Forms and features of quadratic functions
Current time:0:00Total duration:8:29
I have a function here defined as x squared minus 5x plus 6. And what I want us to think about is what other forms we can write this function in if we, say, wanted to find the 0s of this function. If we wanted to figure out where does this function intersect the x-axis, what form would we put this in? And then another form for maybe finding out what's the minimum value of this. We see that we have a positive coefficient on the x squared term. This is going to be an upward-opening parabola. But what's the minimum point of this? Or even better, what's the vertex of this parabola right over here? So if the function looks something like this, we could use one form of the function to figure out where does it intersect the x-axis. So where does it intersect the x-axis? And maybe we can manipulate it to get another form to figure out what's the minimum point. What's this point right over here for this function? I don't even know if the function looks like this. So I encourage you to pause this video and try to manipulate this into those two different forms. So let's work on it. So in order to find the roots, the easiest thing I can think of doing is trying to factor this quadratic expression which is being used to define this function. So we could think about, well, let's think of two numbers whose product is positive 6 and whose sum is negative 5. So since their product is positive, we know that they have the same sign. And if they have the same sign but we get to a negative value, that means they both must be negative. So let's see-- negative 2 times negative 3 is positive 6. Negative 2 plus negative 3 is negative 5. So we could rewrite f of x. And so let me write it this way. We could write f of x as being equal to x minus 2 times x minus 3. Now, how does this help us find the zeroes? Well, in what situations is this right-hand expression, is this expression on the right hand going to be equal to 0? Well, it's the product of these two expressions. If either one of these is equal to 0, 0 times anything is 0. 0 times anything else is 0. So this whole thing is going to be 0 if x minus 2 is equal to 0 or x minus 3 is equal to 0. Add 2 to both sides of this equation. You get x is equal to 2 or x is equal to 3. So those are the two zeroes for this function, I guess you could say. And we could already think about it a little bit in terms of graphing it. So let's try to graph this thing. So this is x equals 1. This is x equals 2. This is x equals 3 right over there. So that's our x-axis. That, you could say, is our y is equal to f of x axis. And we're seeing that we intersect both here and here. When x is equal to 2, this f of x is equal to 0. When x is equal to 3, f of x is equal to 0. And you could substitute either of these values into the original expression. And you'll see it's going to get you to 0 because that is the same thing as that. Now, what about the vertex? What form could we write this original thing in order to pick out the vertex? Well, we're already a little familiar with completing the square. And when you complete the square with this expression, that seems to be a pretty good way of thinking about what the minimum value of this function is. So let's just do that right over here. So I'm just going to rewrite it. So we get f of x is equal to x squared minus 5x. And I'm just going to throw the plus 6 right over here. And I'm giving myself some real estate because what I need to do, what I want to think about doing, is adding and subtracting the same value. So I'm going to add it here, and I'm going to subtract it there. And I can do that because then I've just added 0. I haven't changed the value of this right-hand side. But I want to do that so that this part that I've underlined in this magenta color, so that this part right over here, is a perfect square. And we've done this multiple times when we've completed the square. I encourage you to watch those videos if you need a little bit of a review on it. But the general idea is this is going to be a perfect square if we take this coefficient right over here. We take negative 5. We take 1/2 of that, which is negative 5/2, and we square it. So we could write this as plus negative-- what's negative 5/2 squared? So I could write this-- negative 5/2 squared. Well, if we square a negative number, it's just going to be a positive. So it's going to be the same thing as 5/2 squared. 5 squared is 25. 2 squared is 4. So this is going to be plus 25/4. Now, once again, if we want this equality to be true, we either have to add the same thing to both sides. Or if we're just operating on one side, if we added it to that side, we could just subtract it from that side. And we haven't changed the total value on that side. So we added 25/4, and we subtracted 25/4. So what is this part right over here? What does this become, the part that I've underlined in magenta? Well, this is going to be-- the whole reason why we engineered it in this way is so that this could be x minus 5/2 squared. And I encourage you to verify this. And we go into more detail about why taking 1/2 the coefficient here and then squaring it, adding it there and then subtracting there, why that works. We do that in the completing the square videos. But these two things, you can verify that they are equivalent. So that's that part. And now we just have to simplify 6 minus 25/4. So 6 could be rewritten as 24/4. 24/4 minus 25/4 is negative 1/4, so minus 1/4, just like that. So we've rewritten our original function as f of x is equal to x minus 5/2 squared minus 1/4. Now, why is this form interesting? Well, one way to think about it is this part is always going to be non-negative. The minimum value of this part in magenta is going to be 0. Why? Because we're squaring this thing. If you're taking something like this-- and we're just dealing with real numbers-- and you're squaring it, you're not going to be able to get a negative value. At the minimum value, this is going to be 0. And then it obviously could be positive values, as well. So if we want to think about when does this thing hit its minimum value-- well, it hits its minimum value when you're squaring 0. And when are you squaring 0? Well, you're squaring 0 when x minus 5/2 is equal to 0, or when x is equal to 5/2 if you just want to add 5/2 to both sides of that equation. So this thing hits its minimum value when x is equal to 5/2. And then what is y, or what is f of x, when x is equal to 5/2? f of 5/2-- and once again, you could use any of those forms to evaluate 5/2. But it's really easy in this form. When x is equal to 5/2, this term right over here becomes 0. 0 squared, 0. You're just left with negative 1/4. So another way to think about it is our vertex is at the point x equals 5/2, y equals negative 1/4. So x equals 5/2. That's the same thing as 2 and 1/2. So x equals 5/2. And y is equal to negative 1/4. So if that is negative 1, 1/4 would be something like that. So that right over there is the vertex. That is the point-- let me make it clear-- that's the point 5/2 comma negative 1/4. And what's cool is we've just used this form to figure out the minimum point, to figure out the vertex in this case. And then we can use the roots as two other points to get a rough sketch of what this parabola will actually look like. So the interesting-- or I guess the takeaway from this video is just to realize that we can rewrite this in different forms depending on what we're trying to understand about this function.