- Екі айнымалысы бар теңсіздіктердің графигін салу
- Екі айнымалысы бар теңсіздіктердің графигін салу
- Теңсіздіктер графигі
- Екі айнымалысы бар теңсіздікті графикке қарап жазу
- Теңсіздіктер жүйесінің графиктері
- Графиктік тәсілмен шешілетін теңсіздіктерге шолу (х-у жазықтығы)
y≤4x+3 сияқты екі айнымалысы бар сызықтық теңсіздіктердің графигін салып үйрен. Видео авторлары: Сал Хан және CK-12 Foundation.
Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. On our xy coordinate plane, we want to show all the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. Try to draw a little bit neater than that. So that is-- no, that's not good. So that is my vertical axis, my y-axis. This is my x-axis, right there. And then we know the y-intercept, the y-intercept is 3. So the point 0, 3-- 1, 2, 3-- is on the line. And we know we have a slope of 4. Which means if we go 1 in the x-direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line. We could even go back in the x-direction. If we go 1 back in the x-direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like-- this is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So let's think about what it means to be less than. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all these where y ix less than 4x plus 3? So let's think about what this means. Let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to-- let's plot this one first. When x is equal to 0, y is less than 3. So it's all of these points here-- that I'm shading in in green-- satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x-- let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So it's going to go below the line. So if you were to do this for all the possible x's, you would not only get all the points on this line which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3. Because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x over 2 minus 6. So a good way to start-- the way I like to start these problems-- is to just graph this equation right here. So let me just graph-- just for fun-- let me graph y is equal to-- this is the same thing as negative 1/2 minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis. And our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1/2. Oh, that should be an x there, negative 1/2 x minus 6. So my slope is negative 1/2, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. If I go 2 to the left, if I go negative 2, I'm going to go up 1. So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1/2 x minus 6. Now, our inequality is not greater than or equal, it's just greater than negative x over 2 minus 6, or greater than negative 1/2 x minus 6. So using the same logic as before, for any x-- so if you take any x, let's say that's our particular x we want to pick-- if you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point-- in fact, you draw an open circle there-- because you can't include the point of negative 1/2 x minus 6. But it's going to be all the y's greater than that. That'd be true for any x. You take this x. You evaluate negative 1/2 or negative x over 2 minus 6, you're going to get this point over here. The y's that satisfy it are all the y's above that. So all of the y's that satisfy this equation, or all of the coordinates that satisfy this equation, is this entire area above the line. And we're not going to include the line. So the convention is to make this line into a dashed line. And let me draw-- I'm trying my best to turn it into a dashed line. I'll just erase sections of the line, and hopefully it will look dashed to you. So I'm turning that solid line into a dashed line to show that it's just a boundary, but it's not included in the coordinates that satisfy our inequality. The coordinates that satisfy our equality are all of this yellow stuff that I'm shading above the line. Anyway, hopefully you found that helpful.