- Forms & features of quadratic functions
- Worked examples: Forms & features of quadratic functions
- Features of quadratic functions: strategy
- Vertex & axis of symmetry of a parabola
- Finding features of quadratic functions
- Features of quadratic functions
- Graph parabolas in all forms
- Interpret quadratic models: Factored form
- Interpret quadratic models: Vertex form
- Interpret quadratic models
- Graphing quadratics review
- Creativity break: How does creativity play a role in your everyday life?
Given a quadratic function that models a relationship, we can rewrite the function to reveal certain properties of the relationship. Created by Sal Khan.
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- Can someone quickly run me through all the tips in all the forms. Your answer would be so much help to others and me! You might get more than just 10 votes.
How to find vertex
Example of that equation
How to find vertex
Example of that equation
How to find vertex
Example of that equation
- Vertex form is a form of a quadratic equation that displays the x and y values of the vertex.
You only need to look at the equation in order to find the vertex.
In this case, the vertex is located at (2,-10).
Explanation: since -2 is in the parenthesis, the quadratic equation shifts 2 units to the right. Since the -10 is the constant, the equation shifts 10 units down.
Standard form is another form of a quadratic equation.
f(x) = ax^2+bx+c
To find the vertex in this form, you need to take negative b and divide it by 2a.
Since b =4 and a=1, -(4)/2(1)= -4/2 = -2.
Now that the x-coordinate of the vertex is known, you can substitute the x value in the equation.
f(x) = (-2)^2+4(-2)+4
f(x) = 4-8+4
f(x) = 0
The vertex is located at (-2,0).
The factored form of quadratic equations is basically the product of the two binomials that led to the quadratic equation. This allows you to see the x-intercepts of the quadratic.
To find the vertex in this form, you must take the average of the zeroes of the equation. In order to find the zeroes, you must put the value of f(x) to zero and solve for both values of x.
(x+2)(x-3) = 0
x+2 = 0 and x-3 = 0
x=-2 and x=3
Now, take the average of the zeroes.
This means that the x value of the vertex is equal to 1/2.
Substitute the value of x into the equation.
So, the vertex is located at (1/2,-25/4)
Hope this helps, and sorry it was so long. I really needed to explain everything to avoid confusion.(21 votes)
- How is it that sometimes, I can just simplify the coefficients and sometimes I can't. Like, in other problems, I could just simplify the 2 out so that it could be t^2-10t but for this one I need to take out the 2 like 2(t^2-10t.. etc
Can I only simplify it down when it's in like standard form or something?(5 votes)
- If it's a function, dividing both sides by 2 will get you 1/2v(t)=t^2-10t. You have to divide both sides by 2, which includes the v(t) on the right. With an equation like 0=2t^2-10t, the 2's go out, because 0 divided by 2 is still 0(3 votes)
- First, I factored v(t)=2t^2-20t to be 0=2t(t-10). This gave me the correct zeros (0,0) and (10,0) which I used to get the axis of symmetry (t=5) which got me the vertex, (5,-50).
I double checked using the Completing the Square method. This is where I'm a little confused and making some assumptions. I set the equation to 0. I made it 0=2t^2-20t. I know that you can only complete the square if the first term is equal to 1. I divided everything by 2 and eventually got 0=(t-5)^2-25. This also got me the correct zeros, however, I was under the impression this step in the process was the equivalent of converting the function to vertex form. It seems this is not the case. This is the assumption I'm making and I'm wondering if it's true. Completing the square and converting to vertex form are not the same process. The main distinction is that when you have the equation set to 0, you can divide everything by the leading coefficient, including zero, but when it is set to f(x) for example, you must factor out the leading coefficient(?), which, in this case, is NOT the GCF. That would be 2t, not just 2. You must factor out the leading coefficient and THEN complete the square, which leads to different numbers than if you set everything to 0 and just divided by 2. In this case it leads to v(t)=2(t-5)^2-50, not (t-5)^2-25, which was a step in my process of solving for the zeros using the Completing The Square method. Please correct anything wrong in my understanding of this.(3 votes)
- The zeros of 𝑣(𝑡) = 2𝑡² − 20𝑡 are all the values of 𝑡 for which 𝑣(𝑡) = 0,
so to find the zeros we solve the equation 2𝑡² − 20𝑡 = 0.
