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In this video, I want to show some examples of graphing

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quadratic functions.

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But before I do that, I want to make one fairly minor

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correction to the end of the last video where we did some

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factoring with grouping.

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When I showed you the proof of why that works, I multiplied

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fx plus g times hx plus j.

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When I multiplied them, I did the first term, I

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said fx times hx.

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When I did it in the last video, I just wrote fhx.

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And we know that that's not true, we have f times h

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times x times x.

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It's fhx squared.

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That times that is fhx squared and then the rest of it I had

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gotten right, it's fx times j, so plus we get fjx and then we

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had g times hx so we could write that as plus ghx.

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And then finally g times j, which gives us gj.

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So I apologize for that error, I had forgotten to write the

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exponent there and I think I'd forgotten to write it in the

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step after that, but it didn't change the argument for the

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proof, I at least didn't mess up there.

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So I just want to make that quick correction in case that

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confused anyone.

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And I also made a small annotation in that video if

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you're watching it on YouTube.

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With that out of the way, let's learn to graph some of

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these quadratic functions.

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So let's say I have the function y is equal to x

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squared minus 2x minus 8.

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And what we're going to do is we're going to rewrite this

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quadratic expression on the right-hand side.

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We're going to write it in intercept form.

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That essentially means just writing it

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in it factored form.

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Then doing that, we're able to figure out the x-intercept,

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where we're intersecting the x-access, and then using that

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we'll be able to figure out the vertex.

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And I'll show you you what the vertex is in a second.

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So let's just factor this quadratic on

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the right-hand side.

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So we get y is equal to-- so what are two numbers where

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their product is a negative 8, so they're going to have

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different signs and then their difference is negative 2?

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Let's see, if we have a negative 4, and plus 2.

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When you take the product, you get negative 8, when you add

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them, you get negative 2.

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So we could rewrite y as being equal to x minus 4

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times x plus 2.

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So this is essentially writing this quadratic function in

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intercept form.

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And now let's figure out where this intersects the x-axis So

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let me label my axis.

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This is, of course, the x-axis, that is the y-axis.

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So when will something intersect the x-axis?

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The x-axis is when y is equal to 0, right?

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So let's set this equal to 0.

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0 is equal to x minus 4 times x plus 2.

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We've seen this multiple times before, this means that either

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x minus 4 or x plus 2 or both of them have to be equal to 0.

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So x minus four is equal to 0 or x plus two is equal to 0.

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So that means x could be equal to 4-- we're just adding 4 to

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both sides of that equation-- or subtracting 2 from both

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sides of this equation, x could be equal to negative 2.

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So these points where we intersect, when x is 4, this

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term is 0 so the whole thing is 0.

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That's how we solved for this.

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So the points 4, 0, and the point 2, 0 are going to both

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be on this parabola.

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So let's graph them.

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So 4, 0-- let me do this in a darker color-- 1, 2 3, 4,

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that's 4, 0.

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And then the other point is negative 2, 0.

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1, 2 negative 2, 0 is right there.

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So that's the two points where we intersect the x-axis.

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And now we're going to determine something called the

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vertex of our parabola.

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And the vertex, just to give you a sneak preview, parabolas

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are the graphs of quadratic functions and so the graphic

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is going to look either like an upward u or a downward u,

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and it's going to be shifted around on this.

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But the vertex is either this minimum point or

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this maximum point.

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And the word parabola just describes this function shape.

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It's the shape of the graph of a quadratic.

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Let me write that word down, these are parabolas.

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So to figure out the vertex, you actually just take the

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x-value that's halfway in between the two intercepts.

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No matter what, the intercepts are going to be equidistant

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from the vertex.

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So if that is the graph, those would be our intercepts and

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then our vertex is going to be an x-value exactly halfway

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between them.

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So to figure out the x-value of our vertex, we just average

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these two x-values of our x-intercepts.

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So let's do that.

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We could say the x for the vertex, let's call that x for

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the vertex-- that's why I wrote a v there-- is going to

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be equal to the average of these.

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So 4 plus negative 2 over 2.

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Which is equal to this 4 minus 2, which is 2 over

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2 is equal to 1.

