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
Graphs of Quadratic Functions Graphs of Quadratic Functions
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- In this video, I want to show some examples of graphing
- quadratic functions.
- But before I do that, I want to make one fairly minor
- correction to the end of the last video where we did some
- factoring with grouping.
- When I showed you the proof of why that works, I multiplied
- fx plus g times hx plus j.
- When I multiplied them, I did the first term, I
- said fx times hx.
- When I did it in the last video, I just wrote fhx.
- And we know that that's not true, we have f times h
- times x times x.
- It's fhx squared.
- That times that is fhx squared and then the rest of it I had
- gotten right, it's fx times j, so plus we get fjx and then we
- had g times hx so we could write that as plus ghx.
- And then finally g times j, which gives us gj.
- So I apologize for that error, I had forgotten to write the
- exponent there and I think I'd forgotten to write it in the
- step after that, but it didn't change the argument for the
- proof, I at least didn't mess up there.
- So I just want to make that quick correction in case that
- confused anyone.
- And I also made a small annotation in that video if
- you're watching it on YouTube.
- With that out of the way, let's learn to graph some of
- these quadratic functions.
- So let's say I have the function y is equal to x
- squared minus 2x minus 8.
- And what we're going to do is we're going to rewrite this
- quadratic expression on the right-hand side.
- We're going to write it in intercept form.
- That essentially means just writing it
- in it factored form.
- Then doing that, we're able to figure out the x-intercept,
- where we're intersecting the x-access, and then using that
- we'll be able to figure out the vertex.
- And I'll show you you what the vertex is in a second.
- So let's just factor this quadratic on
- the right-hand side.
- So we get y is equal to-- so what are two numbers where
- their product is a negative 8, so they're going to have
- different signs and then their difference is negative 2?
- Let's see, if we have a negative 4, and plus 2.
- When you take the product, you get negative 8, when you add
- them, you get negative 2.
- So we could rewrite y as being equal to x minus 4
- times x plus 2.
- So this is essentially writing this quadratic function in
- intercept form.
- And now let's figure out where this intersects the x-axis So
- let me label my axis.
- This is, of course, the x-axis, that is the y-axis.
- So when will something intersect the x-axis?
- The x-axis is when y is equal to 0, right?
- So let's set this equal to 0.
- 0 is equal to x minus 4 times x plus 2.
- We've seen this multiple times before, this means that either
- x minus 4 or x plus 2 or both of them have to be equal to 0.
- So x minus four is equal to 0 or x plus two is equal to 0.
- So that means x could be equal to 4-- we're just adding 4 to
- both sides of that equation-- or subtracting 2 from both
- sides of this equation, x could be equal to negative 2.
- So these points where we intersect, when x is 4, this
- term is 0 so the whole thing is 0.
- That's how we solved for this.
- So the points 4, 0, and the point 2, 0 are going to both
- be on this parabola.
- So let's graph them.
- So 4, 0-- let me do this in a darker color-- 1, 2 3, 4,
- that's 4, 0.
- And then the other point is negative 2, 0.
- 1, 2 negative 2, 0 is right there.
- So that's the two points where we intersect the x-axis.
- And now we're going to determine something called the
- vertex of our parabola.
- And the vertex, just to give you a sneak preview, parabolas
- are the graphs of quadratic functions and so the graphic
- is going to look either like an upward u or a downward u,
- and it's going to be shifted around on this.
- But the vertex is either this minimum point or
- this maximum point.
- And the word parabola just describes this function shape.
- It's the shape of the graph of a quadratic.
- Let me write that word down, these are parabolas.
- So to figure out the vertex, you actually just take the
- x-value that's halfway in between the two intercepts.
- No matter what, the intercepts are going to be equidistant
- from the vertex.
- So if that is the graph, those would be our intercepts and
- then our vertex is going to be an x-value exactly halfway
- between them.
- So to figure out the x-value of our vertex, we just average
- these two x-values of our x-intercepts.
- So let's do that.
- We could say the x for the vertex, let's call that x for
- the vertex-- that's why I wrote a v there-- is going to
- be equal to the average of these.
- So 4 plus negative 2 over 2.
