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### Course: Algebra (all content) > Unit 9

Lesson 12: Quadratic inequalities# Quadratic inequalities: graphical approach

Sal solves a few quadratic inequalities by moving all terms to one side of the inequality and graphing the resulting expression. Created by Sal Khan.

## Want to join the conversation?

- How do you solve this inequality x^2+3x is greater or equal to -2. I've tried doing it and got x is greater or equals to -2 and x is greater or equals to -1 but that is no the right answer. Please help(3 votes)
`x² + 3x ≥ -2`

`=> x² + 3x + 2 ≥ 0`

`=> x² + 2x + x + 2 ≥ 0`

`=> x(x + 2) + 1(x + 2) ≥ 0`

`=> (x + 1)(x + 2) ≥ 0`

`=> x ∈ [-∞, -2]⋃[-1, ∞]`

Stated simply,`x < -2`

or`x > -1`

.(9 votes)

- I dont understand how to graph Y >X^2+4(3 votes)
- Also, when an inequality is solved for y, you can interpret "y >" as "above the line" and "y <" as "below the line". Just be careful, as a negative coefficient will reverse this logic. When in doubt, test points!(4 votes)

- What point is the axis of symmetry?(2 votes)
- the axis of symmetry is the line that goes through the vertex. hope this helps! :-)(5 votes)

- Why is the vertex -6 instead of -b/2a or - 1/2 ?(2 votes)
- Sal tells you at about1:55that -6 is the y-intercept (the point (0, -6). He then explicitly tells you that it is
**not**the vertex. The point is close to the vertex, but it is not the vertex.

You can find the x-value of the vertex can be found using -b/2a = -1/2.

Then, you can find the y-value of the vertex by plugging in x = -1/2(4 votes)

- please explain open interval and closed interval and also modulus of real number(3 votes)
- In quadratic inequalities, when do you use dotted/solid lines?(1 vote)
- Dotted lines: Strict Inequality:
`> or <`

Solid lines: Not Equal/Equal:`≥ or ≤`

Hope this helps!

—CT-2/002-24(4 votes)

- How would you actually graph a quadratic inequality? For instance, would you just shade everything outside of or inside of the roots? This part sort of confused me because the graph at3:30was a bit unclear.(2 votes)
- can you graph a linear inequality? then yeah.. quadratic works too.

http://www.youtube.com/watch?v=6Nms3eftLzg(1 vote)

- If a parabola is supposed to be symetric, how is the first example valid? Shouldn't the minimum be at the center of the two roots?(1 vote)
- If you're referring to Sal's graph, it's not drawn particularly to scale. The minimum point is indeed at the midpoint of the two roots
`x=-3`

and`x=2`

. Therefore the vertex should occur at`x=-1/2`

(at which point`y=-25/4`

). The graph does have a y-intercept of`(0,-6)`

, which was the point Sal was using to graph the curve, but it's not the minimum of the function. Good eye! :)(3 votes)

- why did you plot y=0 when -3,0 ans 2,0 why didnt you take any other why 0 ?(2 votes)
- ok, do i understand this right? when solving a quadratic equation, we're actually defining the x intercepts of the parabola. but when solving a quadratic inequality, we're not really defining a particular parabola... we're constraining the possible parabolas? sort of like when we limit domain and range?(1 vote)
- Exactly, you are putting a constraint on the parabola, where the answer can be and where it cannot be.(2 votes)

