Investment Banker Resume Template: Copy This To Get Into KKR (Or At Least, Something Almost as Good)

Uploaded by TheIntegralCALC on 25.08.2011

Transcript:

Hi, everyone! Welcome back to integralcalc.com. Today, we’re going to be doing another problem

out of our analytic geometry section. And in this particular problem, we’ve been asked

to show whether the graph of the equation, x squared plus y squared minus 6x minus 4y

minus 13 equals zero consists of either a single point or of no points. So, this graph

(they’ve already told us) is either a single point or it’s no point. And often times

with these kinds of problems, the graph will either be a circle, a single point, or of

no point. In this case, they’ve narrowed it down for us to either single point or no

points. What we need to do is get our equation here

into the form given in this formula which is the formula for a circle. So in order to

do that, what we need to do is in the left-hand side of the equation, we need to complete

the square with respect to both x and y. So the first thing we’re going to do is get

our x terms together and our y terms together. So I’m going to group these together, the

x squared and the 6x and I’m going to group my y terms together so y squared and 4y. And

then I’m going to go ahead and move the 13 to the other side so we’ll end up with

negative 13 on the right-hand side. And again, we’re trying to get this into this format

over here. So again, you can see x terms together, y terms together and then p consists of just

a constant value. So like I said, complete the square. Remember that with completing

the square, you want to take the coefficient on the first degree term, the first degree

term meaning x to the first power or y to the first power as opposed to x squared and

y squared. You want to take the coefficient of the first degree term, divide it by 2;

and this is assuming that the coefficient on your second degree term is 1. So the fact

that this is 1x squared and 1y squared, we don’t have some other coefficient on the

front of those terms, means we can proceed with completing the square. If you have for

example 2x squared, you would have to divide through by 2 to make the coefficient on the

x squared term 1. But in this case, that’s already done so we can go ahead and take the

coefficient on the first degree term, negative 6, divide it by 2 so we have negative 3. Then

you want to take your answer and square it. And when we square negative 3, we get 9. So

9 is what we’re going to add to x squared minus 6x to complete the square.

Let’s go ahead and do the same thing for the y term. So we’ll take negative 4, the

coefficient on the first degree term, divide it by 2 we get negative 2. The we take our

answer and we square it. Negative 2 squared is 4 so that’s what we’ll be adding to

these collection of y terms to compete the square. So we’ve got to add these things

to both sides so we’ll get x squared minus 6x plus 9 plus y squared minus 4y plus 4 equals

negative 13. And since we added 9 and 4 to the left-hand side, we have to add them to

the right-hand side as well. Otherwise, we would be changing the equation. So we need

to keep things balanced. The reason we complete the square is because this is going to allow

us to get our equation into the same format as the formula that we’ve provided here

for the equation of a circle. So x squared minus 6x plus 9 can be factored into x minus

3 times x minus 3 which can be written as x minus 3 squared. And now this y squared

minus 4y plus 4 can be factored into y minus 2 times y minus 2 which means we can write

it as y minus 2 squared. Then over here at the right-hand side negative 13 plus 9 plus

4 is zero. So we end up with zero on the right-hand side.

Now, you can see that our equation is at the same format as the formula. We have x minus

3 squared plus y minus 2 squared equals some constant. So now that we’ve got the same

format, we can make an evaluation about the constant p over here. So to make a determination

about whether this equation represents a single point or no point, what we need to look at

is the constant on the right-hand side. So you want to make sure you collected all your

constants on the right-hand side that we have and there are no loose constants here on the

left-hand side. So we’re just looking at zero. And when you look at zero (and we’ll

call it p), if p is greater than zero, then you’re dealing with a circle. If p is equal

to zero, then you’re dealing with a single point and if p is less than zero, you’re

dealing with no points. In this case, p is equal to zero. Our right-hand side is equal

to zero. So that means that the graph of this equation consists of one single point. The

graph is literally just one point. So that’s it. That’s how we can make that

determination. I hope this video helped you guys and I will see you in the next one. Bye!

out of our analytic geometry section. And in this particular problem, we’ve been asked

to show whether the graph of the equation, x squared plus y squared minus 6x minus 4y

minus 13 equals zero consists of either a single point or of no points. So, this graph

(they’ve already told us) is either a single point or it’s no point. And often times

with these kinds of problems, the graph will either be a circle, a single point, or of

no point. In this case, they’ve narrowed it down for us to either single point or no

points. What we need to do is get our equation here

into the form given in this formula which is the formula for a circle. So in order to

do that, what we need to do is in the left-hand side of the equation, we need to complete

the square with respect to both x and y. So the first thing we’re going to do is get

our x terms together and our y terms together. So I’m going to group these together, the

x squared and the 6x and I’m going to group my y terms together so y squared and 4y. And

then I’m going to go ahead and move the 13 to the other side so we’ll end up with

negative 13 on the right-hand side. And again, we’re trying to get this into this format

over here. So again, you can see x terms together, y terms together and then p consists of just

a constant value. So like I said, complete the square. Remember that with completing

the square, you want to take the coefficient on the first degree term, the first degree

term meaning x to the first power or y to the first power as opposed to x squared and

y squared. You want to take the coefficient of the first degree term, divide it by 2;

and this is assuming that the coefficient on your second degree term is 1. So the fact

that this is 1x squared and 1y squared, we don’t have some other coefficient on the

front of those terms, means we can proceed with completing the square. If you have for

example 2x squared, you would have to divide through by 2 to make the coefficient on the

x squared term 1. But in this case, that’s already done so we can go ahead and take the

coefficient on the first degree term, negative 6, divide it by 2 so we have negative 3. Then

you want to take your answer and square it. And when we square negative 3, we get 9. So

9 is what we’re going to add to x squared minus 6x to complete the square.

Let’s go ahead and do the same thing for the y term. So we’ll take negative 4, the

coefficient on the first degree term, divide it by 2 we get negative 2. The we take our

answer and we square it. Negative 2 squared is 4 so that’s what we’ll be adding to

these collection of y terms to compete the square. So we’ve got to add these things

to both sides so we’ll get x squared minus 6x plus 9 plus y squared minus 4y plus 4 equals

negative 13. And since we added 9 and 4 to the left-hand side, we have to add them to

the right-hand side as well. Otherwise, we would be changing the equation. So we need

to keep things balanced. The reason we complete the square is because this is going to allow

us to get our equation into the same format as the formula that we’ve provided here

for the equation of a circle. So x squared minus 6x plus 9 can be factored into x minus

3 times x minus 3 which can be written as x minus 3 squared. And now this y squared

minus 4y plus 4 can be factored into y minus 2 times y minus 2 which means we can write

it as y minus 2 squared. Then over here at the right-hand side negative 13 plus 9 plus

4 is zero. So we end up with zero on the right-hand side.

Now, you can see that our equation is at the same format as the formula. We have x minus

3 squared plus y minus 2 squared equals some constant. So now that we’ve got the same

format, we can make an evaluation about the constant p over here. So to make a determination

about whether this equation represents a single point or no point, what we need to look at

is the constant on the right-hand side. So you want to make sure you collected all your

constants on the right-hand side that we have and there are no loose constants here on the

left-hand side. So we’re just looking at zero. And when you look at zero (and we’ll

call it p), if p is greater than zero, then you’re dealing with a circle. If p is equal

to zero, then you’re dealing with a single point and if p is less than zero, you’re

dealing with no points. In this case, p is equal to zero. Our right-hand side is equal

to zero. So that means that the graph of this equation consists of one single point. The

graph is literally just one point. So that’s it. That’s how we can make that

determination. I hope this video helped you guys and I will see you in the next one. Bye!