WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we will be working more on quadratic equations and functions.
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And we are going to start out by analyzing the graphs of quadratic functions.
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Recall: in previous lessons, we discussed quadratic functions in standard form.
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Today, I am going to introduce a new form, which is the vertex form of a quadratic function.
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And this is a very useful form of the function, and it is given as y = a, times (x - h) squared, plus k.
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h and k are the vertex of the parabola; and x = h is the line that is the axis of symmetry.
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And as you will recall, the vertex is the maximum point for a downward-opening parabola;
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and it is the minimum point for an upward-opening parabola.
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So, by having a function in this form, we already have information about what the graph is going to look like.
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So, let's start out with just a very simple function, y = x².
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OK, so looking at it in this form, we have just an a that is 1, and h and k are 0.
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So, the vertex is going to be right here; so h equals 0 and k equals 0, so the vertex is (0,0).
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Now, drawing some points, x and y: when x is -1, y is 1; when x is 1, y is 1; when x is 2, y is 4; and when x is -2, y is also 4.
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So, it is a pretty simple graph.
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Now, by starting out with this graph, we can look at what happens--the significance of h and k on the shape of the graph.
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So, we are starting out with just this graph y = x².
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Now, let's look at a situation where y = x² + 2.
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All right, so now, I still have an h that equals 0; but here, k equals 2; therefore, the vertex is at (0,2).
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So, my vertex is right here; now let's look at some points, x and y.
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When x is -1, then x² will be 1, plus 2--that gives me 3.
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When x is 1, again, y is 3; when x is 2, y is 6; and when x is -2, that is -2 squared is 4, plus 2 is 6.
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So, look at what happens with this graph: when x is 1, y is 3; when x is -1, y is 3; when x is 2, y is going to be up right about here at 6, and right here.
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OK, so as you can see, the graph here is shifted starting right up 2; so, the vertex is going to be right here;
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and when x is 1, y is 3; when x is 2...there we go.
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So, the vertex is right here; OK, so what this is going to give me is a graph that is shifted up by 2.
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So, you see what k does to the graph: it is the same basic graph, but it is shifted up by 2.
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If k were to be -2, then the graph would be shifted down by 2.
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Now, h also has an effect on the graph; so let's look at what h could do to the graph.
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Let's take the example y = (x - 1)².
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Here, I have h = 1, k = 0; so the vertex is at (1,0); now let's find some points.
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The vertex is over here at (1,0): some points x and y for this function: when x is -1, then that is going to give me -1 and -1 is -2, squared is 4.
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When x is 0, 0 minus 1 is -1; squared is going to give me 1.
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I already have the vertex at (1,0), so let's pick 2: 2 - 1 is 1, squared is 1.
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One more point: 3 - 1 is 2, squared is 4; that is going to give me (-1,4), (0,1), (1,0), which is the vertex, (2,1), and then finally (3,4).
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So, I have this graph right here; this graph is y = (x - 1)².
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You can see, what happened is: the graph here is shifted to the right by 1.
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h shifts to the right or left; if this had been -1, the graph would have shifted over to the left.
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So again, k is going to shift the graph either up or down; h is going to shift to the left or right.
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And we will talk about a in a minute.
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So again, vertex form is very useful, because it gives you a lot of information.
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It gives you the vertex, which is (h,k), and it also gives you the axis of symmetry.
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And recall that the axis of symmetry is the line at x = h, which bisects this parabola into two symmetrical halves.
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For example, if I had a parabola here, a downward-facing one, and the vertex is right here, the axis of symmetry would be right here
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at the line x = h, where the vertex is some point (h,k).
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Now, let's talk more about the coefficient a; what the coefficient a tells you is how steep or flat the parabola is.
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For example, if I had one parabola, and let's say it had a = 1, and it looked like this, for example:
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and then I had a very similar parabola, but this time with a = 2; well, as it says here,
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if the absolute value of a is greater than 1, the parabola is narrower.
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I have a = 1; and if I have a = 2, it is going to be a narrower parabola, meaning that, for every change in x, y changes more dramatically.
