Algebra 1 (Eureka Math/EngageNY)
- Finding the vertex of a parabola in standard form
- Graphing quadratics: standard form
- Graph quadratics in standard form
- Quadratic word problem: ball
- Quadratic word problems (standard form)
- Forms & features of quadratic functions
- Worked examples: Forms & features of quadratic functions
- Features of quadratic functions: strategy
- Finding features of quadratic functions
- Warmup: Features of quadratic functions
- Features of quadratic functions
- Graph parabolas in all forms
Different forms of quadratic functions reveal different features of those functions. Here, Sal rewrites f(x)=x²-5x+6 in factored form to reveal its zeros and in vertex form to reveal its vertex. Created by Sal Khan.
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- The vertex from what I have understood from the video is the lowest point in a parabola. But what does the roots imply?(27 votes)
- The roots are the x-intercepts, where the parabola crosses the x-axis. If the parabola opens up and it's vertex is below the x-axis then it crosses the x-axis in two places and has two (real) roots. If the vertex is on the x-axis then the parabola has one root. If the vertex is above the x-axis (and the parabola opens up) then that parabola has no real roots... it still has roots, but they are complex numbers.(66 votes)
- Is there a difference between quadratic equation, quadratic, and quadratic function(21 votes)
- Good question!
quadraticrefers to the degree of a polynomial such as x² - 4x + 3
To be quadratic, the highest power of any term must be 2 (the x is squared). If there is
no equals sign, but it has a quadratic term, then it is a
x² - x - 5 is a
So are the following:
a² + 8a - 6
g² - 16
If there is an equals sign, we call it a
An example is
x² - 4x + 3 = 0
Another example of
16t² - 8t = 3
x = y²
1 + 4/5x - x² = 0
If we are defining how two variables are related, we can be talking about a
function. There are limits on what we can call a function, though, because there has to be one unique value for the dependent variable for any value of the independent variable in order to be a function.
y = x² - 8x + 15 is a
quadratic functionthat describes every point on a parabola
f(x) = x² - 4x + 3 is also a
quadratic functionthat describes every point on another parabola
quadratic equationlooks similar, but is not a quadratic function: it is not a function at all, in fact.
x = y² - 8y - 20
It describes a parabola that is lying on its side on a coordinate grid, but most values of x for this parabola that have any value at all will have two different values of y, so it cannot be a function. It would still be classified as a
quadratic equation(67 votes)
- Is there a easy trick for knowing if whether a parabola is a mininum or a maxinum?(11 votes)
- If you look at the first term before the parenthesis, there would be a number. If its negative, it has a maximum. If its positive, then it has a minimum.(25 votes)
- I obtained the vertex by solving the original equation over the average of the roots of x and got the same value - this is to say that I added the roots and divided by 2 then used the result as x. Is this a legitimate way to find the vertex for all cases? Does anyone know why this works?(14 votes)
- Yes, since a parabola is symmetric about the line of symmetry, you can find it by finding the mid point of the two zeros; however, unless you also need to find the roots for some other reason, it is certainly not the most efficient way to find the vertex.(11 votes)
- I have a quadratic equation that I cannot figure out.
I know the parabola points downward because a is a negative, but I cannot seem to find the axis of symmetry and the vertex. I have the y-intercept, -3.
I keep getting 2 for the axis and -7 for the vertex. None of these answers fit on the graph. Could someone explain this to me, please? Thank you.(5 votes)
- Hi apandafamily. You are correct that the y-intercept is -3. However, the rest of your solution is incorrect so let's take a look at how to solve this. First, I like to factor out the negative sign so I can more easily find the roots and minimize errors in my calculation. Let's rewrite our equation.
y=-x²-4x-3 then becomes y=-(x²+4x+3). Now when we factor, this becomes
y=-(x+3)(x+1). Therefore, the roots are 0=-(x+1) so x=-1 and 0=-(x+3) so x=-3. Now take a look at the roots on a number line. The axis of symmetry will have to lie between -3 and -1 so it will be -2. If you substitute -2 into either the original equation or even our modified equation, y=-1*(x²+4x+3), you'll get y=-1*((-2)²+4*(-2)+3)=-1*(4-8+3)=-1*-1=1.
