WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we are going to start talking about quadratic equations and inequalities.
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And the first thing we are going to review is graphing quadratic functions.
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Recall from Algebra I that a **quadratic function** has the form f(x) = ax² + bx + c.
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So, that is the quadratic function in standard form, where a does not equal 0.
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And if a were to equal 0, this section would drop out; and this is actually what makes it a quadratic function--the x².
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Otherwise, you would just end up with a linear function, which we have described earlier on in the course.
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So, let's first talk about an example of a quadratic function.
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It could be something such as f(x) = x² + 4x - 3, or f(x) = -x² + 1.
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And here, in this second example, b is actually equal to 0; and that is allowed, so this drops out.
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But as I just discussed, a cannot equal 0.
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We are going to be working with the graphs of these functions today, and first just talking about the general shape of these curves.
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The curve of a graph of a quadratic function is a shape called a parabola, and they may open upward;
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a parabola can open upward; it could open downward; and you will see varying shapes, such as wider parabolas and narrower parabolas.
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And as we get deeper into discussing this topic, you will be able to tell, just from looking at the quadratic function,
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roughly what the shape is going to be and which direction it is going to curve in.
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So, getting started with some properties of parabolas: a parabola has an axis of symmetry, and that axis is described by the equation x = -b/2a.
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And what the axis of symmetry is: it is a vertical line that divides the parabola into two symmetric halves.
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And the parabola intersects this axis of symmetry at what is called the vertex.
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Recall that the x-coordinate of the vertex is given by x = -b/2a.
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To illustrate this with an example: if I were looking at the function f(x) = -x² - 2x + 1,
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then first let's look for the axis of symmetry; and the axis of symmetry is up at -b/2a.
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Well, here a is equal to -1, because recall that standard form is ax² + bx + c, so a is -1, b is -2, and c is 1.
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So, if I want -b/2a, that is going to give me -(-2)/2(-1).
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So, looking at that, that will be a negative and a negative, to give me a positive, over -2, which equals -1.
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Therefore, the axis of symmetry is going to be a vertical line right here at -1; so, this is the axis of symmetry.
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Now, this point where x equals -1 is the x-coordinate of the vertex.
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However, I need to find the y-coordinate; so in order to do that, I am going to go back and look for f(-1).
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So, I am replacing my x terms with -1; that is going to give me -1 squared, is 1, so that is -1.
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And then, that gives me -2 times -1, which will become a positive, so that is + 2, + 1.
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So, this is 3 minus 1; so that equals 2.
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What I did is found the x-coordinate for the vertex, substituted that in for my x-value, and found that f(-1) = 2.
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So, the y-value for the vertex is going to be 2; the vertex is going to be at the point (-1,2).
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So, that is (-1,2) right here.
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OK, now, to further graph this, let's go ahead and find some points.
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And it is best to pick some points around the vertex, because if I pick points way out here, my graph is not going to be as accurate.
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I want to get a general shape of the curve in the area of the vertex.
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OK, so my vertex is right here; pick some x-values, and then let's go ahead and determine what the y-values are, based on that.
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When x is -2 (you can go ahead and work this out for yourself), if you worked the whole thing out, you would find that y is 1.
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OK, when x is 0, you can see that these two are going to drop out, and y is going to be 1.
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When x is 1, that is going to give me -1, minus 2, plus 1; that is going to come out to -2.
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So, one more point, maybe: when x is 2, working this out, this is going to give me 2 squared; that is -4.
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And then, 2 times 2, with a negative, is also -4; plus 1--that is going to equal -7; so here we get -7.
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Go ahead and plot out the rest of these points.
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When x is -2, y is 1; when x is 0, y is 1; when x is 1, y is -2.
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And then, this one is way down here, so we will leave that off.
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But looking at how the axis of symmetry can help you with graphing:
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here I have one point on the left, and I am going...I have a couple points on the right.
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Because I know that the axis of symmetry divides this into two symmetric halves, I can actually add another point right over here.
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So, because this is symmetrical, I have a point here on one side of the vertex; I have another one here.
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Looking at the axis of symmetry, if I have a point right here, which is an x-value that is 1, 2 away from the vertex,
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so on this side it is 1, 2 away from the vertex at -3, and then the y is 1, 2, 3, 4 down;
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here I would have it, again, compared to the vertex, 1, 2, 3, 4 down.
