College Algebra: Write Quadratic Equation w Vertex

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So suppose that someone gives you a regular just quadratic equation in just sort of generic form, ax^2 + bx + c, and you want to write it in standard form, so that would be a(x - h), blah, blah, blah. Suppose that you're given that thing. How can you do that? Well, we just saw that you could complete the square; that's one method, but let me show you an easier method. So this method would be the following. Let me show you what we have. We're going to be given the function in sort of just generic form--ax^2 + bx + c. There's the quadratic. Yuck! Our fantasy is to write it like this. Our wish would be to write it this way. a(x - h)^2 + k, where this is the vertex, so the vertex is at (h, k). So the question is, how can I go from here to here? Well, in the last lecture we were actually talking about the notion of completing the square to get from here to here. In fact, if you complete the square, just in general, using this exact equation if you're really, really careful with the algebra, you'd actually see what h and k are. And like we saw before, h has to be equal to , and then k is just whatever the function is there, so k would be just f(h).
So you can actually go from here to here just by sort of plugging these things in. Let me demonstrate that for you with some examples. Let's take a look at the first one. Suppose that we have the following: f(x) = x^2 + 4x + 1. First of all, let's just find h and k. What would we get? What I would see is h would equal -b. Now what's the b here? Well, a is a 1, an invisible 1, b is a 4, and c is a 1. So I'm going to plug that in here. So, a -b, so that's a -4, all divided by 2 times a. Well, that's 2 times 1, which is 2, so I see a -2. So h equals -2. So the vertex is going to be located in the x direction. I'm going to move -2 over. Now, what about the y? Well, to find k all I have to do is plug in -2 into the thing. So what do I get? I plug in a -2^2, that's going to be 4, and then I have a 4 times -2. Well, that's a -8. -8 + 4 = -4 + 1 = -3. So I see (-2, -3). So how can I write this? Well, there's no coefficient, so I just put a 1 there, it's an invisible 1. x - h. h is -2. So minus a -2 is a 2^2, and then I add k, so that's now plus a -3. So now I've written this thing in standard form, just by plugging this in here.
Let's try another example. Just a little substitution and all of a sudden--whammo--we're in business. Let's take a look at the next example: f(x) = x^2 - 8x + 5. To do this one, all you have to do--here's the secret--a is 1, b is -8, and c is +5. So first let's find h. So h equals what? Well, it equals -b, so that's going to be -(-8), which would then be 16 over 2a. Well, that's 2 times 1, which is 2, and so I see that's going to be a value of 8, and what's k? Well, k is just f evaluated at 8, so I plug 8 into here. What's that going to be? That's going to be 8^2 and then I subtract 8 times 8. So it's 64 - 64, that's a zero, plus 5, so I get a 5. So I see the vertex of this quadratic--of this polynomial--is going to be at (8, 5). So therefore, can I write it in standard form? Absolutely--a here is just a 1, so I have a 1 times x, and then I have x - h, and what's h? h is just 8 here, so I have that squared, and then plus k. So, plus 5. Voila! There is this equation written in standard form just by using this little conversion and plugging it into here. Wait, stop everything! Did you find the little typo that I made? Well, I don't know how little it was, but if you look down here the formula is h = , and look what I wrote here. -b would be just 8. I wrote 16. What was I thinking? Well, I don't know, but the answer should be 8. And 8/2 would be 4. So this should have been a 4. So h should be 4, and then what would k be? Well, it would be f(4), so in fact, this should be a 4 here. And so now what is f(4)? Well, I go back to the function and I plug in 4. So 4^2 is 16 minus 8 times 4 is 32, so it's 16 - 32 = -16 + 5 = -11. So, in fact, this number should be -11. So the actual answer should be x - 4, and then -11. Whew, major typo. Caught it just in time. Always get to think about your work and make sure someone's looking over your shoulder and checking it for you.
Anyway, let's go back to the lecture already in progress. I'll try one last example just to show you that I have an ample supply of paper. This example will be less rosy, because in fact, f(x) here is going to equal -x^2 + 4x + 2. Why do I say it's less rosy? Well, because that negative in front of the x^2 means this is a sad faced parabola. This is going to be one that's going down. It's not good, but it's okay, because we'll be able to understand it. So now the role of a, by the way, is being played by -1. B is being played by 4, and c is being played by 2. So let's find h. That's the x value for the vertex. -b, so that's -4, all divided by 2 times a, that's 2 times -1, which is a -2. -4 divided by -2 is just equal to 2. And k, what do I do? I just look at f, evaluate it at this point, 2. So I plug in 2 into here and see what I get. I take 2 and square it, that gives me a 4. So the negative sign is in front, so I see a -4, and then here I have a plus, and then this is going to give me an 8. So I have -4 + 8, which gives me a 4, and then +2 is a 6. And so I see that the vertex is located at (2, 6). So therefore, what does this thing look like? Well, now there is an a that's not 1. So I've got to put that negative sign way out in front because I have the a there, so I have a minus sign, (x - h)^2 and then I add k, which is 6. So there that negative sign tells me this is still a sad faced parabola. In fact, now that's exactly this quadratic, exactly this parabola, but now written in standard form.
So all we have to do to write them in standard form, to write it in this form, is just to find the h, , find the k, plug that value back into the function, and put the h and the k right into standard form with the coefficient of a sitting out in front. That's all there is to it. Try these and see what you think.
Relations and Functions
Quadratic Functions- the Vertex
Using the Vertex to Write the Quadratic Equation Page [2 of 2]