So in summary, the roots of the polynomial are x=-1 and x=-3 with the vertex at (-2,1).(14 votes)
- Is there a formula for finding the vertex of a parabola? Is there a video on that topic?(3 votes)
Here is a video on finding the vertex of a parabola. https://www.khanacademy.org/math/algebra/quadratics/solving_graphing_quadratics/v/finding-the-vertex-of-a-parabola-example(6 votes)
- Thanks for the help, but I am still confused as to why you used completing the square to achieve the minimum point/vertex? Also is this the same for finding the maximum value?(1 vote)
- Completing the square puts a quadratic into what is known as the vertex form, so you can read the vertex directly from the completed square.
Yes it would be the same for max value, it would just have a negative in front of the x^2 term. So if you started with f(x) = - x^2 + 5x - 6, you would get - (x-2)(x-3) for factoring and -( x - 5/2)^2 + 1/4, the vertex would be at (5/2,1/4) instead of the minimum of (5/2, - 1/4).(5 votes)
- Where the positive 25/4 at5:35went to? Sal at5:20wrote (x-5/2)^2 but did not write 25/4, only he wrote - 1/4 at5:55for what is left.(2 votes)
- In order to "complete the square" (for x^2 -5x), Sal ADDED 25/4 to the right side of the equation. Those 3 terms then were then able to be written as ( x - 5/2 ) ^ 2.
But in order not to change the equation, he also had to SUBTRACT 25/4 from the right hand side of the equation which also included the constant term of + 6.
The sum of 6 and - 25/4 produced the - 1/4 that Sal wrote.
So the equation x^2 - 5x + 6 became the same as
( x - 5/2 ) ^ 2 + ( 6 - 25/4) which equals
( x - 5/2 ) ^ 2 - 1/4.
Hope this makes things a bit clearer!(3 votes)
- If I have f(x)=3(x+1)^2-7 or something. Can I distribute the 3? Then take the square root of three when its multiplied by one to get x^2+9-7? Standard form is easiest for me and I am wanting to know if I can convert it.(2 votes)
- you have to follow the order of operations. PEMDAS. So exponents are before multiplication, so you have to square the parenthesis there.
As a side note, this form is acually sper useful. Are you familiar with graph transformations? like x^2 +2 moves it up 2 and (5x)^2 shrinks the graph horizontally by a factor of 5. If this is not familiar to you I can give a quick explanation.
Also also, this form allows you to do some quick algebra to find the 0s or roots. agai, I can totally explain this.(2 votes)
- Why is y = f(x) and not x = f(x)?(1 vote)
- f(x) is read as f of x, and it means what is the value of the function at x, this is the y value, so y=f(x).(4 votes)
I have a function here defined as x squared minus 5x plus 6. And what I want us to think about is what other forms we can write this function in if we, say, wanted to find the 0s of this function. If we wanted to figure out where does this function intersect the x-axis, what form would we put this in? And then another form for maybe finding out what's the minimum value of this. We see that we have a positive coefficient on the x squared term. This is going to be an upward-opening parabola. But what's the minimum point of this? Or even better, what's the vertex of this parabola right over here? So if the function looks something like this, we could use one form of the function to figure out where does it intersect the x-axis. So where does it intersect the x-axis? And maybe we can manipulate it to get another form to figure out what's the minimum point. What's this point right over here for this function? I don't even know if the function looks like this. So I encourage you to pause this video and try to manipulate this into those two different forms. So let's work on it. So in order to find the roots, the easiest thing I can think of doing is trying to factor this quadratic expression which is being used to define this function. So we could think about, well, let's think of two numbers whose product is positive 6 and whose sum is negative 5. So since their product is positive, we know that they have the same sign. And if they have the same sign but we get to a negative value, that means they both must be negative. So let's see-- negative 2 times negative 3 is positive 6. Negative 2 plus negative 3 is negative 5. So we could rewrite f of x. And so let me write it this way. We could write f of x as being equal to x minus 2 times x minus 3. Now, how does this help us find the zeroes? Well, in what situations is this right-hand expression, is this expression on the right hand going to be equal to 0? Well, it's the product of these two expressions. If either one of these is equal to 0, 0 times anything is 0. 