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So x is going to be equal to 1 and what is y when

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x is equal to 1?

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So the y for the vertex is going to be equal to 1

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squared-- I'll just write that as 1-- minus 2 times 1, so

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minus 2, minus 8.

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So what is this?

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This is 1 minus 10, which is equal to negative 9.

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So the vertex is going to be at the point 1, negative 9.

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So we go one and then we go down nine.

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So x is 1 and then y is negative 9.

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1, 2, 3, 4, 5, 6, 7, 8, 9.

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So it's right there.

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And so if we were to graph this parabola, the line of

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symmetry is on the x-value of the vertex.

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So, I want to be doing it in a darker color than that.

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So the graph should be symmetric around x is equal to

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1, and the graph will look something like this.

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I'll try my best to draw it.

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Bam!

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It'll intersect the x-axis there, it'll be symmetric,

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it'll look just like that on the other side.

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And it'll go bam, just like that.

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I think you get the general idea.

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And then the other point that you may or may not be

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interested in is the actual y-intercept.

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If x is equal to 0, you immediately see that y is

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equal to negative 8.

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So the point 0, negative 8 should also be on this graph.

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Let's see if I drew it right.

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0, negative eight right there should also be on that graph.

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Let's do another one of these.

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I'm kind of taking up some of the space for the next

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problem, so let me clear this.

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Let's do another one.

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All right.

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So let's say we have y is equal to 2x squared

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plus 6x plus 4.

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So here, the thing that immediately jumps out at me is

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that I can factor a 2 out of everything.

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So I could rewrite this as y is equal to 2 times x squared

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plus 3x plus 2.

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That saves me the pain of having to do the factoring by

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grouping that we talked about in the last video.

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And this we can factor in a pretty straightforward way.

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This is 2 times 1 is 2, 2 plus 1 is 3, so this is x plus 2

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times x plus 1, right?

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2 plus 1 is 3, 2 times 1 is 2.

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So that's what y is equal to.

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So if we wanted to know the x-intercepts, we figure out

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where this expression is equal to 0, so we said 0 is equal to

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2 times x plus 2 times x plus 1.

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Remember, we want to figure out where we intercept the

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x-axis, that's the x-axis there.

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Y is 0 on the x-axis.

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So let's get back to this problem.

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So either of these can be equal to 0.

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Obviously the 2 can't be equal to 0, so we get the situation

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where x minus x plus 2 is equal to 0 or x plus 1 is

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equal to 0.

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I didn't have to write the parentheses there.

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Subtract 2 from both sides of this equation, we get x is

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equal to negative 2.

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Subtract 1 from both sides of this question or x could be

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equal to negative 1.

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So the point negative 2, 0.

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You put negative 2 here, this is obviously going to make y

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equal to 0 and the point negative 1, 0, same argument.

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Those are our x-intercepts.

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Let me graph it.

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We have negative 2, 0 right there.

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Actually let me do this graph a little bit bigger.

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Let's say that this is right here is-- let me label it-- in

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double steps.

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So let's say this is negative 1, let's say this is negative

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2, that is negative 3, 1, 2, this is our 3.

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It's a 1, 2, 3 just like that, negative 1, negative 2,

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negative 3.

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So, I'm using two blocks to represent one.

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So the point negative 2, 0 is on our graph, with our

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x-intercept, so negative 2, 0 is right there.

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And then we have negative 1, 0, which is right there.

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And then we want to figure out the vertex.

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Well, the vertex is going to be the average of those two,

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so the x for the vertex is going to be negative 2 plus

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negative 1 over 2, that's negative 3/2 which is the same

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thing is negative 1.5.

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Now what is y equal to when x is negative 1.5?

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This is going to be a little bit involved.

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I'll use actual 3/2.

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So so we get y is equal to 2 times negative 3/2 squared

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plus 6 times negative 3/2 plus 4.

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And let us figure this out.

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So this is equal to 2 times 9 over 4 minus-- you have that

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negative sign there-- and then we could make this 6 divided

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by 2 is just the same thing is 3/1.

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So this is minus 9, 3 times 3 plus 4.

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This 2/4 the same thing as 1/2, so this becomes 9/2

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minus 9 plus 4.

