would be the point of that vertex.
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What h is helping you figure out is how far left or right the parabola ended up moving from its original spot said 0, 0.
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The k helps you figure out how far up or down that parabola had to move from its original spot.
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We will look at the h and k more as finding a vertex and then use the value of h to determine how wide or narrow it should be.
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Let us see if we can identify some of these key things with the following problem.
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First of all where is this axis of symmetry?
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We will imagine a line straight down the middle of it, if we are to fold it up along this line it should match up with itself, axis of symmetry.
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Now that we have that we can see where it intersects the graph.
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We will say that this is the vertex.
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We will go ahead and say what that location is.
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It is that -4, 2.
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Notice how this parabola is facing up, in the actual equation of it the (a) value should be positive.
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Let us see if we can take a problem written in the special form and put it onto our graph.
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I’m going to write down the general form, just so we can reference it.
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It is y= ax - h² + k.
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The first thing you want to know is where is that vertex?
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It is coming from looking at the h in the k.
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Be very careful notice how in the formula here, it has a list of that as -h.
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What we want to think of is if we had to rewrite this as x -,
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what would that value need to be to make it match up with that 4 we have in the original.
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It would have had to have been -4 and that will help us figure out the vertex a little bit better.
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This tells us our vertex is at -4, 5.
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Let us make that our first point on the graph.
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-1, 2, 3, 4 and 1, 2, 3, 4, 5 right there
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We can also figure out a little bit more information about it.
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If we look at the (a) value out front, it is negative.
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We know that our parabola should be facing down.
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Do you know if it should be wide? Or should be narrow?
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What else can we figure out about this thing?
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If we look at the absolute value of -2, we get 2 and that is greater than 1.
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It is going to be more narrow than the standard parabola of y = x².
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Let us put this in comparison.
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Suppose I were to graph a parabola facing down that was related to x², that would be these points right here.
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I want to write this light because that is not the parabola we want.
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Our parabola is going to be narrower than this one.
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In fact the values will be twice as much.
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2, 3, 4, 5, 6, 7, 8.
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Ours will look more like this, narrower than the normal x².
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A bit of a rough sketch we get an idea what is going on here.
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If you want some more accurate things to put in here what you could do is plot out a few points or even just to add a few additional ones.
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Let us try some more.
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Let us see if we can work backwards by starting with the graph of a quadratic and see if we can make its equation.
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Like before, one of the first important points we want to know is where is the location of that vertex?
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That one is located at 4, 1.
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Now that we have that I know what at least my h and k values are.
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Y = ax -4² + 1.
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It looks like the only thing I do not know right now is the value of a, and because the problems facing up, this should be a positive value.
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It looks like it might be a little wider than our normal parabola of x² so I'm thinking maybe it is a fraction of some sort.
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To be able to figure this out, I’m going to take another point on the parabola that I know it is on there and substitute this value into my equation.
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Then I will be able to solve that a directly.
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Y is 2, x is 2.
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You can see the only unknown we have in here is that a value.
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2 - 4 would be -2².
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Subtracting one from both sides 1 and -2² would give us 4.
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We almost have that a value let us divide both sides by 4.
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Now we can write down the entire equation.
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This graph comes from ¼ (x - 4² + 1).
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That vertex plays a key part in getting a connection between the graph and the equation.
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Now, unfortunately, a lot of quadratic equations and functions are not written in that nice standard form.
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In fact, we might end up with something that is written more like this.
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When it is written like that, how are we supposed to figure out where that vertex is or which direction it is facing?
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There are a couple of things you could do.
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One is you directly take that and end up rewriting it into that nice standard form the a x - h² + k + 4.
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The way you accomplish that is you will go through the process of completing the square.
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Once you have it in your nice standard form then you can go ahead and identify your vertex, which way it is going, is it going be wider.
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You will know everything you need from there.
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If you recall, completing the square can be quite a clunky process with lots of steps involved.
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Fortunately, just like the quadratic formula, there is a way around.
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Maybe not having to use completing a square and still plot out quadratics that look like this, like a x² + bx + c.
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You can package up a lot of these steps of completing the square and get what is known as your vertex formula.
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It is this one right here.
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It does look a little bit odd, but here is how you can interpret it.
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This first value has given you a small formula so you can figure out the x coordinate of your vertex.
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The second part of this, it has that –b/2a and what is that trying to tell you is that you should take the value you found for your x
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and you should put it back into your quadratic.
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You should evaluate it at that value.
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One other thing to note, the vertex can sometimes be thought of as the minimum or the maximum of your quadratic.
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Here it is a minimum point and here it is a maximum point.
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We will use this vertex formula to help us out if our equation is written in this form.
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Let us use completing the square to see how you could potentially use that as well to figure out what your vertex is.
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In this one I have f(x) = 4x² - 4x + 21 and in the method of completing the square first I need to organize things.
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I will get my variables to one side.
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I have f(x) and I’m going to subtract 21 from both sides.
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That looks pretty good.
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We need to make sure that the coefficient in front of x² is 1.
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Now it is 4, let us divide everything by 4, looking pretty good.
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We need to go ahead and find the new number that we need to add to both sides to factor nicely.
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We need to check our term in front of x in order to do that.
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-1 ÷ 2 and then squared.
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That would equal to ¼.
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We need to add ¼ to both sides so that this will go ahead and factor.
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On the right side there, it will factor nicely.
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In fact this factors into x – ½².