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So, it goes up more steeply--each section of the parabola.
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Now, if you have an absolute value of a that is less than 1 (that is a fraction), the parabola is going to be wider.
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So, a = 1/2 might look something like this; for every change in x, the change in y is less dramatic.
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OK, so the first thing that the absolute value of a tells us is how narrow or wide the parabola is.
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A larger number is going to be narrower; a smaller number, especially a fraction, is going to have a much flatter, wider parabola.
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The second thing that a tells us is which direction the parabola opens.
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All of these parabolas that I showed you opened upward, and that means that a is positive.
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If a is greater than 0, the parabola opens upward; if a is less than 0, the parabola opens downward.
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And here, I have drawn a pretty wide parabola, so it would have a fairly small value of a; and it would have a negative value.
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The absolute value of a would be small, and the value of a itself would actually be negative.
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Here, all of these have an absolute value of a that is...these two are larger; this one is smaller; but all of them have a value of a that is positive.
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So they open upward; so if a is positive, the parabola opens upward; if a is negative, it opens downward.
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If the absolute value is larger, you have a narrower parabola; if the absolute value of a is smaller, you have a wider, flatter parabola.
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Between what we just learned about h and k and the vertex,
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and what we learned here about a, you can get a lot of information
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just by looking at the equation about the shape of the graph of the parabola
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before you have even found points beyond the vertex.
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OK, now since this is an important form of the equation, you need to know how to write it in vertex form.
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And you may actually be given the quadratic equation in standard form and be asked to write it in vertex form.
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And the way you go about that, if you are given a quadratic equation in standard form, is to complete the square.
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And that will allow you to write the equation in vertex form.
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Now, it says here that, if a does not equal 1, then you need to factor out a before you go on to complete the square.
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So, we will deal with that in a minute; let's start out with a simpler case where y equals x² + 6x - 8.
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So, I have standard form; and I am going to go ahead and write it in vertex form.
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Let's just recall vertex form: y - a(x - h)² + k.
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So, I am starting out with standard form; and when I complete the square, the first thing I need to do is isolate the variable terms on the right.
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Here, I have my x variable terms, but I also have a constant here.
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So, what I am going to do is add 8 to both sides to get this.
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Now, I have my x variable terms isolated on the right.
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Isolate the x; I should specify x variable terms on the right; the y stays on the left.
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OK, next complete the square; and recall that, in order to complete the square, you need to add b²/4 to both sides.
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So, here let's figure out what this would give me...y + 8, b is 6, so that is going to be plus 6²/4 = x² 6x + 6²/4.
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Doing some simplification: 6² is 36, divided by 4 is simply 9.
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OK, I have completed the square on the right; I am going to do some simplification here, because this is actually y + 15.
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And this is not quite in vertex form, because if I look up here, vertex form would have y isolated on the left.
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So, I am going to actually change this, and I am going to subtract 15 from both sides.
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And this is going to give me, let's see, -8.
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Actually, one more step: before we do that, I have completed the square; so I want to go ahead
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and write this out closer to vertex form before I even go on.
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And this is x² + 6x + 9, and that is a perfect square, so I am going to write it in this form.
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Then, I am going to go ahead and move my 15 over.
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OK, so I have written this in vertex form; and I accomplished that by first isolating the x variable terms on the right,
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then adding b²/4 to both sides; I did that, and I ended up with this right here.
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And I see, on the right, I have a perfect square; so I can rewrite that as (x + 3)².
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And then, finally, this is vertex form.
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And although vertex form has a negative here, it is fine to write it like this.
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If I wanted to, I could have written it as x - -3, but this is actually acceptable, as well.
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Now, it gets a little bit more complicated if the a term is not equal to 1; so let's go ahead and look at that situation.
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If I had something like y = 2x² - 8x + 5,
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I am first going to isolate the x variable terms on the right (that is going to give me y - 5 = 2x² - 8x),
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and then, as it says here, I need to factor a out, since a is not 1, before I complete the square.
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So, this is going to give me 2x² - 4x; now, I complete the square by adding b²/4 to both sides.