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So let's simplify this.

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Minus 9 plus 4 is the same thing as minus 5, right?

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Minus 9 plus 4.

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Minus 5 we can rewrite is equal to minus 10/2.

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9 minus 10, all of that over 2, is minus 1/2.

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So, our vertex of this parabola is the point minus

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1.5 or negative 1.5, negative 1/2.

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This is our vertex, so we can graph it right here.

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Negative 1.5, negative 1/2, our vertex

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is right over there.

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And then we can a graph this parabola.

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If we want to do the y-intercept, we can.

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The y-intercept is going to be here when x is 0,

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y is equal to 4.

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These terms cancel out.

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X is 0, y is 1, 2, 3, 4 right over there.

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And then we can graph this parabola.

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It will look like that, that's my best, it should be

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symmetric around x-coordinate of the vertex just like that.

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Let's do one more of these.

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And, once again, I've used all my real estate up, let me

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clear this out of the way.

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Let's do one more.

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Let's say that I have y is equal to negative x squared

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plus 10x minus 21.

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Once again, let's write this in intercept form, we just

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factor the right-hand side.

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We get y is equal to-- well the first thing I want to do

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is it is factor this negative 1 out-- negative 1 times x

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squared minus 10x plus 21.

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Now what two numbers, when I multiply them, they're

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positive 21?

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So they have to be the same side, and when I add them I

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get negative 10.

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At least in my brain, negative 7 and negative 3 jump out.

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I mean, there's not that many factors of 21, so 7 and 3 are

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probably two that usually jump out because they add to 10.

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So you have y is equal to negative 1 times x minus 3

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times x minus 7.

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Negative 3 plus negative 7 is negative 10, negative 3 times

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negative 7 is positive 21.

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So if you want to find the x-intecepts, the x-intercepts

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are going to be where-- so we could set this whole equation

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equal to 0, negative 1 times x minus 3 times x minus 7 are

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x-intercepts or the x-values that make y equal to 0 are

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either the ones that make x minus 3 equal to 0 or the

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x-values that make x minus 7 equal to 0.

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Add 3 to both sides of this equation, you got x is equal

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to 3, add 7 to both sides of this equation, you get x is

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equal to 7.

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So so we get our two x-intercepts, the point 3, 0--

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1, 2, 3, 0-- and the point 7, 0-- 1, 2, 3, 4, 5, 6, 7-- 0.

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Now, we know that our vertex is going to be right in

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between them.

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So the x for the vertex is going to be 3 plus 7 over 2,

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which is equal to 10/2, which is equal to 5.

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And to the y-value, let's figure it out, y for our

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vertex going to be negative 5 squared, so negative 25 plus

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10 times 5 plus 50 minus 21.

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So what is this?

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This is equal to-- we have a negative 25 and negative 21,

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you add those together you get a negative 46.

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Add that to 50, you get y is equal to 4.

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So the vertex is at 5, 4.

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Let me graph that, the vertex is 5, 4.

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So 1, 2, 3, 4, 5, right in between.

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And then up four, 1, 2, 3, 4.

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So notice something, in this graph over here, the last two

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graphs, actually I think I just cleared them, you saw

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that the vertex was below the x-axis.

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So, when we plotted the x-intercept and then the

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vertex, the upward-forming u-shape was the only option.

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But here, the only option is a downward u-shape.

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This being either the minimum or maximum point and going for

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both of these points and having a u-shape, our u-shape

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is going to have to look something like this.

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Now,-- let me draw that little bit better-- the u is going to

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have to look something like that.

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That's a better shot at it.

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Now notice what was the difference between this

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equation and the previous ones?

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The difference was is that this one had a negative in

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front of the x squared.

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The coefficient in front of the x squared

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was a negative number.

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And that's why we have kind of a downward opening parabola.

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If this was a positive number, we would have an

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upward-opening parabola.

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So that's why this one is kind of a downward shape and its

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vertex represents a maximum point.

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It's the highest point on the parabola.

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In the previous videos, we had an upward opening, or in the

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previous problems, with an upward-opening parabola and

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our vertex was the minimum point.

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