- Which is equal to this 4 minus 2, which is 2 over
- 2 is equal to 1.
- So x is going to be equal to 1 and what is y when
- x is equal to 1?
- So the y for the vertex is going to be equal to 1
- squared-- I'll just write that as 1-- minus 2 times 1, so
- minus 2, minus 8.
- So what is this?
- This is 1 minus 10, which is equal to negative 9.
- So the vertex is going to be at the point 1, negative 9.
- So we go one and then we go down nine.
- So x is 1 and then y is negative 9.
- 1, 2, 3, 4, 5, 6, 7, 8, 9.
- So it's right there.
- And so if we were to graph this parabola, the line of
- symmetry is on the x-value of the vertex.
- So, I want to be doing it in a darker color than that.
- So the graph should be symmetric around x is equal to
- 1, and the graph will look something like this.
- I'll try my best to draw it.
- Bam!
- It'll intersect the x-axis there, it'll be symmetric,
- it'll look just like that on the other side.
- And it'll go bam, just like that.
- I think you get the general idea.
- And then the other point that you may or may not be
- interested in is the actual y-intercept.
- If x is equal to 0, you immediately see that y is
- equal to negative 8.
- So the point 0, negative 8 should also be on this graph.
- Let's see if I drew it right.
- 0, negative eight right there should also be on that graph.
- Let's do another one of these.
- I'm kind of taking up some of the space for the next
- problem, so let me clear this.
- Let's do another one.
- All right.
- So let's say we have y is equal to 2x squared
- plus 6x plus 4.
- So here, the thing that immediately jumps out at me is
- that I can factor a 2 out of everything.
- So I could rewrite this as y is equal to 2 times x squared
- plus 3x plus 2.
- That saves me the pain of having to do the factoring by
- grouping that we talked about in the last video.
- And this we can factor in a pretty straightforward way.
- This is 2 times 1 is 2, 2 plus 1 is 3, so this is x plus 2
- times x plus 1, right?
- 2 plus 1 is 3, 2 times 1 is 2.
- So that's what y is equal to.
- So if we wanted to know the x-intercepts, we figure out
- where this expression is equal to 0, so we said 0 is equal to
- 2 times x plus 2 times x plus 1.
- Remember, we want to figure out where we intercept the
- x-axis, that's the x-axis there.
- Y is 0 on the x-axis.
- So let's get back to this problem.
- So either of these can be equal to 0.
- Obviously the 2 can't be equal to 0, so we get the situation
- where x minus x plus 2 is equal to 0 or x plus 1 is
- equal to 0.
- I didn't have to write the parentheses there.
- Subtract 2 from both sides of this equation, we get x is
- equal to negative 2.
- Subtract 1 from both sides of this question or x could be
- equal to negative 1.
- So the point negative 2, 0.
- You put negative 2 here, this is obviously going to make y
- equal to 0 and the point negative 1, 0, same argument.
- Those are our x-intercepts.
- Let me graph it.
- We have negative 2, 0 right there.
- Actually let me do this graph a little bit bigger.
- Let's say that this is right here is-- let me label it-- in
- double steps.
- So let's say this is negative 1, let's say this is negative
- 2, that is negative 3, 1, 2, this is our 3.
- It's a 1, 2, 3 just like that, negative 1, negative 2,
- negative 3.
- So, I'm using two blocks to represent one.
- So the point negative 2, 0 is on our graph, with our
- x-intercept, so negative 2, 0 is right there.
- And then we have negative 1, 0, which is right there.
- And then we want to figure out the vertex.
- Well, the vertex is going to be the average of those two,
- so the x for the vertex is going to be negative 2 plus
- negative 1 over 2, that's negative 3/2 which is the same
- thing is negative 1.5.
- Now what is y equal to when x is negative 1.5?
- This is going to be a little bit involved.
- I'll use actual 3/2.
- So so we get y is equal to 2 times negative 3/2 squared
- plus 6 times negative 3/2 plus 4.
- And let us figure this out.
- So this is equal to 2 times 9 over 4 minus-- you have that
- negative sign there-- and then we could make this 6 divided
- by 2 is just the same thing is 3/1.
- So this is minus 9, 3 times 3 plus 4.