## Video transcript

Welcome to the presentation
on quadratic inequalities. Before we get to quadratic
inequalities, let's just start graphing some functions and
interpret them and then we'll slowly move to the
inequalities. Let's say I had f of x is equal
to x squared plus x minus 6. Well, if we wanted to figure
out where this function intersects the x-axis or the
roots of it, we learned in our factoring quadratics that we
could just set f of x is equal to 0, right? Because f of x equals 0 when
you're intersecting the x-axis. So you would say x squared
plus x minus 6 is equal to 0. And you just factor
this quadratic. x plus 3 times x
minus 2 equals 0. And you would learn that the
roots of this quadratic function are x is equal to
minus 3, and x is equal to 2. How would we visualize this? Well let's draw this
quadratic function. Those are my very uneven lines. So the roots are x is
equal to negative 3. So this is, right here, x is at
minus 3y0 -- by definition one of the roots is where
f of x is equal to 0. So the y, or the f of
x axis here is 0. The coordinate is 0. And this point here
is 2 comma 0. Once again, this is the x-axis,
and this is the f of x-axis. We also know that the y
intercept is minus 6. This isn't the vertex,
this is the y intercept. And that the graph is going to
look something like this -- not as bumpy as what I'm drawing,
which I think you get the general idea if you've ever
seen a clean parabola. It looks like that with x minus
3 here, and x is 2 here. Pretty straightforward. We figured out the roots, we
figured out what it looks like. Now what if we, instead of
wanting to know where f of x is equal to 0, which is these two
points, what if we wanted to know where f of x
is greater than 0? What x values make f
of x greater than 0? Or another way of saying
it, what values make the statement true? x squared plus x minus 6
is greater than 0, Right, this is just f of x. Well if we look at the graph,
when is f of x greater than 0? Well this is the f of x
axis, and when are we in positive territory? Well f of x is greater than
0 here -- let me draw that another color -- is greater
than 0 here, right? Because it's above the x-axis. And f of x is greater
than 0 here. So just visually looking at it,
what x values make this true? Well, this is true whenever x
is less than minus 3, right, or whenever x is greater than 2. Because when x is greater than
2, f of x is greater than 0, and when x is less than
negative 3, f of x is greater than 0. So we would say the solution to
this quadratic inequality, and we pretty much solved this
visually, is x is less than minus 3, or x is
greater than 2. And you could test it out. You could try out the number
minus 4, and you should get f of x being greater than 0. You could try it out here. Or you could try the number 3
and make sure that this works. And you can just make sure
that, you could, for example, try out the number 0 and make
sure that 0 doesn't work, right, because 0 is
between the two roots. It actually turns out that
when x is equal to 0, f of x is minus 6, which is
definitely less than 0. So I think this will give you a
visual intuition of what this quadratic inequality means. Now with that visual intuition
in the back of your mind, let's do some more problems and maybe
we won't have to go through the exercise of drawing it, but
maybe I will draw it just to make sure that the
point hits home. Let me give you a slightly
trickier problem. Let's say I had minus x squared
minus 3x plus 28, let me say, is greater than 0. Well I want to get rid of
this negative sign in front of the x squared. I just don't like it there
because it makes it look more confusing to factor. I'm going to multiply
everything by negative 1. Both sides. I get x squared plus 3x minus
28, and when you multiply or divide by a negative, with any
inequality you have to swap the sign. So this is now going
to be less than 0. And if we were to factor this,
we get x plus 7 times x minus 4 is less than 0. So if this was equal to 0, we
would know that the two roots of this function -- let's
define the function f of x -- let's define the function as f
of x is equal to -- well we can define it as this or this
because they're the same thing. But for simplicity let's define
it as x plus 7 times x minus 4. That's f of x, right? Well, after factoring it, we
know that the roots of this, the roots are x is equal to
minus 7, and x is equal to 4. Now what we want to know
is what x values make this inequality true? If this was any
equality we'd be done. But we want to know what
makes this inequality true. I'll give you a little bit of a
trick, it's always going to be the numbers in between the
two roots or outside of the two roots. So what I do whenever I'm doing
this on a test or something, I just test numbers that are
either between the roots or outside of the two roots. So let's pick a number that's
between x equals minus 7 and x equals 4. Well let's try x equals 0. Well, f of 0 is equal to -- we
could do it right here -- f of 0 is 0 plus 7 times 0 minus 4
is just 7 times minus 4, which is minus 28. So f of 0 is minus 28. Now is this -- this is the
function we're working with -- is this less than 0? Well yeah, it is. So it actually turns that a
number, an x value between the two roots works. So actually I immediately
know that the answer here is all of the x's that are
between the two roots. So we could say that the
solution to this is minus 7 is less than x
which is less than 4. Because now the other way. You could have tried a number
that's outside of the roots, either less than minus 7 or
greater than 4 and have tried it out. Let's say if you
had tried out 5. Try x equals 5. Well then f of 5 would
be 12 times 1, right, which is equal to 12. f of 5 is 12. Is that less than 0? No. So that wouldn't have worked. So once again, that gives
us a confidence that we got the right interval. And if we wanted to think about
this visually, because we got this answer, when you do it
visually it actually makes, I think, a lot of sense,
but maybe I'm biased. If you look at it visually
it looks like this. If you drive visually and this
is the parabola, this is f of x, the roots here are minus 7,
0 and 4, 0, we're saying that for all x values between these
two numbers, f of x is less than 0. And that makes sense, because
when is f of x less than 0? Well this is the
graph of f of x. And when is f of x less than 0? Right here. So what x values give us that? Well the x values that give
us that are right here. I hope I'm not confusing
you too much with these visual graphs. And you're probably saying,
well how do I know I don't include 0? Well you could try it out, but
if you -- oh, well how come I don't include the roots? Well at the roots, f
of x is equal to 0. So if this was this, if this
was less than or equal to 0, then the answer would be
negative 7 is less than or equal to x is less
than or equal to 4. I hope that gives you a sense. You pretty much just have to
try number in between the roots, and try number outside
of the roots, and that tells you what interval will
make the inequality true. I'll see you in the
next presentation.