We're on problem 39.
And it says which ordered pairs the vertex of f of x is
equal to x squared plus 6x plus 5?
Now, finding vertexes of parabolas, there's a formula,
and negative b over 2a, but you might forget that when you
turn 30 years old.
The whole goal of this isn't just to pass an algebra exam.
It's to learn algebra.
Then when you learn calculus, you'll learn actually a much
easier way to figure out minimum and maximum points.
But there's actually an intuitive way of doing it in
algebra as well.
What we have to do is just write this in that form.
If we could write this function in the form like
this, f of x is equal to A, times x plus or minus
something, I don't know, b, squared, plus C.
Then we know the vertex happens when x is equal to b.
Because if you look at this form of a parabola-- and you
can write any of these in this form-- then you say OK, this
term right here is always going to be positive.
So the minimum point of this parabola is when this term is
equal to 0.
Then the minimum point of this term happens when x is equal
to b, when this term is equal to 0.
So the vertex becomes xb comma C.
So let's do that with this.
You could memorize it.
It ends up being minus b over 2A.
But let's just actually solve it, because I want you to have
the intuition.
f of x is equal to x squared plus 6x, plus 5.
I wrote a gap there for a reason, because we're
essentially going to complete a square.
We want to turn this x squared plus 6x, plus something, into
a perfect square.
Hopefully you have some practice
completing the square.
What number is 1/2 of 6?
Because when we square it, if we do x plus a squared, that
equals what? x squared plus 2ax plus a squared.
So if we map that to that, x squared is x squared.
2ax is equal to 6.
So a would be equal to 3.
And if a is equal to 3, a squared would be equal to 9.
You can't just add a 9 to one side of the equation.
You either have to do it to both sides, or you have to do
something to offset that.
So if we added a 9 here, let's just subtract a 9 as well.
So then we get x squared plus 6x, plus 9, plus 5, minus 9.
So this right here, this part of it, that's equal to x plus
3, squared.
And then this part, 5 minus 9, minus 4.
There, we got it into our format that we needed.
So when does this function achieve a minimum point?
Well, it achieves a minimum point when this part is equal
to 0, or when x is equal to minus 3.
When x is equal to minus 3, this is 0.
So f of x is going to be equal to minus 4.
So that's the vertex, and that's choice A.
You could have done minus b over 2A.
You'd get minus 12 over 2-- no minus 6 over 2,
which is minus 3.
That's something you could memorize.
But it's better, I think, to have the intuition.
Problem 40.
The graph of x over 2, squared, minus y over 2,
squared equals 1 is a hyperbola.
That makes sense.
This is y over 3. x over 2, squared, minus y over 3,
squared is equal to 1.
If this is a plus, we would have an ellipse.
Fair enough.
Which set of equations represents the asymptotes of
the hyperbola's graph?
Once again, there's some equations that people memorize
and formulas people memorize, but you can actually have an
intuition on this.
So let's think about what happens as x approaches
positive or negative infinity.
Because that's what we care about as an asymptote, right?
What does the graph approach when we get to really, really
large values of x, and it turns out, really
large values of y?
So let's just rearrange this a little bit.
Let's put y on the other side.
Let's square this.
So this is x squared over 4.
Let me keep it actually in that format.
Let's keep that as x over 2, squared is equal to 1, plus y
over 3, squared.
Let's subtract 1 from both sides.
So you get x over 2, squared, minus 1 is equal
to y over 3, squared.
And now, let's multiply both sides by 3 squared.
Let's multiply both sides by 9.
Let's do that.
Actually, to simplify this, let me rewrite this.
I wanted to keep in that form.
That's the same thing as x squared over 4, minus 1, is
equal to y squared over 9.
Multiply both sides by 9.
I'm going to switch the sides so we get y here.
So you get y squared is equal to 9 over 4 x
squared, minus 9.
All right, and that's about as far as we can get just with
straight up algebra.
When we think about asymptotes, those are points
that the graphs will never reach but it'll get close to.
So we have to think about what happens with x gets really
large and correspondingly y gets really large.
Well, as x gets really large, so as x approaches infinity--
and you'll learn this is precalculus and calculus,
these are limits, but it's really a very intuitive
concept-- as x gets really large, this minus 9 doesn't
matter much.
This term starts to dominate it.
These terms matter a lot more than this minus 9.
So as x approaches infinity, y squared is approximately equal
to 9/4 x squared.
Now we could take the square root of both sides of this.
So you get y is approximately equal to plus or minus 3/2 x.
Those are the actual asymptotes, y equals 3/2 x and
y is equal to minus 3/2 x.
That's actually choice A.
I won't go into the actual instruction of how you would
think about graphing.
Actually, why not?
Why don't we just do that.
We know what the asymptotes are.
