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## Algebra 1

### Course: Algebra 1 > Unit 14

Lesson 10: Quadratic standard form# Graphing quadratics: standard form

CCSS.Math: , , , ,

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.(42 votes)
- Why does -b/2a (given y = ax^2 + bx + c) give the x-coordinate of the vertex?(17 votes)
- 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`

.(20 votes)

- At3:10when 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?(7 votes)
- 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.(18 votes)

- this is helpful, but what if you can't factor the equation?(6 votes)
- 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.(16 votes)

- 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(4 votes)
- 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.(6 votes)

- 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?(4 votes)
- The x-intercept is when y = 0, so it you can plug in 0 to the equation to find the x-intercept(s).(5 votes)

- what if it doesnt intervene into the x axis?(4 votes)
- If the function doesn't touch the x-axis, there are no real solutions. Ø is what you put.(3 votes)

- (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?(3 votes)- The zero product rule tells us that if ab=0, then either a=0 or b=0. So if you have (x-3)(x-1)=0, your "a" is (x-3) and your "b" is (x-1).

-- What makes x-3=0? Solve the mini equation and you get x=3.

-- What makes x-1=0? Again, solve the mini equation and you get x=1.

-- So, the solutions (the values of "x") that would make the equation = 0 are x=3 and x=1.

Hope this helps.(4 votes)

- 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(3 votes)
- 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!

Keep Asking Questions!(4 votes)

- whats the process of graphing polynomial functions(2 votes)
- Hi lizete g,

This link might help clear your confusion:

https://www.khanacademy.org/math/algebra2/polynomial-functions/graphs-of-polynomials/a/graphs-of-polynomials

Hope that helps!

- JK

#YouCanLearnAnything(6 votes)

## Video transcript

We're asked to graph
the following equation y equals 5x squared
minus 20x plus 15. So let me get my
little scratch pad out. So it's y is equal to 5x
squared minus 20x plus 15. Now there's many
ways to graph this. You can just take three
values for x and figure out what the
corresponding values for y are and just graph
those three points. And three points actually
will determine a parabola. But I want to do something a
little bit more interesting. I want to find the places. So if we imagine our axes. This is my x-axis. That's my y-axis. And this is our curve. So the parabola might
look something like this. I want to first figure out where
does this parabola intersect the x-axis. And as we have already
seen, intersecting the x-axis is the same thing
as saying when it does this when does y equal
0 for this problem? Or another way of
saying it, when does this 5x squared
minus 20x plus 15, when does this equal 0? So I want to figure
out those points. And then I also want to
figure out the point exactly in between, which is the vertex. And if I can graph
those three points then I should be all set with
graphing this parabola. So as I just said,
we're going to try to solve the equation 5x
squared minus 20x plus 15 is equal to 0. Now the first thing
I like to do whenever I see a coefficient out here on
the x squared term that's not a 1, is to see if I can
divide everything by that term to try to simplify
this a little bit. And maybe this will get us
into a factor-able form. And it does look like every
term here is divisible by 5. So I will divide by 5. So I'll divide both sides
of this equation by 5. And so that will give
me-- these cancel out and I'm left with x squared
minus 20 over 5 is 4x. Plus 15 over 5 is 3 is
equal to 0 over 5 is just 0. And now we can attempt to
factor this left-hand side. We say are there two numbers
whose product is positive 3? The fact that their
product is positive tells you they both
must be positive. And whose sum is negative
4, which tells you well they both must be negative. If we're getting a
negative sum here. And the one that
probably jumps out of your mind--
and you might want to review the videos
on factoring quadratics if this is not so fresh-- is
a negative 3 and negative 1 seem to work. Negative 3 times negative 1. Negative 3 times
negative 1 is 3. Negative 3 plus negative
1 is negative 4. So this will factor out as
x minus 3 times x minus 1. And on the right-hand
side, we still have that being equal to 0. And now we can think about what
x's will make this expression 0, and if they make
this expression 0, well they're going to
make this expression 0. Which is going to make
this expression equal to 0. And so this will be true if
either one of these is 0. So x minus 3 is equal to 0. Or x minus 1 is equal to 0. This is true, and you can
add 3 to both sides of this. This is true when
x is equal to 3. This is true when
x is equal to 1. So we were able to figure
out these two points right over here. This is x is equal to 1. This is x is equal to 3. So this is the point 1 comma 0. This is the point 3 comma 0. And so the last one
I want to figure out, is this point right
over here, the vertex. Now the vertex always
sits exactly smack dab between the roots,
when you do have roots. Sometimes you might not
intersect the x-axis. So we already know what its
x-coordinate is going to be. It's going to be 2. And now we just have
to substitute back in to figure out its y-coordinate. When x equals 2, y is going to
be equal to 5 times 2 squared minus 20 times 2 plus 15,
which is equal to-- let's see, this is equal to 2 squared is 4. This is 20 minus 40 plus 15. So this is going
to be negative 20 plus 15, which is
equal to negative 5. So this is the point
2 comma negative 5. And so now we can go
back to the exercise and actually plot
these three points. 1 comma 0, 2 comma
negative 5, 3 comma 0. So let's do that. So first I'll do the vertex
at 2 comma negative 5, which is right there. And now we also know
one of the times it intersects the
x-axis is at 1 comma 0. And the other time
is at 3 comma 0. And now we can check our answer. And we got it right.