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## Algebra 1

### Course: Algebra 1 > Unit 14

Lesson 2: Solving and graphing with factored form# Graphing quadratics in factored form

An example for a quadratic function in factored form is y=½(x-6)(x+2). We can analyze this form to find the x-intercepts of the graph, as well as find the vertex.

## Want to join the conversation?

- What are some real life examples of a parabola (or a quadratic equation)?(34 votes)
- Ballistics with no atmospheric drag (i.e. launching a ballistic missile on the Moon), an orbit with eccentricity of 1, cutting 2 vertex-tangent cones parallel to a slant, astrophysics(9 votes)

- So when you're figuring out the x-values for the intercepts, why doesn't the 1/2 matter?(16 votes)
- The 1/2 will not change the result created by either factor.

Let's say you keep the 1/2 and use it with (x-6)

We use the zero product rule: 1/2 (x - 6) = 0

Distribute: x/2 - 3 = 0

Add 3: x/2 = 3

Multiply by 2: x = 6

Notice: this is the same result you would get if you just started with x-6 = 0

Hope this helps.(32 votes)

- Is infinity a number? Or is it just describing all numbers? I'm so confused about this.(7 votes)
- Infinity is not a number, nor does it describe all numbers. Infinity is simply the concept of an arbitrarily large value that does not obey the rules of arithmetic.(30 votes)

- At5:00in the video, why does Mr. Khan place positive 2 in the x place, if the x value was -2? Was that a mistake or something else?(14 votes)
- Hi Sierra,

Mr. Khan put the positive 2 on the x axis to show the average of -2 and 6(using the slope formula). The average shows where to put the parabola.(1 vote)

- Is there a way to find the sum or the products of the roots?(7 votes)
- You can find the sum of the roots by using -b/a, and the product of the roots, you can use c/a, hope this helps(18 votes)

- I remember learning this like you youngsters but back in the day, it was much harder without Khan Academy. Y'all should be grateful! Peace, LOL (Lots of Love), and Laughter! :) -Linda Roberts(11 votes)
- How sweet :) We're indeed very lucky to have Khan Academy!(3 votes)

- How do you know if a parabola is going to be upward or downward opening?(4 votes)
- If the x^2 term is positive, the parabola will open upwards.

If the x^2 term is negative, the parabola will open downwards.(13 votes)

- what if the quadratic doesn't give me the x-intercepts? is there another way to find the two points?(5 votes)
- A parabola

𝑦 = 𝑎𝑥² + 𝑏𝑥 + 𝑐

has its vertex at (−𝑏∕(2𝑎), 𝑐 − 𝑏²∕(4𝑎))

Then we can pick a couple of 𝑥-values fairly close to the line of symmetry and plug them into the equation to find their corresponding 𝑦-values.

– – –

Example: 𝑦 = −3𝑥² + 6𝑥 − 7

The vertex is (−6∕(2 ∙ (−3)), −7 − 6²∕(4 ∙ (−3))) =

= (1, −4)

Picking 𝑥 = 0 ⇒ 𝑦 = −7,

so (0, −7) is a point on the parabola, and since the parabola is symmetric around 𝑥 = 1, we know that (2, −7) is also a point on the parabola.

Similarly, (−1, −16) and (3, −16) are also points on the parabola, and now I'd say we have enough information to draw a decent representation of the parabola.(10 votes)

- What are some real life examples of a parabola(4 votes)
- A McDonalds sign or throwing a ball in the air, or a banana.(4 votes)

- Can a parabola's
`vertex`

be at an`x-position`

that is not the`average`

of the 2`x-intercepts`

? In other words, if I have a parabola with`x-intercepts`

of**1**and**3**, is there ever a case where the parabola's`vertex`

be at an`x-position`

of**2.5**, rather than**2**?(5 votes)- No, it is not possible for a parabola's vertex to be at an x-position that isn't the average of its x-intercepts (unless the parabola doesn't have any x-intercepts). This is because the vertex must be on the axis of symmetry. And, if the vertex isn't spaced exactly between the two zeros, then it wouldn't be symmetrical.(7 votes)