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So, that is going to reflect it right across here.
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So, using the axis of symmetry, I can find the mirror image point.
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Even if I have just graphed this half, if I have the axis of symmetry, I can reflect across to the other half.
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Just reviewing: an axis of symmetry for a parabola is given by the equation right here; and I found my axis of symmetry.
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And that is a vertical line; the parabola actually intersects that axis at the vertex.
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The x-value of the vertex is also given by this equation; and once I have found that (which was -1),
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I can substitute that in and find the function's value for y at that point (which turned out to be 2).
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The axis of symmetry can help you with graphing.
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Now, we talked a little bit about the vertex; but the vertex can give you either a maximum or minimum value of the function,
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depending on the shape of the curve--whether it opens upward or downward.
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That tells you if you have a maximum value or a minimum value.
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So, investigating this a little bit further: if the coefficient of the x² term is positive, the graph will open upward,
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whereas if the coefficient of the x² term is negative, the graph opens downward.
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So, recall that the standard form is ax² + bx + c, so that is the standard form of the function f(x).
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And if I look at a, if a is greater than 0 (that means that the x² term is positive), the parabola opens upward.
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If this is a negative value--if a is less than 0--the parabola is going to open downward.
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Just by looking at the equation I am given, I will have a general sense of the shape--whether it opens upward or downward.
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Now, if this is positive, and it opens upward, let's give an example.
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If I have a parabola like this, and it opens upward, this tells me that a is greater than 0.
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The vertex is going to be a minimum; and thinking about what that means--
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that means the minimum value that y can have--the minimum value of the function.
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If the coefficient of x² is positive, the graph opens upward, and the function has a minimum value.
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Right here at the vertex, that is as small a value as you will find for y.
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y only gets bigger from there, no matter what x-value you put in.
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Conversely, if I have a parabola that opens downward (here a is less than 0--I have a parabola that opens downward),
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in this case, the vertex is going to give me a maximum value.
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So, right here, y is approximately equal to 1; that is as large as y is going to get.
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No matter what value I plug in for x, y is just going to be smaller than that.
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So again, this gives us more information about the graph, just by looking at what this coefficient is right here-- the coefficient of x².
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So, in order to graph, as I said, once you find the vertex, you look at the equation and say, "OK, it is going to open [upward or downward]."
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And then, look for some points around the vertex to get the shape of the graph.
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Does it open very wide? Is it a very shallow graph, or is it a very steep graph?
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And we will talk more about that later on--about the shape.
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But for now, just focus on the vertex--whether the parabola opens upward or downward,
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and then on finding some x and y values in order to help you make the graph.
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OK, let's practice these with some examples.
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Find the equation of the axis of symmetry, the coordinates of the vertex, and the graph of this function.
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OK, first recall that the axis of symmetry is given by the vertical line at x = -b/2a.
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And then, always just review what standard form is in order to find this: it is ax² + bx + c.
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So here, I have x =, and then my a value is 1; there is no bx term, therefore b must be 0.
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There is no term with a coefficient and then just x; that implies that b is 0.
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And then, here, c is -9; but I really don't need to worry about that right now.
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So, I am going to go ahead and substitute these in; and as soon as I see that b is 0,
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it honestly doesn't matter what a is, because if you take 0 and divide it by anything, you are going to get 0.
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Now, this is the equation for the axis of symmetry; it is a vertical line at x = 0, right here.
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The axis of symmetry is just going to follow this y-axis.
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The coordinates of the vertex: well, I have my x-coordinate right here, which is 0.
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That means that this axis of symmetry is going to intersect this parabola at x = 0, and then at some y-coordinate.
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In order to find the y-coordinate, let's find f(0).
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f(0) is -9; so, let's make this -2, -4, -6, -8, -10; this is the y-coordinate for the vertex.
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The x-coordinate is x = 0, so the coordinates of the vertex are (0,-9); (0,-9) is right there.
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Now, let me go ahead and write that in--that is what that is--that is the vertex, right there.
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Now, in order to graph this, I need to find some other points; so let's find some values for x, and then figure out what f(x) is at those points.
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When x is 1, that would be 1 squared minus 9; you are going to get -8 for y.
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When x is 2, 2 squared is 4, minus 9--that is going to give me 5.
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Let's see: when x is 3, 3 squared is 9, minus 9 is 0.
