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# Interpret quadratic models: Factored form

Given a quadratic function that models a relationship, we can rewrite the function to reveal certain properties of the relationship. Factored form helps us identify the x-intercepts or zeros of the function. Created by Sal Khan.

## Want to join the conversation?

• 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
Example: (x-1)^2+5

How to find the vertex:

1. Look at the part being squared, so in this case it is (x-1).

2.Find the constant term in the part that is being squared. In this case, the constant is -1.

3. Find the opposite of the constant. In this case the opposite of the constant (-1) is equal to 1. This is the x-coordinate.

4. The y-coordinate is the constant term of the entire equation. In this case, the y-coordinate is 5.

In Conclusion, the Vertex of the equation is (1,-5).

Standard Form
Example: x^2+6x+5

How to find the vertex:

To find the vertex from standard form, you must complete the square. You can find how to complete the square here:

Once you complete the square, the equation will be in vertex form (see above!), and from there you can solve for the vertex.

Going back to my example of Standard Form, you will eventually get (x+3)^2 - 4 once you complete the square. From there you will find out that the vertex is (-3,-4). (Once again, you can see above for how to find out coordinates from vertex form.

Factored Form
Example: (x+4)(x-2)

How to find the vertex:

To find the vertex from factored form, you must first expand the equation into standard form. From there, you must complete the square (see above!).

If you are following my example of factored form, you should get x^2+2x-8 once you expand. From there, you can convert that to vertex form, which will be (x+1)^2 - 9. From here, you can figure out that the vertex is located at (-1,-9). (As Always, details on how to do this are above!)

You're Welcome! :)
• How do you put a quadratic function into factored form when the function only has two numbers, like h(t) = -5t^2 + 45t?
• Factor out a t like this: t(-5t-45). The roots would be t=0 OR t=-9
• it's not often you find a video without any comments whatsoever
• Isn't this problem a little violent? That helicopter is going to be traveling at the maximum velocity achieved in flight when it hits the ground.
• You are right ofc but if it helps your imagination, just add an additional velocity function describing the descent :)
• I gotta be honest with you Sal, you never do any hard equations going through this and then when I get to the unit test or the practice units the questions are completely different and usually start with fractions which lets face it, everybody hates fractions. You need to start pulling up with stuff similar to what is going to be asked in the actual unit tests/ practice units. It would be helpful for sure in my opinion.
• you can just use the quadratic formula... right?
• Hi Mark,
Typically speaking, it is much faster and easier to find the roots of a quadratic equation using factoring; however, not all quadratic functions can be easily factored, especially when they involve complex solutions. In these cases, it may be best to use the quadratic formula.

Thanks!
Thomas
• Can't you also use the Quadratic formula for this problem?
• Yes, you can. There are multiple ways to factor.
• I understand how to find what the problem asks me to find (vertex, zeros, y-intercept) but I have trouble putting the problem into the form the problem asks (factored, vertex,standard). For example I put the factored form as (t-8)(t+4)=0 but the correct way to write it was h(t)= -3(t-8)(t+4). How can I improve my answers?
• So in this equation h(t)=-3t^2+24t+60 if t=0, then
h(0)=-3*0+24*0+60
h(0)=60
This is the maximum point of the graph right, because that's the maximum value that the graph can take. But maximum point of every graph, especially this graph, doesn't have 0 as the x value. What am I missing? I think I've muddled up something with the vertex form but I can't figure out what. It would be a huuuuuuge favour if someone can sort this out for me please :)
(1 vote)
• To get the function in vertex form, first factor out −3, then complete the square, and finally distribute the −3 back in.
ℎ(𝑡) = −3𝑡² + 24𝑡 + 60
= −3(𝑡² − 8𝑡 − 20)
= −3((𝑡 − 4)² − 36)
= −3(𝑡 − 4)² + 108