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Interpret quadratic models: Vertex form | Algebra I | Khan Academy


4m read
·Nov 10, 2024

We're told that Taylor opened a restaurant. The net value of the restaurant, in thousands of dollars, two months after its opening is modeled by ( v(t) = 2t^2 - 20t ). Taylor wants to know what the restaurant's lowest net value will be. Let me underline that, and when it will reach that value.

So, let's break it down step by step. The function which describes how the value of the restaurant, the net value of the restaurant, changes over time is right over here. If I were to graph it, I can see that the coefficient on the quadratic term is positive, so it's going to be some form of upward-opening parabola. I don't know exactly what it looks like; we can think about that in a second.

So, it's going to have some point right over here which really is the vertex of this parabola, where it's going to hit its lowest net value. And that's going to happen at some time ( t ). If you could imagine that this right over here is the ( t )-axis, so my first question is: is there some form, is there some way that I can rewrite this function algebraically so it becomes very easy to pick out this low point, which is essentially the vertex of this parabola? Pause this video and think about that.

All right, so you could imagine the form that I'm talking about is vertex form, where you can clearly spot the vertex. The way we can do that is actually by completing the square. So the first thing I will do is actually let me factor out a 2 here, because 2 is a common factor of both of these terms. So, ( v(t) ) would be equal to ( 2(t^2 - 10t) ) and I'm going to leave some space because completing the square, which gets us to vertex form, is all about adding and subtracting the same value on one side. So, we're not actually changing the value of that side but writing it in a way so we have a perfect square expression, and then we're probably going to add or subtract some value out here.

Now, how do we make this a perfect square expression? If any of this business about completing the square looks unfamiliar to you, I encourage you to look up completing the square on Khan Academy and review that. But the way that we complete the square is we look at this first-degree coefficient right over here; it's negative 10. We say, "All right, well let's take half of that and square it." So half of negative 10 is negative 5, and if we were to square it, that's 25.

So if we add 25 right over here, then this is going to become a perfect square expression. You can see that it would be equivalent to this entire thing if we add 25. Like that is going to be equivalent to ( (t - 5)^2 )—just this part right over here. That's why we took half of this and we squared it.

But as I alluded to a few seconds ago, a few minutes ago, you can't just willy-nilly add 25 to one side or to an equation like this. That will make this equality no longer true. In fact, we didn't just add 25. Remember, we have this 2 out here; we added ( 2 \times 25 ). You can verify that if you redistribute the 2, you'd get ( 2t^2 - 20t + 50 + 2 \times 25 ). So, in order to make the equality—or in order to allow it to continue to be true—we have to subtract 50.

So just to be clear, this isn't some kind of strange thing I'm doing. All I did is add 50 and subtract 50. You're saying, "Wait, you added 25, not 50." No, look, when I added 25 here, it's in parentheses, and that whole expression is multiplied by 2. So I really did add 50 here, and then I subtract 50 here to get back to what I originally had.

When you view it that way, now ( v(t) ) is going to be equal to ( 2 ) times this business, which we already established is ( (t - 5)^2 ), and then we have the minus 50. Now, why is this form useful? This is vertex form. It's very easy to pick out the vertex; it's very easy to pick out when the low point is.

The low point here happens when this part is minimized, and this part is minimized. Think about it: you have two times something squared, so if you have something squared, it's going to hit its lowest point when this something is zero. Otherwise, it's going to be a positive value. So this part right over here is going to be equal to zero when ( t = 5 ).

So, the lowest value is when ( t = 5 ). Let me do that in a different color; I don't want to reuse the colors too much. If we say ( v(5) ) is going to be equal to ( 2 \times (5 - 5)^2 - 50 ). Notice this whole thing becomes 0 right over here, so ( v(5) ) is equal to negative 50.

That is when we hit our low point in terms of the net value of the restaurant. So, ( t ) represents months. We hit our low point; we rewrote our function in vertex form, so it's easy to pick out this value, and we see that this low point happens at ( t = 5 ), which is at time five months.

Then, what is that lowest net value? Well, it's negative 50, and remember the function gives us the net value in thousands of dollars. So it's a negative 50 thousand dollars is the lowest net value of the restaurant.

And you might say, "How do you have a negative value of something?" Well, imagine if, say, the building is worth fifty thousand dollars, but the restaurant owes a hundred thousand dollars; then it would have a negative fifty thousand dollar net value.

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