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So, let's try these points; and that is going to give me: when x is 1, y is -8; when x is 2, y is going to be...actually, it should be -5 right here.
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When x is 3, y is 0; 3 is about here.
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OK, now recall that the axis of symmetry can help me to graph--it can save me work,
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because here, I have some points on one side of the axis of symmetry, and that means that on the other side...
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I can just reflect across to get points that are mirror-image--approximately here, and then here at...this is here,
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and then here is at (3,0), so I am going to go over here to (-3,0) and put a point.
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Therefore, I was able to cut down my work by knowing that the axis of symmetry divides this into two symmetrical halves.
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So, checking, I did find the equation of the axis of symmetry, x = 0.
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I found the coordinates of the vertex, and that was (0,-9).
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And I graphed by finding some points, and then reflecting across, using the axis of symmetry.
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Again, find the equation of the axis of symmetry, the coordinates of the vertex, and the graph.
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The equation of the axis of symmetry is given by x = -b/2a.
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And here, I have a = 1, b = -4, because f(x) in standard form is ax² + bx + c.
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So, I am going to go ahead and find this; the axis of symmetry is x = -b (that is negative -4), over 2 times a (which is 1).
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Therefore, x equals...that negative and negative gives me...positive 4 over 2, which equals 2.
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So, this is the axis of symmetry; I'll go ahead and form an x-y axis.
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The axis of symmetry is going to be right here at x = 2, and there is going to be a vertical line passing through that point.
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My next thing that I need to do is find the coordinates of the vertex.
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And I have the x-coordinate (that is x = 2), but I need to find the y-coordinate.
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And the y-coordinate is going to be given for f(2), which equals 2 squared, minus 4, times 2, plus 4.
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So, that gives me 4 minus 8, plus 4, so that is 4 + 4 is 8, minus 8 is 0.
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So, this equals 0; so the y-coordinate f(2) for the vertex is 0.
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So, the vertex is at (2,0); that is where the axis of symmetry is going to intersect with the graph.
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Then, to finish graphing this, I just need to find some points--some x-values and some corresponding y-values.
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So, let's start out with some points around the vertex.
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When x is 1, that is going to give me 1 squared (is 1), minus 4, plus 4; so these two are going to cancel, and I am just going to get 1 right here.
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OK, I already have 2; that is my vertex; how about 3?
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When you go ahead and plug this in and figure it out, you are going to get 3 times 3 (that is 9), minus 3 times 4 (is 12), plus 4.
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So, that is 9 minus 12 (is going to give me -3), plus 4: that is going to give me 1.
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OK, when x is 4, if you work this out, you will find that y is 4; and then another easy point--when x is 0, these drop out; you get 4 here also.
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When x is 1, y is 1; when x is 3, y is 1; when x is 4, y is 4; when x is 0, y is 4.
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Now again, I could have used the axis of symmetry; I could have found even fewer points,
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and then just used symmetry to find the points on the other side of that axis.
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So, I completed the tasks that I was asked to do, which are to find the axis of symmetry (and that is right here at x = 2);
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to find the vertex (the vertex is here at (2,0)), and to graph this (which is a parabola opening upward,
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with a minimum value--the vertex here is a minimum, because this opens upward--at (2,0)).
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In this next one, we are asked to determine whether the function has a maximum or a minimum,
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Find the maximum or minimum, and state the domain and range.
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So, we don't even have to graph this one; but we first need to look at it and determine if it has a maximum or a minimum.
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And the way we figure that out is by seeing what this coefficient of x² is--what a is.
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Here, a equals -1; recall that, when a is less than 0--when a is negative--the parabola opens downward.
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Now, just sketching this out to think about it: if the parabola opens down--it faces downward--
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the vertex is going to give me a maximum value; that is as large as y is going to get; y only gets smaller.
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So here, the vertex is a maximum value--it doesn't get any larger.
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If this were to open upward, then the vertex would be a minimum value.
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So, here the parabola opens downward; my vertex is a maximum.
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Now, I am asked to find that maximum or minimum--in other words, to find the vertex.
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And recall that the vertex is given by -b/2a.
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So, I have that my a is -1; here, b is 12; so let's find the vertex.
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It is going to be a vertical line at x = -b, over 2 times -1.
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So, that is going to give me x = -12/-2, or x = 12/2, or x = 6.
