Hi! We show you're using Internet Explorer 6. Unfortunately, IE6 is an older browser and everything at MindBites may not work for you. We recommend upgrading (for free) to the latest version of Internet Explorer from Microsoft or Firefox from Mozilla.
Click here to read more about IE6 and why it makes sense to upgrade.

College Algebra: Using Intercepts to Graph Lines

Preview

Like what you see? Buy now to watch it online or download.

You Might Also Like

About this Lesson

  • Type: Video Tutorial
  • Length: 11:06
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 120 MB
  • Posted: 06/27/2009

This lesson is part of the following series:

Trigonometry: Full Course (152 lessons, $148.50)
Trigonometry: Algebra Prerequisites (60 lessons, $69.30)
College Algebra: Graphing Linear Equations (4 lessons, $4.95)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

Thinkwell
Thinkwell
2174 lessons
Joined:
11/14/2008

Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

Thinkwell lessons feature a star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...

More..

Recent Reviews

This lesson has not been reviewed.
Please purchase the lesson to review.
This lesson has not been reviewed.
Please purchase the lesson to review.

One of the most basic objects in mathematics is that of the line and if you think about it, the equation of the line is going to resemble to some extent a linear type thing where you just have an x. Here's the generic form of a line, y = mx + b. Now the unknowns here, the variables, are the x and the y. The m represents the slope. This is the slope of the line, which is the pitch, rise over run, and b represents the y intercept. That's where the line will cross the y-axis. These have to be numbers if you actually are thinking of a particular line. If you're thinking of a line very specifically, these have to be numbers and these are the unknowns. These are the variables. These are the things that change and as x changes, y will change in a suitable way. We take x and we multiply it by the pitch, the rise over run, and then we add to it the y intercept. If you think of the graph of this, you can plot the points by plotting the x, y points on an axis. So let me draw some axis here and what you can do is plot a point. If you give me x I can plug this in, figure out what y is and then I go over x in this direction where this is positive and that's negative and then once I go to that x, I will then plot the y value, either going up if it's positive or down if it's negative.
Let's look at an example. Suppose we're looking at y = 5x - 10. How would I proceed here? First of all, there's a number of ways of graphing this. I notice that this is in the appropriate slope y intercept form. Since it's in the slope y intercept form, the y intercept must be -10 and the slop must be 5. Just using that I can figure out what the graph of this would like because all I do, because I know this line's going to intercept the y-axis at -10, I go down to -10. Let's see if I can do that for you. -10, that's the y intercept and then the slope is 5. Now technically that's 5 over an invisible 1 so and since slope is being thought of as rise over run so I have to go five units and then one over. Starting at this point, I'm now going to rise 5 and go over 1. If I do that again from this point, then I'd be 2 units over. If you think about it, you think of a line being straight. Only two points are required in order to know the graph of that line perfectly. That's to say that if I know 2 points, that will completely determine the line connecting them. Just draw the straight line to it. Take a ruler and connect those two points with a straight line. If I have two points, there's only one line that goes between it. There it is. Two points are all you need but sometimes you want a little extra one just to make sure you're doing OK. That's a way to visualize this algebraic expression that represents the y. Here's a visual. Now that's the real slick nice way.
We can also do it by just lifting of the intercepts. If I think about the y intercept, what does that mean? The y intercept is where this line crosses the y-axis and if you cross the y-axis, what's the x value? Well the y-axis, notice that every single point on the y-axis happens when x = 0. That means that x = 0 and if you let x be 0, if you plug in x = 0 the y value you get will give you the y intercept. Let's try that here. If I let x be 0 and plug in 0 for x, what value do I get for y? I get -10 and look. We've already said that is the y intercept so that confirms that observation.
How would I find the x intercept, where the line crosses the x-axis? Well, the x intercept will happen whenever y = 0. You see y has to equal 0 in order for me to be on this line because remember, if y were like one that would be one unit up from the x-axis. If y were to equal -3, that would be 3 units down from the x-axis. The x-axis itself lives on the line y = 0. If we set y = 0, that should give us the x intercept. Let's do that. If I put 0 in for y, then what do I see? What I see is 5x - 10 = 0. What must x be? x must be 2 and look, that's actually what we discovered when we were doing this out by graphing it using the slope intercept form. We saw that in fact the x intercept is 2 and that confirms it by setting y = 0 and solving. That is an easy way of finding the intercept and of course two points terminal line.
In fact, let's use that principle to look at a more exotic phrase for a line, 3x + 2y = 12. Suppose I want you to graph that. Now that is a line and why? Because I only have x appearing to the first power and I only have y appearing to the first power and there's no xy or anything complicated, just a lone x, a lone y, some constants and I'm adding them together. That means it's a line. How could you solve this? Well, one way to solve it is to convert it back to this form and read off the slope and read off the y intercept. You see right now I don't know what the slope or y intercept is. Another different way would be just to find the x intercept and the y intercept and remember that two points determine a line. Let's try it both ways really fast and so you can see both operations in practice. Let's find the intercept. Let's find the y intercept. That's when we set x = 0 and solve for y. If I set x = o, the y intercept must be what? Well, it must be 6. If you want to find the x intercept, I set y = 0. If you set y = 0, that terms goes away and what must x be? x must be 4. That's all you need to know to graph this thing because if you want to graph it, what would it look like? Well, I see that the y intercept is going to be y = 6. This thing is going to cross at y = 6. It crosses the y-axis at y = 6 and the other x intercept is at x = 4. Therefore I know this point is on the line and this point is on the line. Two points determine a line so in fact that must uniquely determine the line. It must be this line right here. If you look at that and you think about slope a little bit, what I see is the slope of this thing must be negative because I'm falling rather than climbing as I go out. In fact, if I write it in this form the slope should be negative and the y intercept is 6. If you look at this, you could actually figure out what the slope would be, rise over run. How far did I run over? Actually I dropped here. That's -6. I'm going down so that's a drop 6 and then I go over 4. It looks like the slope would be - which is equals -. If you wanted to write this equation in this form, just by looking at the picture I see what it should be. It should be y = -x + the y intercept which we saw was 6. That should be the answer. Is it? Let's take this and solve it for y. If I solve for y I bring this over to the other side and I see 2y = -3x + 12. I just brought that term to this side. Now divide through by 2. y = -x = is 6. We get the same answer. You can see all these different ways of thinking about lines and you can see they're all equivalent. If you're given a line, you can find the y and x intercepts to graph it or you can convert it to slope intercept form and read off the information and then go -3 down and 2 over and so forth and you graph it.
The Straight Line
Graphing Linear Equations
Using Intercepts to Graph Lines Page [1 of 2]

Embed this video on your site

Copy and paste the following snippet: