If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Forms & features of quadratic functions

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.

## Want to join the conversation?

• The vertex from what I have understood from the video is the lowest point in a parabola. But what does the roots imply? •   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. •   Good question!
The word `quadratic` refers 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 `quadratic expression`.
x² - x - 5 is a `quadratic expression.`
So are the following:
a² + 8a - 6
g² - 16
If there is an equals sign, we call it a `quadratic equation`.
An example is
x² - 4x + 3 = 0
Another example of `quadratic equation` is
16t² - 8t = 3
and,
x = y²
and
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 function` that describes every point on a parabola
f(x) = x² - 4x + 3 is also a `quadratic function` that describes every point on another parabola
The next `quadratic equation` looks 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`
• Is there a easy trick for knowing if whether a parabola is a mininum or a maxinum? • 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? • I have a quadratic equation that I cannot figure out.
y=-x^2-4x-3
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. • 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).
• Is there a formula for finding the vertex of a parabola? Is there a video on that topic? • 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).
• Where the positive 25/4 at went to? Sal at wrote (x-5/2)^2 but did not write 25/4, only he wrote - 1/4 at for what is left. • 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!
• 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. • 