## Video transcript

- [Instructor] We're asked
to graph the equation y is equal to one-half times x minus six times x plus two. So like always, pause this video and take out some graph paper or even try to do it on
a regular piece of paper and see if you can graph this equation. All right now, let's work
through this together. There's many different ways that you could attempt to graph it. May be the most basic is
try out a bunch of x values and a bunch of y values and
try to connect the curve that connects all of those dots. But let's try to see if
we can get the essence of this graph without
doing that much work. The key realization here without
even having to do the math is if I multiply this out, if I multiplied x minus
six times x plus two, I'm going to get a quadratic. I'm going to get x-squared, plus something, plus something else. And so this whole thing
is going to be a parabola. We are graphing a quadratic equation. Now a parabola you might remember can intersect the x-axis multiple times. So let's see if we can find out where this intersects the x-axis. And the form that it's in, it's in factored form already, it makes it pretty straightforward for us to recognize
when does y equal zero? Which are going to be the
times that we're intersecting the x-axis. And then from that, we'll
actually be able to find the coordinates of the vertex and we're going to be able
to get the general shape of this curve which is
going to be a parabola. So let's think about it. When does y equal zero? Well to solve that we just
have to figure out when, if we want to know when y equals zero, then we have to solve for when does this expression equal zero? So let's just solve the equation. One half times x minus six, times x plus two is equal to zero. Now in previous videos we've
talked about this idea. If I have the product of multiple things and it needs to be equal to zero, the only way that's going to happen is if one or more of these things are going to be equal to zero. Well one half is one-half, it's not going to be equal to zero. But x minus six could be equal to zero. So if x minus six is equal to zero, then that would make this equation true. Or if x plus two is equal to zero, that would also make this equation true. So the x values that
satisfy either of these would make y equal zero
and those would be places where our curve is
intersecting the x-axis. So what x value makes
x minus six equal zero? Well you could add six to both sides, you're probably able to
do that in your head, and you get x is equal to six. Or you subtract two from both sides here and you get x is equal to, these cancel out, you get x is equal to negative two. These are the two x values where y will be equal to zero. You can substitute it back
into our original equation. If x is equal to six, then this right over here is
going to be equal to zero, and then y is going to be equal to zero. If x is equal to negative two, then this right over here is
going to be equal to zero, and y would be equal to zero. So we know that our parabola
is going to intersect the x-axis at x equals
negative two right over there, and x is equal to six. These are our x-intercepts. So given this, how do we
figure out the vertex? Well the key idea here is to recognize that your axis of
symmetry for your parabola is going to sit right between
your two x-intercepts. So what is the midpoint between, or what is the average
of six and negative two? Well, you could do that in your head. Six plus negative two is
four divided by two is two. Let me do that. So I'm just trying to find the
midpoint between the point, let's use a new color. So I'm trying to find the midpoint between the point negative two comma zero and six comma zero. Well the midpoint, those are just the average of the coordinates. The average of zero and zero
is just going to be zero, it's going to sit on the x-axis. But then the midpoint
of negative two and six or the average negative
two plus six over two. Well let's see, that's four over two, that's just going to be two, so two comma zero. And you see that there. You could have done that
without even doing the math. You say okay, if I want to go right in between the two, I want to be two away from each of them. And so just like that,
I could draw an axis of symmetry for my parabola. So my vertex is going to sit
on that axis of symmetry. And so how do I know what the y value is? Well I can figure out, I can substitute back
in my original equation, and say well what is y equal
when x is equal to two? Because remember the vertex
has a coordinate x equals two. It's going to be two comma something. So let's go back, let's see what y equals. So y will equal to one-half times, we're going to see when x equals two, so two minus six, times two plus two. Let's see, this is negative four, this is positive four. Negative four times four is negative 16. So it's equal to one-half
times negative 16, which is equal to negative eight. So our vertex when x is equal to two, y is equal to negative eight. And so our vertex is going
to be right over here two common negative eight. And now we can draw the general shape of our actual parabola. It's going to look something like, once again this is a hand-drawn sketch, so take it with a little
bit of a grain of salt, but it's going to look
something like this. And it's going to be symmetric
around our axis of symmetry. That's why it's called
the axis of symmetry. This art program I have,
there's a symmetry tool, but I'll just use this and there you go. That's a pretty good sketch
of what this parabola is or what this graph is going
to look like which it is, an upward-opening parabola.