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That gives me the equation; there is going to be a line at x = 6--that is the axis of symmetry.
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So, the axis of symmetry is at x = 6.
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And that is a maximum value; so the axis of symmetry is x = 6.
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The next thing I am asked to find is the domain and the range.
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Well, looking at the domain, I could say that x is really any real number.
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I could say it is one, a thousand, negative two hundred...so the domain is going to be all real numbers.
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It may be tempting to say, "Oh, the range is also all real numbers."
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But if you think about it, it is actually not: now, let's go a little bit deeper and find this maximum point--let's find the vertex.
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I found that x equals 6, so I know that my x-coordinate is 6; but let's find the y-coordinate, which is going to be at f(6).
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So, that f(6) equals -6 squared, plus 12, times 6, plus 8, which is going to equal...-6 times -6 is 36; 12 times 6 is 72.
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6 times 6 is 36, and that is negative, so it is -36, plus 72, plus 8.
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6 squared; take a negative; plus 12 times 6; plus 8; this is going to give me -36 + 72 + 8, which actually comes out to 44.
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So, my vertex is actually going to be at the point (6,44); and what that is telling me is that this is the largest y is ever going to get.
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Now, y can be an infinite number of smaller values; but this is the maximum it is going to get.
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Therefore, the range is actually all real numbers, where y is less than or equal to 44.
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So, it might be tempting to just look at this and say "all real numbers" for y, but that, in fact, is not true,
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because we have a maximum here that y cannot be greater than.
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Therefore, answering all of this, I determined that I have a vertex that is a maximum.
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I found what the maximum is, and it was x is 6 and y is 44; so the maximum is y = 44.
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And the domain and the range--the domain is all real numbers; the range is all real numbers such that y is less than or equal to 44.
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OK, Example 4: Determine whether the function has a maximum or a minimum.
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Find the maximum or minimum, and state the domain and range.
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Recall standard form, and that I need to look at the coefficient of x--I need to look at a; and here, a equals 3.
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And since that is greater than 0--since a is positive--this tells me that this parabola opens upward.
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And since it opens upward, just to help visualize it, the vertex right here is going to be a minimum.
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It is going to be the smallest value that y can achieve for any x.
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They want me to find that minimum value; so I am going to recall that the x-coordinate of the minimum value is defined by -b/2a.
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I know what a is; b is -12; so, just figuring that out, that is negative, times -12, over 2 times a (which is 3).
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That is going to give me...a negative and a negative is a positive 12, divided by 2 times 3, which is 6; so this is going to give me 12/6; that is 2.
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OK, so the x-coordinate for the minimum is 2; to figure out the y-coordinate, I am going to figure out f(2),
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which is going to be 3 times 2 squared, minus 12 times 2, plus 7.
00:29:06.600 --> 00:29:19.000
This is going to give me 3 times 4, minus 12 times 2 (is 24), plus 7; so this is 12 minus 24, plus 7.
00:29:19.000 --> 00:29:25.100
So, that is going to give me -12 + 7, which is -5.
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Therefore, the minimum value is going to be y = -5; so this is my minimum.
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Now, looking at this, in reality, actually, this is going to be at x = 2, y = -5; it is actually going to be more like this.
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We weren't asked to graph it, but just to help visualize it: -5, let's say, is down here; this is actually going to be down here, in this quadrant.
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OK, now, I found that I have a minimum; I have found that the minimum occurs at y = -5.
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What is the domain? Looking, I can make x anything I want--any real number--so the domain is going to be all real numbers.
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What is the range? Well, there is a limit on what that can be, because I just said that the minimum value for y is -5, right here.
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That is the minimum value; therefore, y can be greater, but it can never be less than that.
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So, actually, the range is all real numbers, where y is greater than or equal to -5.
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The domain is all real numbers; the range is more restricted--it is all real numbers greater than or equal to -5.
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OK, so we found that we had a minimum, and we know that because a is positive.
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We found that using the formula -b/2a to get the x-coordinate, which is 2.
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Then, we found f(x) using 2, so f(2), to tell me that y equals -5.
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Since y is -5, then there is a limit here on the range.
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The domain is all real numbers, whereas the range is only those real numbers greater than or equal to -5.
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That concludes this session of Educator.com, where we covered graphing quadratic functions.
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I will see you next lesson!