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# Interpret quadratic models: Vertex form

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.

Vertex form
How to find vertex
Example of that equation
Etc.

Standard
How to find vertex
Example of that equation
Etc.

Factored
How to find vertex
Example of that equation
Etc.
• Vertex form is a form of a quadratic equation that displays the x and y values of the vertex.
f(x)= a(x-h)^2+k.
You only need to look at the equation in order to find the vertex.
f(x)= 2(n-2)^2-10
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.
Example: x^2+4x+4
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.
(x+a)(x+b)
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.
-2+3/2= 1/2
This means that the x value of the vertex is equal to 1/2.
Substitute the value of x into the equation.
(1/2+4/2)(1/2-6/2)
(5/2)(-5/2)
-25/4
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.
Also, don't solicit votes like that.
• 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?
• 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
• 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.
• 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
• couldn't you just use the formula -b/2a to find the x-value of the vertex, then plug in said value to find the y-value as well?
• Yes, it is really quick
• 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.
• 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.
• In the practice, due to the question involving square roots of non-perfect squares, the question said: Round to two decimal places. So when i took the square root 2, I rounded it. However, I got it wrong. The right answer was basically where, only the FINAl ANSWER was rounded and not when the square root was taken. So basically, whenver it says to round, you just round the final answer?
• Yes, that is what you have to do, otherwise you are rounding rounded answers which cause more and more errors. Do not round until the end.
• Does anyone know why sometimes we can divide the entire equation by let's say 2 in this case, and other he factors out?
• Maybe it has to do with the difference of an equation and with solving for the x intercepts. If you have y=2x^2+8x+6, you have to factor out a 2 because if you divided by 2, you would also have to divide y/2. If you are trying to find the x intercepts, where y=0, you would have 2x^2+8x+6=0, so you can divide by 2 because 0/2=0.