Main content

## Algebra 1

### Course: Algebra 1 > Unit 14

Lesson 11: Features & forms of quadratic functions- 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?

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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.(12 votes)**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:

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:completing-square-quadratics/v/solving-quadratic-equations-by-completing-the-square

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! :)(21 votes)

- How do you put a quadratic function into factored form when the function only has two numbers, like h(t) = -5t^2 + 45t?(8 votes)
- Factor out a t like this: t(-5t-45). The roots would be t=0 OR t=-9(2 votes)

- it's not often you find a video without any comments whatsoever(7 votes)
- 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.(2 votes)
- You are right ofc but if it helps your imagination, just add an additional velocity function describing the descent :)(5 votes)

- 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.(3 votes)
- 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?(2 votes)
- Can't you also use the Quadratic formula for this problem?(2 votes)
- Yes, you can. There are multiple ways to factor.(2 votes)

- 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(4 votes)

- In the second rewrite of the equation, we have t^2-8t-20 = 0.

However, when we have a positive coefficient to the x^2 term, then isn't the parabola going to face up(like a smile) ? Yes, when you equate for the zeros in the same equation you get 10 and -2, but what does the positive squared term mean over here ?(2 votes)- You give hints on how to answer your own question. By setting it equal to 0 (the x intercepts are where y=0), you no longer have the original function, you only have a way to find the two x intercepts. So f(t)=t*2-8t-20 does open upward, but t^2-8t-20=0 only is good for the x intercepts, not the full parabola.(2 votes)

- you can just use the quadratic formula... right?(2 votes)
- 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(2 votes)

## Video transcript

- [Instructor] We're
told that Rodrigo watches a helicopter take off from a platform. The height of the helicopter,
in meters above the ground, t minutes after takeoff is modeled by, and we see this function right over here. Rodrigo wants to know when the helicopter will land on the ground. So pause this video and see
if you can figure that out. All right, now let's
think about this together. So let's just imagine
actually what the graph of this function looks like. And it'll also help us
imagine what's going on with the helicopter. So our horizontal axis,
this is t, time in minutes, and then our vertical axis is height. So height as a function of time. And maybe I just write it like this. I'll just write height, and this is given in
meters above the ground. Now I don't know exactly
what the graph looks like, but given that I have
a negative coefficient on my quadratic term, I know that it is a downward
opening parabola like that. And it says that the helicopter
takes off of a platform. So however high the platform is, then it takes off and it's
going to do something like this. I don't know exactly what
the graph looks like, but probably something like this. Now, if they asked us
what is the highest point of the helicopter, and at
what time does it happen, then we'd wanna figure
out what the vertex is of this parabola. But that's not what they're asking. They're asking when does the
helicopter land on the ground? That's this time right over here. So if we wanted to find the vertex, we would wanna put this into vertex form, but here we wanna figure out when does that function equal zero. We want to find a zero of this
quadratic right over here. So the best way that I
can think about doing it is try to factor it, try to set them this thing equal to zero, and then factor it and
then see what t values make that equal to zero. So let me do that. So I say negative three t
squared plus 24t plus 60, remember we care when our height is equal to zero, equal zero. So let's see maybe the
first thing I would do just to simplify the second
degree term a little bit. Let's just divide both
sides by negative three. If we did that, this would become t squared
24 divided by negative three is negative eight, negative eight t. 60 divided by negative
three is negative 20, and then zero divided by negative three is of course still zero. And now can I think of
two numbers whose product is negative 20? So they would have different signs in order to get a negative product, and who sum is negative eight. So let's see what about
negative 10 and two, and that scene seems to work. So I could write this as t
minus 10, times t plus two is equal to zero. And so in order to make
this expression equals zero, either one of these
could be equal to zero. So either t minus 10 is equal to zero, or t plus two is equal to zero. And of course on the left here, I can add 10 to both sides. So either t equals 10, or I could subtract two
from both sides here, t is equal to negative two. So there's two places where the function is equal to zero. One at time t equals negative two, and one at time t is equal to 10. Now we're assuming we're
dealing with positive time here. We don't know what the
helicopter was doing before the takeoff. So we wouldn't really think about this. So what we really care about is that t is equal to 10 minutes. That's when the helicopter
is right over there. And actually we know at t equals zero, these two terms become zero. We know it takes off at 60 meters. It goes up. If we figured out the vertex, we would know how high it went, but then it starts going back down, and in 10 minutes after takeoff, it is back at zero, back on the ground.