Do you want to go ahead and not solve the problem?
Happy New Year!
That was not what I was anticipating.
I was going to ask if you wanted to go ahead and tell us some of the intercepts.
No.
No, not none of them.
Because the goal is to find as many points of interest.
Can you tell us how one might find an intercept if one with an integer?
One might set the equation equal to zero.
Okay?
So it has, let's say, zero equals x to the fourth minus 8x squared.
How many intercepts or solutions will that produce?
Three.
Three. Why not four?
Because we can't find minus zero.
Okay.
Fair enough.
So what are those four, Chris?
I'm sorry.
They're zero and plus or minus two would be two.
Okay, so that's going to be three distinct points, right?
Does that make to zero a double root?
Yeah.
I don't know if it's technically a double root, but it certainly feels right to call it a double root.
So zero, what?
Like if we're going to have a point of interest, then we need to know the full point.
So zero, zero, zero, zero.
We're just pointing back in the function to get the value.
That's one.
The other one with an integer, you said?
Two root two.
Two root two.
No, no, two root two.
Oh, oh, sorry.
It's a two root two.
I'm just going to call that one 18.
Another one with two root two.
I think we're going to run out of space.
So why don't we get them to put that in the function here?
Oh, yeah.
That's fine.
That's fine.
I forgot about that.
So the other one was what?
Negative root eight.
Negative root eight.
That was zero.
Okay.
So if we were to put those on the graph, you'd get one right here, zero zero.
This is a 10 by 10 graph.
Let's say it's going to be a little squished.
That's okay.
I'll give you five plus.
What were the axes?
Okay.
I didn't want you to.
I want to do it again.
So I know you have one intercept right here.
And what is the spirit of eight, Chris?
The spirit of eight.
So first, we did a two point three, it's less than three, one by two.
That's five.
Yeah.
Large.
It'll be a little closer to three and two.
It's closer to nine and it's four.
So we're looking at roots that are...
Actually you don't know what it's going to be.
It's always one and two and three.
And then the same roots on the opposite side.
So those are our three intersects.
What about our y-intercept?
00 already found it.
It's right here.
We're going to call this 5 and this 10.
So our y-intercept, we're going to figure it out.
So that gives us a rough idea of what's going on with the graph.
And we just have to fill in the blanks beyond that.
Now, the next step is to figure out the number of critical points.
How do we do that?
We're going to take a derivative.
Okay.
So there we go.
Now, let's find the root points.
Where are we going to go with that?
Now, in fact, we're on an x.
What are you saying about that?
Zero.
That one solution is when x equals to zero.
So this right here is going to be a critical point.
That's going to slow down to zero at that point.
And whether it's curving up or down, you don't know you have to calculate the second root.
But we're going to be saving it.
At that point, the graph is flat.
So if we factor out zero, we're left with zero equals four x squared minus 16.
Okay, that's pretty solid.
What are the other solutions for that?
Because you factor out the tensor.
I hope I factor out an x.
Okay.
If we factor out an x, we're left with that equation, which has two solutions.
What are those solutions made of?
Two and eight and two.
So the problem is remember to draw it on our graph.
The y value is four of those x values.
So we have to take two and negative two and plug them back into our original equation to figure out where those other points are going to be.
My scale is a little off. We're going to have to drop this thing down a little bit.
But right here is negative 16.
And so we're going to have critical points here.
Now already you can probably figure out the next and half of this graph.
We can continue together and find the other points of interest.
QoI, points of inflection.
If we take the derivative again, what's the whole problem?
Peter, what's the question?
Thank you so much for Peter's statement.
It's 12x squared minus 16.
Okay, so then Peter, how do we find the points of inflection?
So that equals zero, which produces what?
You don't know it.
I got this, right?
I got plus one is two over three.
Anyone else?
So what is two over three?
Heck, what's root three?
1.71.
Yeah, 1.71, 1.72.
Half of that is 2.86.
So what's two over backwards?
Between one and two, that's about right.
It's a little bit of one, but it's not quite half the two over two.
That's about one.
Yeah, that's what we've already drawn in the graph.
But between one and two, we can easily search.
Is it probably closer to one or closer to two?
Closer to one.
Closer to one, because two is pretty close to one, right?
So we have points that are two over root three and negative two over root three.
So what we can do now is to see if we get a value of this to get y-byes to correspond.
Was anybody able to do that?
I got it.
Yeah, I also didn't even get to that part of the product.
Oh, I never, I didn't find all the y-byes.
Can we take a little less than ten?
It's a little less than ten.
It'd be nice if we could get y-byes.
It's true you could just sketch it.
But if you didn't identify, this is exactly where it is.
That's one more point where we can still put down a connected dot scan.
One more point where our connected dot scan is kind of important.
The root three is actually going to work very well with our original y equation.
We have something to the fourth with something squared.
So root three is going to turn into a nine and a three.
I may not be that bad.
Let's go ahead and try this.
Y equals two over root three to the fourth minus eight times two over root three squared.
What's two over root three and four?
What's two over root three and four?
That's sixteen minus.
Sixteen minus.
Where?
Now, minus two over root three squared is going to be four thirds, right?
Times eight gives me four.
How are we going to finish this little amount?
Not the problem.
You mean for common denominator?
Common denominator.
Do you have a negative three times three?
Sixteen minus ninety-six.
So sixteen minus ninety-six thirds, which is negative eighty.
Third.
Ninth.
Ninth.
So negative eighty-ninths.
What about the negative side of that?
Is that going to happen?
No.
It's all squared into four.
So it's going to be eighty-ninths also, which is a little less than nine.
A little less than nine.
A little less than nine.
A little less than nine.
A little less than nine.
A little less than nine.
So if here is our nine, a little less than nine, so we're looking somewhere along this line.
And I should probably always solve that.
And we decided that these were kind of close to one.
So we're looking at points kind of like here.
And here.
For our points of view.
Here is one day we are down with me.
So you know, you do this.
What is this graph book like when you're done?
W.
It's big fat W.
Some channels, sweeps are here.
And some points.
And some collections.
Some collections.
Now it goes.
And I grab the book.
Let's.
And.
How about the text.
Ready for you.
I have someone in my house.
You already have to talk to me.
I know.
Oh.
I got five.
Thank you.
