### Online College Courses for Credit

#### FREE EDUCATIONAL RESOURCES PROVIDED by SOPHIA

##### Are you a student?
Free Professional Development
+
3 Tutorials that teach Graphing Parabolas

# Graphing Parabolas

##### Rating:
(3)
• (2)
• (1)
• (0)
• (0)
• (0)
Author: Colleen Atakpu
##### Description:

This lesson covers graphing parabolas.

(more)

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*

No credit card required

29 Sophia partners guarantee credit transfer.

312 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 27 of Sophia’s online courses. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

## Video Transcription

Today we're going to talk about graphing parabolas. Remember, a parabola is just what we call the shape of a graph from a quadratic equation. So we're going to do some examples graphing parabolas both from quadratic equations written in standard form and in vertex form.

So for my example I'm going to graph a quadratic equation that is written in standard form. I know that the graph of a quadratic equation looks like a parabola or a U shape. And I also know that that U shape is going to be upward facing because the coefficient in front of my x squared term is positive.

So to graph the points to create my parabola, I'm going to start by identifying the y-intercept. I know the y-intercept is the value of y when x is equal to 0. So if I substituted 0 in for the x values here, then I would have that y is equal to negative 3 when x is 0. So I'm going to start by plotting my y-intercept at negative 3 on my y-axis.

The next point I'm going to find on my parabola is the vertex. To find the x-coordinate of the vertex, I can use the formula negative b over 2 times a. And as I said, that's equal to the x-coordinate of our vertex.

My a and b number are going to come from my equation. a is the coefficient of my x squared term. And b is the coefficient of my x term. So a is 1 and b is 2.

So substituting those values into my formula, I have negative 2 over 2 times 1, which is negative 2 over 2, which would equal negative 1. So I know the x coordinate of my vertex is negative 1.

And to find the y-coordinate I can substitute negative 1 into my equation for x. So I'll have y is equal to negative 1 squared plus 2 times negative 1 minus 3. Simplifying this, negative 1 squared will be positive 1. 2 times negative 1 is negative 2. And then I'll bring down the minus 3. And then simplifying from left to right, I have 1 plus negative 2, which would give me negative 1 minus 3 will give me a negative 4. So I found that the y-coordinate of my vertex is negative 4.

So I'll plot that on my graph. x is negative 1. And y is negative 4.

So now that I have my vertex, I know that the vertex lies on the axis of symmetry, which is an imaginary vertical line that goes through the vertex. And the axis of symmetry is a line of reflection. So all of the points that I find on one side of my parabola can then be reflected to the other side to find more points that I can connect to create my parabola.

So I'm going to continue to find two more points on this side of my axis of symmetry. And then I'll reflect them on the other side. So I'm going to go ahead and find what the value of y would be when x is equal to 1. And I'll do that again by substituting 1 into my equation for x.

So substituting 1 into my equation for x, I'll have 1 squared plus 2 times 1 minus 3. Simplifying this, 1 squared is 1. 2 times 1 is 2. Bring down the minus 3. And then simplifying, 1 plus 2 is 3 minus 3 will give me 0. So the value of y is 0 when x is 1. So I'll plot that point x is 1 and y is 0.

And then I'll find one other point. I'll find the value of y when x is equal to 2. So substituting 2 into my equation for x, I'll figure out the value of y. Now simplifying this, 2 squared will give me 4. 2 times 2 is another 4. Bringing down my minus 3, I have y is equal to 4 plus 4, which is 8 minus 3 will give me 5. So I found that the value of y is 5 when x is 2. So I'll plot the point x is 2 and y is 5.

So now that I have some points on my parabola, I can go ahead and connect them with a line. And I know it's going to be a curved line, like a U shape. And again, because my axis of symmetry is a line of reflection, I can reflect these points and the connecting line on the other side.

So this point is one away from the axis of symmetry. So it will also be one away on the other side. This point here is two away from my axis of symmetry, so it will also be two away on the other side. Whoops. Sorry. This point is one way. No, two away. This is correct.

For this final point, this point is one, two, three away from my axis of symmetry. So the point on the other side will also be three away. One, two, three, which will put me above negative four. And now I can connect these points, which gives me the shape of my parabola for this quadratic equation.

For my second example, I'm going to graph another quadratic equation, except for this time the equation is written in vertex form. So I'm going to start by identifying the vertex.

I know that the x-coordinate of the vertex is going to be positive 2 and the y-coordinate of the vertex will be negative 3. So I'm going to plot a point at 2 negative 3. And this will be the vertex of my parabola.

Because I know that the vertex is on the axis of symmetry-- so we know that if we imagine a vertical line through this vertex point, that this will cut my parabola in half. So if I can plot the points on one side of my axis of symmetry, I can just reflect them to the other side.

So I'm going to go ahead and pick two x values. I'm going to use positive 1 and 0 for my x values, substitute them into my equation to find their corresponding y values, and then plot those points. So first using an x value of positive 1, I'm going to substitute that into my equation.

So I'll have 1 minus 2 squared minus 3. Simplifying in my parentheses, 1 minus 2 is negative 1. Negative 1 squared will give me a positive 1. And 1 minus 3 will give me a y value of negative 2. So I know that when x is 1, y is negative 2. So I can plot a point here.

And then I'm going to use the x value of 0 to find its corresponding y value. I'll substitute it into my equation. Simplifying in my parentheses, 0 minus 2 is negative 2. Negative 2 squared is positive 4. And 4 minus 3 will give me a y value of positive 1. So I know that when x is 0, y is 1.

So I can see by connecting the points, my parabola will look something like this. And so now I can reflect these two points over my x-axis symmetry to find the other half of my parabola. See that this point is one away from my axis of symmetry. So I'll put this point one away also. And this point is two away from my axis of symmetry. So I'll go two away in the other direction. And now I can graph the other side of my parabola.

So let's go over our key points from today. In the standard form of a quadratic equation, the constant c is the y-intercept of the parabola. The vertex of a parabola can be calculated using the a and b coefficients in the standard form or by identifying the values for h and k in the vertex form. And the vertex of a parabola lies on the parabola's axis of symmetry. And as such, points plotted on one side of the parabola can be reflected across the axis to represent more points on the parabola.

So I hope that these key points and examples helped you understand a little bit more about graphing parabolas. Keep using your notes, and keep on practicing, and soon you'll be a pro. Thanks for watching.

Formulas to Know
X-Coordinate of Vertex