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### Course: Алгебра 1 > Unit 7

Lesson 2: Екі айнымалысы бар теңсіздіктердің графигін салу- 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)

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# Екі айнымалысы бар теңсіздіктердің графигін салу

y≤4x+3 сияқты екі айнымалысы бар сызықтық теңсіздіктердің графигін салып үйрен. Видео авторлары: Сал Хан және CK-12 Foundation.

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## Видео транскрипті

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.