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But you have to be careful on the left; let's see what happens.
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This is going to give me x² - 4x, and then b²/4 is actually going to be (-4)²/4.
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Now, I can't just add that to the right, because in reality, what I am adding is not just (-4)²/4.
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But it is actually (-4)²/4, times 2; that is what I need to add to the left.
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Otherwise, the equation won't be balanced.
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So, let me go ahead and simplify this (-4)²/4; that equals 16/4, which equals 4.
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Now, what I am really adding to both sides is actually 4 times 2; 4 times 2 equals 8.
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This is 2 times 4; so that is what I need to add to this side, as well.
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So, this is y - 5 + 8; it is going to give me y + 3 = 2 times x² - 4x + 4.
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Now, I am almost there, because I have a perfect square, which is (x - 2)².
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So then, I can rewrite this in vertex form.
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And as you see, it was more complicated, because I started out with a leading coefficient that is other than 1.
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In that case, I have to first go ahead and isolate my x variables, and then factor that out, so that this becomes 1.
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Once I have done that, I can complete the square; but I have to be very careful, on the left,
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that I am not just adding b²/4, but that I am adding b²/4,
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times what I factored out, to keep the equation balanced.
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And now, you see, I have vertex form; and here a equals 2; h equals 2; and k equals -3.
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OK, finally, if you are given the vertex and another point on the graph, you can write the equation of the graph in vertex form.
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For example, if I am given that the vertex equals (2,11), and I am also told that a point on the graph
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is at (1,6), then let's look at what I have.
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Well, I have y, because I have a point on the graph; so this is h; this is k; this is x; this is y.
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So, I have y; I have x; I have h; and I have k.
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The only thing that I am missing is a, but since I have everything else, I can solve for a.
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So, I am going to go ahead and do that: y is 6; 6 equals a times...x is 1, minus h squared, plus k.
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OK, so this gives me 6 = a, and this is -1 squared plus 11, so 6 equals a times 1, plus 11; 6 = 11 + a; therefore, a equals -5.
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Now that I have solved for a, I can go ahead and write this in vertex form.
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I have y = -5(x -...I know that h is 2) + k, which is 11.
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So again, if I am given vertex form and a point on the graph, then I can write this, substituting in everything I have, and then solving for a.
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And here, I found that a equals -5; so now, I have h, k, and a, and all I need to write an equation in vertex form is h, k, and a.
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So, if you are either given the quadratic equation in standard form, or you are given
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the vertex and a point on the graph, either of those is enough to rewrite the equation in vertex form.
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OK, practicing this with some examples: here I am given a quadratic equation in standard form, and I am asked to write it in vertex form.
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Remember that the first step is to isolate the x variable; I am going to do that by subtracting 7 from both sides.
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OK, so I have y - 7 = x² -4x; next, I am going to complete the square.
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And remember, you do that by adding b²/4 to both sides.
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And here, b is -4; so this is going to be (-4)²/4 = x² - 4x + (-4)²/4.
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Simplifying: this gives me 16/4; simplifying further, I get simply 4.
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OK, so I have -7 + 4; that is going to give me y - 3 = x² - 4x + 4.
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And what I have on the right is a perfect square, so I am going to rewrite that as y - 3 = (x - 2)².
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One more step to getting this in vertex form is to move the constant over to the right.
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And recall that vertex form is y = a(x - h)² + k.
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So, here I have a = 1; h is 2; and k equals 3; and the vertex is at (2,3).
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And I accomplished that by isolating the x variables on the right, completing the square
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by making sure I added b²/4 to both sides, and then once I had that completed,
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I could rewrite this in this form, the (x - h)² form, and then moving the constant over to the right.
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OK, here: find an equation in vertex form for the parabola with vertex at (-4,3), and passing through the point (-2,15).
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So, first I am looking at what I have and at what I need.
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Vertex form: y = a(x - h)² + k.
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So, to write in vertex form, I need a, h, and k.
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What do I have? Well, I have the vertex; this is h and k; and I have a point--this is x and y.
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What I am missing is a; but since I have everything else, I can solve for a.
