College Algebra: Find Equation's x & y-Intercepts

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Suppose someone gives you an equation and you want to get this sort of visual image of what it looks like and sort of how does it behave and what does it look like and what's the relationship between these unknown variables. So graphing is a really powerful way of doing that. But the question is "how do you graph one of these algebraic expressions" that you're given. The answer is, well, you have a sense of what some of these things look like. For example, a vague idea of what something looks like if it's a line or if it's going to be a parabola--if there's like a square kind of term in there and so forth.
Another way--once you have a vague sense of what the general shape is--the one way to get it sort of solidified in your mind is to look for where that curve is going to cross the axis. These are called x intercepts and y intercepts. So, what's an x intercept? An x intercept--take a look at this function. Here's a curve in the plane and the x intercepts--those are places where the curve crosses the x-axis. So for example, we have x intercepts. One right here, and one right here. So these are x intercepts because, in fact, this is where the function crosses the x-axis. A y intercept is where the function crosses the y-axis. For example--it's right here. One little point of fact is that, well, there could be many, many intercepts. Not just two here, one here and so forth. So you have to be careful.
Now how would you find these intercepts--if you wanted to? Well, you see, one way to find the intercepts is to think what it means. For example, what does a y intercept mean? Well that's where the curve crosses the y-axis. So what would happen there? Well, the y-axis is where all the x values are 0. In fact, to find the y intercept what you would do is you would let x = 0 and see what the y value equals. To find the y intercept, we let x be 0. How would you find the x intercepts? Well notice the x intercepts happen at a location where y is 0. Because in fact, this is along the x-axis and all these points have the property that y has a value of 0. Because here is the value of y = 1, y = 2, 3, y = -1. This is where y = 0. So if we find out for what values of x does y = 0, that will actually locate these points.
Let me try to actually show you this in practice with some specific examples. So, for example--let's graph 2x + 5y = 12. By just finding the x intercepts and y intercepts and knowing whatever we know about this. Well, if you look at this--I see an x and y and they're both appearing to the first power. So, in fact, I guess that this is going to be some kind of line. Because as we saw earlier if we just x and y to the first powers--usually we get a line. , in fact, this is a line in standard form. This is called standard form, where you have something x plus something y equals a number.
How would we graph this? Well I'm going to try to first find the x intercepts and the y intercepts. So, let's find the x intercepts first--x intercepts. What do I do? I want to find out where this thing crosses the x-axis. What I have to do is let y = 0. So let's let y = 0. If I let y = 0 that's going to lead to the x intercepts. If I put 0 in for y, what do I see? I see 2x + 0 = 12 and so I see that x = 6. So that's the x intercept. So this function--whatever it does--is going to cross the x-axis at a value of 6--one, two, three, four, five, six. It's going to cross right there. What about the y intercept? Well to find the y intercept I do the related thing. I let x be 0. If I let x be 0, then this term goes away and I see 5y = 12 and so I see that y = ^12/[5]. So 12 over 5--what is that? That's about 2 ^2/[5]. So, in fact, here are my axes again--where does it cross the y-axis? At y = ^12/[5] which is about 2 and a little bit.
So given those two facts and our intuition that, in fact, this is actually describing a line, we can actually sketch the graph of this. Let me put that information over here. So we have the x intercept, it says, is at 6. The y intercept is at y = ^12/[5]. So what does the graph look like? Well, if I go over 6 here, one, two, three, four, five, six. Let's say that this is 6 right here, two, three, and ^12/[5] is 2 and ^2/[5]. So it's almost 2^1/[2]. It's right around here. So that is ^12/[5] and this is 6. Knowing it passes through those two points and that it's a straight line, I can graph a very accurate picture of it. The graph would look like this. A straight line that goes through these two intercepts. You can see I can get a really neat picture of it fast just by finding the intercepts.
Let's try another example. Suppose I look at the equation y = x^2 - 4x + 3. So first let me find the y intercept. To find the y intercept, I let x be 0, because I want to see where it crosses the y-axis. If I let x be 0 then this term goes away and that term goes away and it's real easy to see, y is just 3. Pretty easy. So that is the y intercept. Where does this curve cross the y-axis? It's going to cross the y-axis at one, two, three, right there. That's where it's going to cross the y-axis. Now what about the x-axis? Where is it going to cross the x-axis? So, that's the x intercept. To find the x intercept what do I do? Well, I'm going to now let y be 0. I want to see where it crosses the x-axis so I set y = 0 and solve for x. Well, that's actually a little quadratic isn't it? So that's going to require a little quadratic gymnastics here. I've got a quadratic equation and I hope this factors. I'll put an x and an x. This plus tells me that they're both going to be the same sign and they're both going to be minus, and how about 3 and 1. -3x, -1x is -4x. This looks great. So, if I solve this I see either (x - 3) = 0, which means x = 3 or (x - 1) = 0, which means that x = 1.