Let's actually think about how you would graph this.
So if you were to draw the-- it's all about learning, not
just about assessing-- that's the x- and y-axis.
And they tell us their asymptotes, 3/2 x
and minus 3/2 x.
So 3/2 x, the slope of 3/2 is 1.5.
So it's a slope that will look something like-- let me get
the line tool out-- so x might look like, so plus 3/2 x would
look something like that.
Minus 3/2 x would be something like that.
Now we have to figure out does this hyperbola go from there
and there, or does it go from here and here?
There were just say, OK, when y is equal to 0 we get x over
2 is equal to plus or minus 1.
You actually get x would be there.
So you know that this thing has x-intercepts.
You can set y is equal to 0.
You can't set x is equal to 0.
If you get x is equal to 0, then you get minus y squared
is equal to something.
You get y squared is equal to negative 1.
You get some imaginary solution.
So you know that it has x-intercepts.
You know that these are the asymptotes.
So if you want to draw the graph, it would look
something like that.
I know that's not a great drawing, but hopefully it
serves the purpose.
Anyway, the choice was A.
And hopefully you learned a little bit about asymptotes.
Next problem, 41.
Which of the following represents a hyperbola?
Let me copy those.
No, which of the following represents a parabola, they
want to know.
That's too small.
I don't think you can see that.
I'll get a bigger version of it.
All right, that's the bigger version.
There you go.
Which of these represents a parabola?
Well, we could just go one by one.
This is a circle.
That's a circle.
Then you could graph if you like.
This is an ellipse, where you just kind of weight the y's
and the x squareds.
This is an ellipse.
If this was a negative, this would be a hyperbola, ellipse.
This one is a hyperbola because you have
this negative here.
This one might not look like a traditional parabola,
but it really is.
Because if I were to divide both sides of this by 4p, I'd
get x is equal to 1 over 4p times y squared.
So it's a sideways parabola.
It looks something like this.
If that's the x- and y-axis, the graph will look
something like this.
It's going to be a sideways parabola.
But it's a parabola nonetheless.
So the choice is C.
We have time for at least one more, problem 42.
They give the following equation.
Let me paste it here.
All right, and they say what is the standard form of the
equation of the conic section above?
And essentially they want us to complete the square with
the x's and the y's so we get that x minus something
squared, over some m plus or minus y, minus something
squared, over some constant.
So let's do that.
So let's group the x and the y terms. So you get 4x squared,
minus 16x, plus something.
That's how we're going to complete the square.
Then you get plus minus 5y squared.
Let me just minus, and I'll do plus 5y squared.
We have a minus here, so it becomes plus 30y.
And then out here you have a minus 9 is equal to 0.
Just to simplify it, let's factor a 4 out of this part.
So you get 4 times x squared, minus 4x, plus something,
minus-- let's factor a 5 out-- 5, times y squared, plus 6y,
probably plus something-- and let's add the 9 to both
sides-- is equal to 9.
All right, so let's think about it.
If we wanted to complete the square here, what do we put
here to make this a perfect square?
If this was a plus 4, then this would
be x minus 2 squared.
So we're adding a 4 here.
But we're really adding a 4 times a 4.
We're really adding a 16 if you think about it.
If I were to redistribute this, I just added 16.
So if I add 16 to one side of the equation, I have to add 16
to the other side of the equation.
So I'll add 16 there.
To make this a perfect square, let's see, 1/2 of 6 is 3, so 3
squared is 9.
But I'm really adding a 5 times a 9.
I could redistribute the 5.
I really added a 45.
Oh, no, no, no.
I added a minus 5 times 9.
I caught myself.
I added a minus 5 times 9.
So I'm really subtracting 45 from that side.
So I have to subtract 45 from that side.
And so this simplifies to 4 times x minus 2 squared--
right, that's x minus two squared-- minus 5 times-- this
is y plus 3 squared-- is equal to-- let's see, 9 plus 16 is
25, 25 minus 45.
Let's see, have I made a mistake someplace?
25 minus 45, no, that's right.
25 minus 45 is minus 20.
And now we can divide both sides by minus 20, if I did
this correctly.
We'll divide both sides by minus 20.
So you get x minus 2 squared.
What's 4 divided by minus 20?
It's minus 1/5.
Let me just do that.
So 4 over minus 20, minus 5, times y plus 3 squared, over
minus 20 is equal to 1.
So this is equal to, let's see this becomes a positive, and
this one becomes a negative.
So let's put this one first. So 5 over 20 is the same thing
as y plus 3 squared, over 4-- right, that's this
one-- plus, no, sorry.
We have a minus.
Minus-- and this becomes 1/5-- x minus 2 squared, over 5.
That is equal to 1.
We haven't made a careless mistake.
That is choice B.
See you in the the next video.