Graphing quadratics
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Ex 3: Graphing a quadratic function
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Example: Roots and vertex of a parabola
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Example: Parabola vertex and axis of symmetry
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Graphs of Quadratic Functions
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Graphing parabolas in standard form
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Parabola Focus and Directrix 1
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Focus and Directrix of a Parabola 2
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Vertex of a parabola
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Graphing parabolas in vertex form
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Graphing parabolas in all forms
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Parabola intuition 3
Focus and Directrix of a Parabola 2 Finding the focus and directrix of a parabola
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- In the last video, we showed that if you had a line, which
- we'll call a directrix-- we draw the directrix.
- That's our directrix and it has the equation y is equal to k.
- And you have a point, that's our focus, and it's the
- coordinate a comma b.
- -- that the locus of all points in the xy plane that are
- equidistant to this focus and this directrix has a shape that
- looks something-- we know that this was a point-- that
- looks something like this.
- And the actually equation, we actually just took an arbitrary
- xy in the coordinate plane, and we said that xy has to satisfy
- the condition that its distance to this focus is equal to--
- that this distance right here is equal to this distance here.
- And we set that up using the distance formula to here, and
- then setting that equal to the distance to the line,
- to the directrix.
- And then we did a bunch of algebra and we got this
- equation right here.
- For all of the points in the xy plane that satisfy the
- conditions that its distance to the focus is equal to the
- distance to this line.
- And we hopefully satisfied ourselves that this is
- in fact a parabola.
- And although it looks a little bit hairier than most parabolas
- we see, I think if you look at it closely, you'll realize
- that it is indeed a parabola.
- For example, the classic parabola is y is
- equal to x squared.
- How is this the same thing as this?
- Well in this case, this coefficient out here
- is just equal to 1.
- So in this case, 1 over 2 b minus k is equal to 1, right?
- Let me delete that little thing there, because it
- looks like a squared.
- All right, the coefficient in front of the x squared
- term is equal to a 1.
- What is a equal to?
- Well this is the same thing-- actually, let
- me rewrite it that way.
- I could rewrite this equation right here as y minus 0 is
- equal to 1 times x minus 0 squared.
- And now hopefully you see that this has the same pattern as
- this, although the y's and the x's are just sitting
- on different sides of the equation.
- So here you see that the 1/2 over b minus k, that's the
- coefficient in front of the x minus a squared term.
- So that's where we got that that has to be equal to 1.
- a has to be equal to 0 in this situation.
- This curve right here is equal to x squared, y is equal
- to x squared, a is 0.
- And then b plus k over 2, this is going to be equal to that.
- And keep in mind, I actually swapped the left and
- the right hand sides.
- But this is y minus something, this is y minus 0.
- So this something has to be equal to 0.
- So you get b plus k over 2 is equal to 0.
- And now we should be able to use this information to figure
- out the actual coordinates of the focus and the directrix of
- what I would call the classic parabola, y is equal
- to x squared.
- So keep in mind what we did.
- In the last video, we said what is the equation of the line
- that is equidistant between this focus and this directrix?
- And we got this equation, which we said, hey,
- that's a parabola.
- Now we're going the other way around.
- We say we have a parabola, and we're saying that this parabola
- is the set of all points that's equidistant between some
- focus and some directrix.
- What is that focus and that directrix for this
- particular parabola?
- That's what we're trying to do.
- And the way we did that was, we pattern matched this formula to
- this one, to essentially set up these variables, which
- essentially define a focus and a directrix, right?
- b is the y-coordinate of the focus, k defines the horizontal
- line of the directrix.
- So if we can solve for b and k, we know what the focus
- and the directrix are.
- And for a, although a is pretty easy.
- So let's do that.
- So if we know that 1 over 2 b minus k is equal to 1--
- let me do that over here.
- I'm going to do it in particular for this case right
- here, but you don't want to do that every time, so then we'll
- do a more general formula that we can just plug in next time
- you see a new parabola.
- So if we multiply both sides of this equation by 2 b minus k,
- you get 1 is equal to 2 times b minus k.
- Actually I should redivide both sides by 2, so you get 1/2
- is equal to b minus k.
- Fair enough.
- Let's see, this equation right here, b plus k
- over 2 is equal to 0.
- That means that b plus k must be equal to 0, right?
- This numerator has to be equal to 0.
- So let me write that down.
- b plus k is equal to 0.
- Two linear equations of to unknowns.
- Let's add them up.
- The k's cancel out.
- You get 2b is equal to 1/2.
- Divide both sides by 2, you get b is equal to 1/4.
- And now what's k?
- Well, we can just resubstitute back.
- Let's use this equation.
- We get 0 is equal to 1/4 plus k.
- You can figure that out by inspection, but you can
- subtract 1/4 from both sides.