0 times anything else is 0. So this whole thing is going to be 0 if x minus 2 is equal to 0 or x minus 3 is equal to 0. Add 2 to both sides of this equation. You get x is equal to 2 or x is equal to 3. So those are the two zeroes for this function, I guess you could say. And we could already think about it a little bit in terms of graphing it. So let's try to graph this thing. So this is x equals 1. This is x equals 2. This is x equals 3 right over there. So that's our x-axis. That, you could say, is our y is equal to f of x axis. And we're seeing that we intersect both here and here. When x is equal to 2, this f of x is equal to 0. When x is equal to 3, f of x is equal to 0. And you could substitute either of these values into the original expression. And you'll see it's going to get you to 0 because that is the same thing as that. Now, what about the vertex? What form could we write this original thing in order to pick out the vertex? Well, we're already a little familiar with completing the square. And when you complete the square with this expression, that seems to be a pretty good way of thinking about what the minimum value of this function is. So let's just do that right over here. So I'm just going to rewrite it. So we get f of x is equal to x squared minus 5x. And I'm just going to throw the plus 6 right over here. And I'm giving myself some real estate because what I need to do, what I want to think about doing, is adding and subtracting the same value. So I'm going to add it here, and I'm going to subtract it there. And I can do that because then I've just added 0. I haven't changed the value of this right-hand side. But I want to do that so that this part that I've underlined in this magenta color, so that this part right over here, is a perfect square. And we've done this multiple times when we've completed the square. I encourage you to watch those videos if you need a little bit of a review on it. But the general idea is this is going to be a perfect square if we take this coefficient right over here. We take negative 5. We take 1/2 of that, which is negative 5/2, and we square it. So we could write this as plus negative-- what's negative 5/2 squared? So I could write this-- negative 5/2 squared. Well, if we square a negative number, it's just going to be a positive. So it's going to be the same thing as 5/2 squared. 5 squared is 25. 2 squared is 4. So this is going to be plus 25/4. Now, once again, if we want this equality to be true, we either have to add the same thing to both sides. Or if we're just operating on one side, if we added it to that side, we could just subtract it from that side. And we haven't changed the total value on that side. So we added 25/4, and we subtracted 25/4. So what is this part right over here? What does this become, the part that I've underlined in magenta? Well, this is going to be-- the whole reason why we engineered it in this way is so that this could be x minus 5/2 squared. And I encourage you to verify this. And we go into more detail about why taking 1/2 the coefficient here and then squaring it, adding it there and then subtracting there, why that works. We do that in the completing the square videos. But these two things, you can verify that they are equivalent. So that's that part. And now we just have to simplify 6 minus 25/4. So 6 could be rewritten as 24/4. 24/4 minus 25/4 is negative 1/4, so minus 1/4, just like that. So we've rewritten our original function as f of x is equal to x minus 5/2 squared minus 1/4. Now, why is this form interesting? Well, one way to think about it is this part is always going to be non-negative. The minimum value of this part in magenta is going to be 0. Why? Because we're squaring this thing. If you're taking something like this-- and we're just dealing with real numbers-- and you're squaring it, you're not going to be able to get a negative value. At the minimum value, this is going to be 0. And then it obviously could be positive values, as well. So if we want to think about when does this thing hit its minimum value-- well, it hits its minimum value when you're squaring 0. And when are you squaring 0? Well, you're squaring 0 when x minus 5/2 is equal to 0, or when x is equal to 5/2 if you just want to add 5/2 to both sides of that equation. So this thing hits its minimum value when x is equal to 5/2. And then what is y, or what is f of x, when x is equal to 5/2? f of 5/2-- and once again, you could use any of those forms to evaluate 5/2. But it's really easy in this form. When x is equal to 5/2, this term right over here becomes 0. 0 squared, 0. You're just left with negative 1/4. So another way to think about it is our vertex is at the point x equals 5/2, y equals negative 1/4. So x equals 5/2. That's the same thing as 2 and 1/2. So x equals 5/2. And y is equal to negative 1/4. So if that is negative 1, 1/4 would be something like that. So that right over there is the vertex. That is the point-- let me make it clear-- that's the point 5/2 comma negative 1/4. And what's cool is we've just used this form to figure out the minimum point, to figure out the vertex in this case. And then we can use the roots as two other points to get a rough sketch of what this parabola will actually look like. So the interesting-- or I guess the takeaway from this video is just to realize that we can rewrite this in different forms depending on what we're trying to understand about this function.