Graphing equations Graph the following equations.
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Question

Graphing equations Graph the following equations.

c. x² + 2x + y² + 4y + 1 = 0

Accepted Answer

Welcome back, everyone. In this problem, we want to sketch the graph of the equation X2 minus 6 X + Y2 + 4Y + 9 equals 0. Here we have possible choices or possible results for our graph in answer choices A. B See, OK. And D. Now, let's go back to our problem statement, OK? And let's make sure we understand exactly what we want here, OK? But we're going to sketch the graph of our equation, and we know that an equation like this gives us a circuit. So we can complete the square for our equation, and that is going to help us figure out our center and radius for our circle. So let's go ahead and do that. Now the coefficient of X2 and the coefficient of Y2 are both positive and 1. So that's how we know the graph of the function will be a circle. So taking our equation and completing the square, OK. Then for our X term, OK, if we, here we have X2 minus 6 X + Y2 + 4 Y + 9 equals 0. And if we complete the square 4 or X terms, OK, that's going to be X2 minus 2 X multiplied by 3, OK, plus 3 squared. Likewise, for our Y terms, OK, Y square + 4 Y. That's going to be + Y2 + 2 Y multiplied by 2 + 22. OK. Remember we were already adding 9 and know that we've introduced 3 squared and 2 squared into our equation, we'll have to subtract those terms. So that's minus 32 minus 2 squad from our equation. So, all of this equals 0, OK? Now, here, if we do go ahead and complete the square for our X term it's going to be X minus 3 all squared, OK? And for the Y terms it's going to be Y + 2 all squared. And no, we're going to have 9 minus 9 minus 4, OK, which is going to leave us with negative 4. So all of that equals 0. So we can rewrite this then as X minus 32 + Y + 22 equal to 4, and then we can rewrite 4 as 22, OK? For us to get it in the general form of a circle. No, when we have it written like this, OK, this tells us then that the center of our circle is going to be equal to 32 based on the constant terms here for our X and Y values or X and Y expressions respectively, and then our radius is going to be equal to the square root of 2 squad or just 2. So that means for our equation, its center is 32 and its radius is 2. So if I were to do that sketch, OK. Then if these are the X and Y axis, OK. Then our center will be at 3-2, probably somewhere down here. And our circle then is going to have a radius of 2. So that means our circle is probably just going to touch the tip of the X axis, OK? Because it's Y coordinate, it's the Y coordinate of the center is negative 2, OK, and the radius of the circle is too. So, basically, this is what we expect our circuit to look like. Let's go ahead and look at our answer choices and see which one satisfies those conditions. Now if we look at our first circle for answer twice A, notice the center is at 32, so we know that can't be the correct answer. If we go down to answer twice B, again, the center is 32, so that's that can't be right. If we go to answer choice C, OK, well, no, we noticed that the center is 3-2, but our radius is not 2 units because notice here it goes from -2, sorry, the when it's completely vertical, sorry, it goes from -2 to 1, OK? And this shows us that the radius would be 3, not 2. When we go to answer choice D, notice that the center is at 32 and the radius is 2 units. Therefore, that tells us that D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Graphing equations Graph the following equations.

c. x² + 2x + y² + 4y + 1 = 0

Key Concepts

Conic Sections

Completing the Square

Graphing Techniques

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