- So this tells us that k is equal to minus 1/4.
- So let's draw this.
- So if this is my coordinate axis-- I'll do it in
- a different color.
- That's the x-axis, that's the y-axis.
- We're dealing with y is equal to x squared, which you should
- hopefully be reasonably familiar with at this point.
- It's kind of the classic U-shaped parabola that
- looks just like that.
- Let me write that down.
- This is y is equal to x squared, which we could have
- rewritten as y minus 0 is equal to 1 times x minus 0 squared.
- Same thing, that's this line.
- It's equidistant from a focus and a directrix, where
- this is the focus.
- And how do I know that the focus is right there, instead
- of right here or right there?
- Well one, you could kind of think about it.
- But the other thing to realize is, what is the vertex
- of this parabola?
- And I think we've gone over that already.
- But the vertex are the coordinates given by this
- x value and this y value.
- Or what x values make this whole term 0 and make
- this whole term 0?
- And that's essentially 0,0.
- So this is the vertex of this parabola.
- This is the vertex, 0,0 is the vertex.
- I'll say v for vertex.
- In an upward opening parabola, you can view that as
- the minimum point.
- If it was a downward opening, it would be the topmost
- point, or the maximum point.
- That's the vertex.
- And we saw from when we solved this, when we just did our
- pattern matching, that the x-coordinate of our focus is
- equivalent to, if we just pattern match right here, x
- minus a squared, here you have x minus 0 squared.
- This is always going to be the same as this.
- So the x-coordinate of our focus is always going to
- be the same thing as the x-coordinate of our vertex.
- So we figured out that a was equal to 0.
- So the coordinate here is 0, and then what's b?
- This was the coordinate 0,b.
- We figured out that b is equal to 1/4, 0, 1/4.
- And then the directrix is the line y is equal to k.
- In this case, we figured out k is equal to minus 1/4.
- So our directrix is going to be right down here.
- Right down here, like that.
- And it's going to be line y is equal to minus 1/4.
- And that first of all looks about right.
- And it also gels with what we know.
- Because this point right here, our vertex, is equidistant,
- it's 1/4 below our focus and it's 1/4 above our directrix.
- So at least that little reality check holds.
- But it's was a particular circumstance, and you don't
- want to have to go through this whole situation every time you
- have to figure out the focus or the directrix of a parabola.
- So let's see if we can come up with a more general solution.
- And actually what I'm going to do is, I'm going
- to erase all of this.
- Just because I want to reuse that, and I haven't committed
- that formula to heart yet.
- Let me erase all of this, and erase all the work
- we did down here.
- So let's do a general form.
- And actually, I've abused this so much, let me rewrite it.
- So I'll put the y's on the left hand side, instead
- of keep telling you that I've switched them.
- So this is the same thing as-- I'll do it in-- the color
- choosing is the hard part.
- I'll do it in this light, off-white color.
- So you get y minus b plus k over 2 is equal to 1 over 2
- times b minus k, times x minus a squared.
- This is the locus of all points that are between the directrix,
- y is equal to k-- well, they're equidistant to the directrix,
- y is equal to k, and the focus a comma b.
- So let's say that I have a parabola.
- And I'm going to try to use different letters, so
- we don't get confused.
- I give you the parabola y minus y1.
- And let's say this parabola has a vertex at x1 comma y1.
- So this parabola I'm drawing has a vertex right there.
- So its formula will be y minus y1 is equal to some constant,
- let's make that a capital A.
- This capital A is different than that lowercase a.
- This is what I was going to embark on in the previous
- video, and then I realized that I was trying to load that video
- up too much and probably confusing you.
- So some constant factor, some type of scaling factor, times
- x minus the x value of our vertex.
- So x1, all of that, squared.
- So if you're given this parabola, or if you can get a
- parabola to this form, how do you figure out the a's, the
- b's, and the k's so you know the coordinates of the
- directrix and the focus?
- So the easy thing is to figure out a, because you just
- do a pattern match.
- This is going to be equal to that.
- So you know that a is equal to x1.
- So the x-coordinate of our focus is the same as the
- x-coordinate of our vertex.
- You can pattern match that this scaling factor right here is
- going to be the same thing as this 1 over 2 b minus k.
- So let's write that down.
- A is equal to 1 over 2 b minus k.
- And then we have, finally, this.
- Let me do another color.
- I'll do a magenta.
- The b plus k over 2, we have y minus that thing.
- We have y minus y1.
- So that's going to be equal to y1.
- So you have y1 is equal to b plus k over 2.
- And now we have two equations with two unknowns.
- Remember, this is going to be a given.
- The A is going to be given, and the y1 is going to be given.
- So now we can solve for b and k in terms of the
- numbers that we have.
- So let's see if we can do that.