- This 2/4 the same thing as 1/2, so this becomes 9/2
- minus 9 plus 4.
- So let's simplify this.
- Minus 9 plus 4 is the same thing as minus 5, right?
- Minus 9 plus 4.
- Minus 5 we can rewrite is equal to minus 10/2.
- 9 minus 10, all of that over 2, is minus 1/2.
- So, our vertex of this parabola is the point minus
- 1.5 or negative 1.5, negative 1/2.
- This is our vertex, so we can graph it right here.
- Negative 1.5, negative 1/2, our vertex
- is right over there.
- And then we can a graph this parabola.
- If we want to do the y-intercept, we can.
- The y-intercept is going to be here when x is 0,
- y is equal to 4.
- These terms cancel out.
- X is 0, y is 1, 2, 3, 4 right over there.
- And then we can graph this parabola.
- It will look like that, that's my best, it should be
- symmetric around x-coordinate of the vertex just like that.
- Let's do one more of these.
- And, once again, I've used all my real estate up, let me
- clear this out of the way.
- Let's do one more.
- Let's say that I have y is equal to negative x squared
- plus 10x minus 21.
- Once again, let's write this in intercept form, we just
- factor the right-hand side.
- We get y is equal to-- well the first thing I want to do
- is it is factor this negative 1 out-- negative 1 times x
- squared minus 10x plus 21.
- Now what two numbers, when I multiply them, they're
- positive 21?
- So they have to be the same side, and when I add them I
- get negative 10.
- At least in my brain, negative 7 and negative 3 jump out.
- I mean, there's not that many factors of 21, so 7 and 3 are
- probably two that usually jump out because they add to 10.
- So you have y is equal to negative 1 times x minus 3
- times x minus 7.
- Negative 3 plus negative 7 is negative 10, negative 3 times
- negative 7 is positive 21.
- So if you want to find the x-intecepts, the x-intercepts
- are going to be where-- so we could set this whole equation
- equal to 0, negative 1 times x minus 3 times x minus 7 are
- x-intercepts or the x-values that make y equal to 0 are
- either the ones that make x minus 3 equal to 0 or the
- x-values that make x minus 7 equal to 0.
- Add 3 to both sides of this equation, you got x is equal
- to 3, add 7 to both sides of this equation, you get x is
- equal to 7.
- So so we get our two x-intercepts, the point 3, 0--
- 1, 2, 3, 0-- and the point 7, 0-- 1, 2, 3, 4, 5, 6, 7-- 0.
- Now, we know that our vertex is going to be right in
- between them.
- So the x for the vertex is going to be 3 plus 7 over 2,
- which is equal to 10/2, which is equal to 5.
- And to the y-value, let's figure it out, y for our
- vertex going to be negative 5 squared, so negative 25 plus
- 10 times 5 plus 50 minus 21.
- So what is this?
- This is equal to-- we have a negative 25 and negative 21,
- you add those together you get a negative 46.
- Add that to 50, you get y is equal to 4.
- So the vertex is at 5, 4.
- Let me graph that, the vertex is 5, 4.
- So 1, 2, 3, 4, 5, right in between.
- And then up four, 1, 2, 3, 4.
- So notice something, in this graph over here, the last two
- graphs, actually I think I just cleared them, you saw
- that the vertex was below the x-axis.
- So, when we plotted the x-intercept and then the
- vertex, the upward-forming u-shape was the only option.
- But here, the only option is a downward u-shape.
- This being either the minimum or maximum point and going for
- both of these points and having a u-shape, our u-shape
- is going to have to look something like this.
- Now,-- let me draw that little bit better-- the u is going to
- have to look something like that.
- That's a better shot at it.
- Now notice what was the difference between this
- equation and the previous ones?
- The difference was is that this one had a negative in
- front of the x squared.
- The coefficient in front of the x squared
- was a negative number.
- And that's why we have kind of a downward opening parabola.
- If this was a positive number, we would have an
- upward-opening parabola.
- So that's why this one is kind of a downward shape and its
- vertex represents a maximum point.
- It's the highest point on the parabola.
- In the previous videos, we had an upward opening, or in the
- previous problems, with an upward-opening parabola and
- our vertex was the minimum point.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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