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.

## Algebra 1

### Course: Algebra 1>Unit 14

Learn how to graph any quadratic function that is given in standard form. Here, Sal graphs y=5x²-20x+15. Created by Sal Khan.

## Want to join the conversation?

• Is it just me or is the Vertex Formula left out of this video? Maybe it is left out to avoid relying on memorization but it seems like some of the "Graphing quadratics: Standard Form" Practice/Mastery Challenges seem to rely on it for equations that are difficult to factor. I didn't see it in the "Graphing Quadratics: Vertex Form" section either but I did not miss it there like I did in the Standard Form challenges. • Why does -b/2a (given y = ax^2 + bx + c) give the x-coordinate of the vertex? • It's derived from a key concept in calculus, called the derivative. A derivative is a measure of how quickly a function changes; for example, velocity (or speed) would be a derivative of position. The derivative of a function `ax^2 + bx + c` ends up being `2ax + b`, for reasons you might learn later. At the vertex of a parabola, the derivative is 0, so we can set up the equation `2ax + b = 0` and solve for `x` to get `2ax = - b`, then `x = -b/2a`.
• this is helpful, but what if you can't factor the equation? • There are multiple ways that you can graph a quadratic.
1) You can create a table of values: pick a value of "x" and calculate "y" to get points and graph the parabola.

2) If the quadratic is factorable, you can use the techniques shown in this video.

3) If the quadratic is not factorable, you can use the quadratic formula or complete the square to find the roots of the quadratic (the x-intercepts) and then find the vertex as shown in this video.

4) You can convert the equation into vertex form by completing the square.

All these techniques are covered in various videos in the lessons on quadratics.
• At when Sal factors out the five from the equation, why does he not put the newly factored equation in brackets with a five on the outside? • In order to get rid of the 5, Sal divided both sides of the equation by 5, giving (5x^2 - 20x + 15) / 5 = 0/5, which simplifies to x^2 - 4x + 3 = 0. You don't need to include the 5 as a factor in the left-hand side because you divided both sides by it.
• Are you always supposed to switch your x values to positive? In the video, he switched -3 and -1 to 3 and 1. If not please tell me why he did it there. Thanks • hey, Jeremy
He didn't exactly switch his x values to positive. the -3 and the -1 that he got were not his x values. the -3 and -1 were numbers that he got when he factored the equation into a binomial. In order to find the x values he used that binomial and made it equal 0, and his x values that he found were 3 and 1. The -3 and -1 were not x values, they were just what he used to factor in a way that he teaches in previous factoring videos. you can check them out if you are still confused.

I think that was the question you were asking, I hope that helped, if not, hopefully this will:

(this basically explains how to get the 3 and 1 for the x values and it explains why -3 and -1 do not work for x values.)

the reason that Sal made the x values positive 3 and 1 is because they are the only two x values that would make his equation equal zero.
When you have:

(x-3)(x-1) = 0

x cannot equal -3 or -1 because if x was -3 then this would happen:

(-3 -3) (-3 -1) =0
-6 * -4 = 0
24 DOES NOT EQUAL 0

so -3 doesn't work as an x, and the same thing would happen with -1:

(-1 -3) (-1 -1) = 0
-4 * -2 = 0
8 DOES NOT EQUAL 0

Both of these solved binomial equations show that -3 and -1 cannot be x values for the parabola. Therefore, 3 and 1 are the only possible x values.
• How do you find the x-intercept of the vertex when it can't be as easily estimated like he did in this specific equation? • up vote or ur sad • Is there a more simple way because this is pretty time consuming and I have to take a timed test on this and i don't want to waste time if i don't have to • For the detailed explanation, Sal takes over 4 minutes, but once you understand the process, the question Sal demonstrated can be done in less than 60 seconds (and less than 30 if you really understand the process):
y=5x² - 20x + 15
5x² - 20x + 15 = 0
5(x² - 4x + 3) = 0
5(x - 3)(x - 1) = 0,
giving x intercepts of x=1 and x=3.
The x value of the axis of symmetry is 2 since 2 is equidistant from (that is, in the middle of) x=1 and x=3.
The value of f when x=2 is 5(2 - 3)(2 - 1) = 5(-1)(1) = -5.
Now you have all you need to graph.

Keep studying!
• what if it doesnt intervene into the x axis? • (x-3)(x-1)=0
if (x-3)(0)=0
then 0=0 not x-3=0

Where does that zero go? Anything times zero is zero. How does that zero just disappear? 