So what I see is that there are two x intercepts, not just one. One at 3 and one at 1. So there are two x intercepts and one y intercept. Now, if I look at this thing I also remember that since I see a squared on the x this is going to be some sort of parabola, some sort of curve that looks like this. So how would that go? Well given these intercepts we should be able to sketch this pretty well. Because what do I know? I have a y intercept at y = 3, so I have, let's see, one, two, three. So it's going to pass through the y-axis at 3. What about the x intercept? There's two of them. One's at 3 and one's at 1. So we're going to have one at 1, 2, 3. So, in fact, our graph goes through these points. Goes through the point here on the y-axis at 3 and these two points on the x-axis. So, since it's a parabola, I can sort of bend this curve and make it look sort of like a parabola-like shaped thing. That gives mergers and acquisitions an amazingly accurate sketch of the graph. Just knowing those intercepts and knowing that roughly it should be a parabola like this, gives me an amazingly accurate sketch of this.
Let's try one last one just to really drive this home. Then I'll give you a chance to try some yourself. So, let's look at the equation x = -y^2 + 4y + 12, and I want to find the intercepts. So first let's find the x intercept. What do you do for the x intercept? If you want to find the x intercept, you want to see where it hits the x-axis, we let y equal 0. We always let the opposite variable equal 0. So we let y = 0. That's pretty easy, because if these all are 0, it's easy to see that x has to be 12. So that's pretty easy. So x = 12. There's only one x intercept and it's at x = 12. What about the y intercepts now? Well, here I'm going to let x = 0. If I let x = 0 then notice I have a quadratic equation and -y^2 + 4y + 12 = 0. Okay, so let's see if this factors or not. So let's see, I'll put a -y here. Now you might be tempted to write a -y here, but actually that's not a good temptation because that would multiply to give me a plus y^2. So I just put a y here and that product would be a -y^2. By the way, if you are nervous about dealing with these negative in front of the y^2, why don't you just multiply everything through by -1. That way it'll be a positive here and you can factor like you're always use to doing. In fact, maybe I'll just do that right now to show you that if this is a little disturbing to you, you don't have to worry about that.
Okay, I'm going to need the same sign here. What is that same sign going to be? Well, it looks like it may be positive, but it's hard to tell because of this thing here. So, it's a little tricky, a little tricky. Let's see if we can figure this out now. So I want to put in two numbers here that are going to multiply to give 12, but then combine to give 4. Let's see, I think 6 and 2 are going to be good. Now how should I put them in though? See 6 and 2, I think I should put the 6 here and the 2 here. Then if I put a plus sign here and plus sign here, I think I'm in good shape. Let's see, 6y and then -2y is a 4y and this times this is 12. If you had a little trouble with that negative sign here and worry about it just multiply everything through by -1, that becomes a positive and these become negatives and you can factor more directly.
Anyway these two things multiply to give 0. So either this equals 0, which means y is 6 or this equals zero, which means y is -2. So, in fact, there are two y intercepts. One is at -2 and one is at y = 6. Now, I also see a square here, so I know this is going to be a parabola. Since it's a y squared, I know the parabola is going to open up either like this or like this. Let's see if we can figure out which way it's suppose to go. So let's see, we have an x intercept at 12. We'll put 12 right here. Suppose that's 12--so it crosses the x-axis at 12. What else do we see? We see that the y intercepts are -2 and 6. So -2 is over here and 6 would be up here. So, in fact, this is a parabola somehow that's suppose to open either like this or like this and pass through these points. So, it seems like, in fact, the parabola probably is going to go like this and look something like that.
Look how beautiful. We can sort of estimate or guess what the picture of that equation looks like just by finding the intercept. Okay, you try these finding intercept type things and seeing if you can sketch roughly what a graph would look like. It's real important to be able to, again, to get a visual for an algebraic type expression. Enjoy.
Relations and Functions
Graphing Equations
Finding the x and y Intercepts of an Equation Page [2 of 2]