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I am going to substitute in here for y; that is 15; I don't know a; I know that x is -2; I know that h is -4; and k is 3.
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So, solving: a negative and a negative is actually a positive, so this gives me 15 = a...
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-2 plus 4 is 2, squared; so then, this is 15 =...this is actually...
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I am down to here; I am going to subtract 3 from both sides to get 4a = 12.
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Isolate a by dividing both sides by 4 to give me a = 3.
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So, now that I have a, I can write in vertex form: I have y = 3 times x, minus h (is -4).
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And I can either write this as x + 4 or x - -4, plus k, which is 3.
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This is vertex form; and I accomplished that by recognizing that I had everything I needed except for a,
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substituting in, solving for a, and then going back and writing this in vertex form.
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And here, a equals 3; h equals -4; and k equals 3 (vertex form).
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OK, again, I am asked to write the quadratic equation in vertex form.
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But notice this time: this is with the standard form; a does not equal 1.
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Recall that I am going to have to do an extra step before completing the square in this situation.
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So, the first step is to isolate the x variables on the right.
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Now, the extra step I have to take is to factor out a.
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Now that I have done that, I can complete the square; and that is accomplished by adding b²/4 to both sides.
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So, I need to add b²/4 to both sides.
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And here, we have b² is actually -3, but I need to be careful,
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because I am not just adding b²/4; I actually have factored out the 4;
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and if I don't add that back in on the left, my equation won't be balanced.
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So, this is actually 4 times b²/4.
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x - 3x + (-3)²/4: now, I already have 4 times b²/4 here, so I needed to balance it by having that on the left side, as well.
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This is going to give me y - 20 + 4, times 9/4, equals 4 times x, minus 3x, plus 9/4.
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OK, this is going to give me y - 20; and the 4's cancel here, which is kind of convenient;
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so that is going to give me...these will cancel; I will just get 9.
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And -20 + 9 is -11, so this is y - 11 equals 4 times...and this actually is a perfect square, and you can check it out for yourself.
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This equals (x - 3/2)²; this is a perfect square.
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Now, I am almost in vertex form; I just have to move this 11 over to the right by adding 11 to both sides.
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And this is vertex form; so again, vertex form is y = a(x - h)² + k.
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Here, I have a = 4, h = 3/2, and k = 11; so the vertex is going to be at (3/2,11).
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This was a little more complicated, because I did have a coefficient here, an a, that did not equal 0.
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So, right after I isolated my x variable terms, I factored that out.
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And then, when I complete the square, I need to make sure that I add 4 times b²/4 to the left, since that is what I am adding to the right.
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And then, I just went on and did that; I got a perfect square on the right.
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And this is going to end up...this is actually all x²'s, and this is going to end up giving me (x - 3/2)² as a perfect square.
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And then, I just move the 11 over to the right.
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OK, now find an equation in vertex form of the parabola with vertex at (2,-3) and passing through the point (4,-19).
00:28:33.100 --> 00:28:44.300
OK, vertex form: y = a(x - h)² + k.
00:28:44.300 --> 00:28:53.200
Now, I look at what I am given; and I am given the vertex, so that is h and k, and a point (that is x and y).
00:28:53.200 --> 00:28:59.200
Since I have all of these, I can find a; I don't know what a is--that is my unknown.
00:28:59.200 --> 00:29:16.900
Well, I have y = -19; I don't know a; x is 4; h is 2; and k is -3.
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4 minus 2 gives me 2, squared, minus 3; so that is -19, equals a, times 4, minus 3;
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-19 equals 4a - 3; so I am going to add 3 to both sides to give me -16 = 4a; therefore, a equals -4.
00:29:38.500 --> 00:29:50.800
Now that I have found a, I just go back in here and write this as y = -4x, and then I have h as 2, and k is -3.
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So, here I have vertex form; I know the vertex is (2,-3).
00:29:55.600 --> 00:30:01.900
And now I know that a equals -4; so this is going to be a parabola that opens downward.
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OK, this concludes this session of Educator.com.
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And I will see you again next lesson.