- So if we multiply both sides of this equation times b minus k,
- you get b minus k times A is equal to 1/2.
- Divide both sides by A, you get b minus k is
- equal to 1 over 2A.
- And on the right hand side, let's use this other
- formula right here.
- If we multiply both sides by 2-- I'll use a different
- color-- you get 2 times y1 is equal to b plus k.
- Two equations with two unknowns.
- Let's take this and bring it down here, so you get b plus
- k is equal to 2 times y1.
- And so you add them, and you get 2b is equal to-- the k's
- cancel out; I'm just adding these two equations--
- 1 over 2A plus 2 y1.
- Or b-- which is, remember, what was b?
- That was the y-coordinate of our focus-- is equal to, divide
- everything by 2, is equal to 1 over 4A plus y1.
- Which is interesting.
- It tells us that we take the y-coordinate of our vertex
- and we add 1 over 4A to it.
- Which is exactly what happened last time, right?
- I actually erased what I did last time.
- But last time, when we had y equal x squared,
- this value was 0.
- The scaling factor was 1.
- So we saw that the y-coordinate of our focus was 1/4.
- So this is gelling with what we know.
- So that's our b.
- And let's see if we can solve back for k.
- So we know that-- let's do it up here.
- We know that 2 y1 is equal to b, which is 1 over
- 4A plus y1 plus k.
- We can subtract y1 from both sides, so this goes away and
- we just have a y1 there.
- Subtract 1 over 4A from both sides and you get y1 minus
- 1 over 4A is equal to k.
- So then we are done.
- So this is interesting.
- Right now these might look a little, these little
- hairy formulas.
- But if we actually graph it, I think they'll become a
- little bit more intuitive.
- So once again, we had the parabola y minus y1 is equal to
- A times x minus x1 squared.
- And so the graph of that parabola will look
- something like this.
- I don't know where it is relative to the x- and y-axis.
- It will look something like this.
- It'll be this U shape.
- It could actually be downward pointing, but I'll just assume
- an upward pointing parabola for now.
- That's this formula.
- Its vertex, right there, its vertex is the
- point x1 comma y1.
- Right?
- What y value makes this expression equal 0?
- It's y1.
- What x value makes that equal 0?
- It's x1.
- Now notice, what is this directrix?
- It's the line ok, y equal to k, but it's 1/4 less than
- this value right here.
- So you literally just go down 1/4-- not just 1/4, 1/4
- times this scaling factor.
- So it's 1/4 A.
- This distance is 1/4 A.
- And then you get your directrix, right?
- It's y1 minus 1/4 A is equal to k. y1 minus
- 1/4 A is equal to k.
- Or 1 over 4A.
- So your directrix is going to be right there.
- Let me write this.
- It's y is equal to y1, which is just this
- level, minus 1 over 4A.
- But this is the intuitive part.
- Just remember, you're going 1 over 4A below the vertex.
- And then the focal point, the x value of the focal point is
- going to be x1, the same as the x value of the vertex.
- And then the y value is just b, and we just go
- 1 over 4A above it.
- Which makes sense because we need to be equidistant from
- the directrix and the focus.
- So this is y1 plus 1 over 4A.
- But the intuitive part is, we just took 1 over 4A above it.
- So in general, if I gave you-- so just think
- about this a second.
- If I gave you an equation, an arbitrary equation now.
- Say, y minus 1 is equal to 2 times x minus 3 squared, we can
- graph this and draw the focus fairly-- and actually
- it's foci, if there was more than one focus.
- Let's see, that's the x-axis, that's the y-axis.
- Where's its vertex?
- The vertex is at the point 3 comma 1.
- So we go 1, 2, 3 comma 1.
- So that's its vertex.
- It's an upward opening parabola, because this
- is a positive number.
- Actually, we don't even have to know that.
- If you just draw the focus and the directrix, you might be
- able to figure that out.
- But where's the focus point?
- The focal point is going to be 1 over 4A above this point.
- So 1 over 4 times 2 is equal to 1/8.
- So the focal point is going to be really close.
- It's going to be right up here.
- It's going to be 1/8 above our vertex.
- And then our directrix is going to be 1/8 below it.
- So it's going to be right there.
- The directrix is going to look like this.
- The directrix is going to be there, the
- focal point is there.
- And then the graph of this parabola is going to look
- something like this.
- So anyway, hopefully I didn't confuse you too much.
- And the whole point of doing all of this is just to realize
- that it's pretty easy to find the focus and the directrix of
- a parabola if you have its equation in this form.
- You just take whatever is multiplying times the x minus
- whatever squared term.
- And you essentially divide 1 by 4 times this value.
- And then you know the distance from the vertex to the focus,
- and you know the distance from the vertex to the directrix,
- and you're all done.
- See you in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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