At its vertex, however, we don't know what 𝑣(𝑡) is, so we can't set it equal to zero.
This means that when dividing by 2, we actually get
𝑣(𝑡)∕2 = 𝑡² − 10𝑡
After completing the square, we then have
𝑣(𝑡)∕2 = (𝑡 − 5)² − 25
Now we can multiply the 2 back to the right-hand side, which gives us
𝑣(𝑡) = 2(𝑡 − 5)² − 50 ⇒ 𝑣(5) = −50(6 votes)
- So - How do we find the zeros of 2t^2 - 20t?
I am completely lost now. I thought I understood how to do this but the 2 at the beginning is messing everything up for me.(3 votes)
- You have to start off with a function y=2t^2-20t or all you can do is factor. Since 20 is even, it is easy to factor out 2t to get y = 2t(t-10), and y=0 for zeroes. Thus, either 2t=0 or t-10=0.(2 votes)
- At about4:50, I understand how he zeroes out the -5, but wouldn't the -50 become -100 (2x-50)?(1 vote)
- Because of the order of operations, you do the exponent, parenthesis, multiplication, and division, then addition and subtraction. The first half is zero, so when you do subtraction it will be 0-50. That will equal -50, so that is the answer. I hope this helps.(2 votes)
- how do you know when to use factored or vertex form when solving a equation?(1 vote)
- Factored form is best to find roots assuming that there are roots. You could then easily find axis of symmetry and by substitution find the vertex. Vertex form allows you to solve by taking the square root. If you want to find the solutions (x intercepts), factored form is easiest since it will generally give whole numbers. However, solutions are not always whole numbers, so not easily factorable.(1 vote)
- Im extremely confused when it comes to completing the square, how are you getting these numbers?(1 vote)
- Why does it have to be a subtraction sign if the answer is positive? Like why can't you put (t+3)^2 instead of (t-3)^2(0 votes)
- [Instructor] We're told that Taylor opened a restaurant. The net value of the restaurant in thousands of dollars, t months after its opening is modeled by v of t is equal to two t squared minus 20t. Taylor wants to know what the restaurant's lowest net value will be, 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. And 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 in that value, and that's going to happen at some time t, if you can 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 re-write 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 can imagine the form that I'm talking about is vertex form, where you can clearly spot the vertex. And 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 two here, because two is a common factor of both of these terms. So v of t would be equal to two times t squared minus 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, and we say all right, well let's take half of that and square it. So half of negative 10 is negative five, 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. And 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 minus five squared, 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, or a few minutes ago, you can't just willy nilly add 25 to one side of an equation like this, that will make this equality no longer true. And in fact we didn't just add 25. Remember we have this two out here, we added two times 25. You can verify that if you redistribute the two, you'd get two t squared minus 20t plus 50, plus two 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 was add 50 and subtract 50. You're saying wait, you added 25, not 50. No look, when I added 25 here, it's in a parenthesis, and then the whole expression is multiplied by two, so I really did add 50 here, so then I subtract 50 here to get to what I originally had. And when you view it that way, now v of t is going to be equal to two times this business, which we already established is t minus five squared, 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. And so this part right over here is going to be equal to zero when t is equal to five. So the lowest value is when t is equal to five. Let me do that in a different color, don't wanna reuse the colors too much. So if we say v of five is going to be equal to two times five minus five, trying to keep up with the colors, minus five squared minus 50. Notice this whole thing becomes zero right over here. So v of five 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, so we hit our low point, we rewrote our function in a form, in vertex form, so it's easy to pick out this value, and we see that this low point happens at t equals five, which is at time five months. And 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 negative $50,000 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 $50,000, but the restaurant owes $100,000, then it would have a negative $50,000 net value.