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You can see that we are starting to build that similar form on the right side.
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To see all of our other terms we need to go ahead and clean this up a bit and get that f(x) all by itself again.
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I’m going to go ahead and combine the 21 and the 1 since they do have the same denominator down here.
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That will be f(x) -20 ÷ 4.
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Let us multiply both sides by 4.
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Let us go ahead and add 20.
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That was quite a bit of work, but at least now we can see what is going on here.
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These two values here, we can get rid of our vertex.
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I'm at ½ , 20.
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For my (a) value out here we can see that it should be facing up and the absolute value of 4 is larger than 1, it is going to be a narrow parabola.
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Notice how we could have also figured out lots of information about our a value from the original equation out here.
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This value here is the same a value as this one here.
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You could get rid of that information at the very beginning.
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Let us see if we can go ahead and graph another quadratic.
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This one is f(x) = - x² + 2x +5.
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If we are going to graph this one, there are few things that we can do.
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One, we could go through the process of completing the square, so we can figure out where its vertex is.
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Or since it is in this form we might as well just use the vertex formula and get it directly.
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That formula is –b ÷ 2a and once we find it we will put it back into the original problem.
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Negative the value b is 2 ÷ 2 × a.
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I have -2 ÷ -2 = 1.
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This tells me that the x value of my vertex is 1.
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We will borrow that and evaluate it back into the function.
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What happens when we evaluate this at 1, -1² + (2 × 1) + 5.
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1² is 1 that is -1 + 2 + 5 we have 6.
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We have the location of the vertex.
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It looks like it is at 1, 6.
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Let us see what else we can say about this.
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What we need to know about a value to see which direction it is facing and looks like it is negative.
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I know that this one is facing down.
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Let us see what else we have figured out about it.
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When it is written in this form, this last number will give us the y intercept.
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I know it will go through 5.
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Since the absolute value of our (a) term is exactly 1, then it is exactly as wide as our parabola x².
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Let us do that.
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1, 1 will go out to 1, 2, 3, 4.
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Now we can sketch out a nice graph of this parabola.
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That vertex form can help out if you have your quadratic in this particular form.
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Let us go ahead and sketch a graph of 1/3 x² + 3x – 6.
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This looks like another good one we should use our vertex formula on.
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-b ÷ 2a then we will play it back in, -b ÷ 2a.
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Negative our value of b is 3 ÷ 2 × -1/3.
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This will be -3 ÷ -2/3.
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Negative divided by negative will be a positive, we have 3 ÷ 2/3, which is the same as 3 × 3/2 = 9/2.
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That will give me the x value of my vertex, something 9/2.
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Now I have to take that and evaluate that by putting it back into the original function.
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Take a little bit of work, so (-1/3 9/2²) + (3 × 9/2) and then we will subtract 6.
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9/2² will be 81/4 and I have 3 × 9/2 that will 27/2.
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I can do a little bit of simplifying here.
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3 goes into 81 27 times.
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(-27 ÷ 4) + (27 ÷2) – 6
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That will to do a little bit of work if I'm going to get a common denominator.
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I have to multiply by this fraction, top and bottom by 2 so 54/4.
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We need a common denominator down here 24.
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Slowly combining these together I have 27/4 - 24/4 = ¾ .
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Quite a bit of work but now I know the location for sure of my vertex.
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Let us see if we can now plot this out.
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I need ½ , 2/2. 3/2, 4/2, 5/2, 6/2, 7/2, 8/2, and 9/2.
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We will go up ¾
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Not all the way up to 1 but just ¾ of that, right about here.
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It looks like this particular parabola is facing down and because the 1/3 is going to be a fairly wide parabola.
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How large that will make it?
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I know it crosses through the y axis at -6.
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We can put that value on there as well.
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I have a good idea on what this looks like.
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There is our sketch of the quadratic.
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Let us do one last one just to make sure that we have all of these techniques down.
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On this one, we want to find the maximum or the minimum value of our function.
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This is just the same as finding the vertex.
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We will go ahead and graph it out so you can see what that vertex is.
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-b ÷ 2a, ½ (-b) ÷ 2a.
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We will need both of those.
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Starting off with trying to find the x coordinate.
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Negative the value b2 ÷ 2 × a.
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-2 ÷ - 4 will give us ½.
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I would need to go ahead and put this back into the original.
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-2 × ½² + 2× ½ + 4
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A little bit of simplifying but not too bad -2 × ¼² + 1 + 4
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-2 × ¼ = -1/2, 1 + 4 = 5 and now I can find a common denominator and put these two together.
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I will get 9/2 for the second half of my vertex.
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This vertex is located at ½, 9/2.
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Let us put that on the graph.
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½ for x and I have 1, 2, 3, 4, 5, 6, 7, 8, 9/2.
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There is where our vertex is located.
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The (a) value for this one is negative, so I know that the parabola is facing down.
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It looks like it has a y intercept of 4.
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It goes through that spot.
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Be just a little bit more narrow than your average parabola because the absolute value of 2 is greater than 1.
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Now that we sketched that out and from the sketch since it is facing down I can say that this vertex here is a maximum.
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The maximum value of our function is that 9/2 that comes directly from the vertex which is located at ½ and 9/2.
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A lot of different ways that you could go about actually sketching the graph of the quadratic and it is actually not so bad.
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If it is in a nice standard form you can simply get rid of the vertex and its (a) value.
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But if it is not in that form of then feel free to use the vertex formula.
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