Expert: Sherry Wallin - 9/21/2009

Question
QUESTION: I need help with the STEPS (preferrably very detailed steps) on graphing the following equation (and yes, this is all the information my Prof. gave us for the question: "Graph:" and then this equation):

x2 (x to the 2nd power)+20-y2 (y to the 2nd power)-7x-2y+21-8x+ 6y-2 = 2-4y-5x-2y2 (2y to the 2nd power)

So far I have the following, and I'd like to know if I'm correct:

x2-7x-8x -y2-2y+6y +20+21-2 = 2-4y-2y2-5x

This math class is called "Mathematics - A Modeling Approach" and it's about using math in the real world and is not about any particular kind of math - it jumps around from Algebra II to Calculus to Trigonometry, to I don't know what. And I'm having trouble trying to figure out (from teacher's notes handed out or from my own notes taken in class) what in the world formula I'm supposed to use, and what kind of graph I'm supposed to be making.  Thank you SO much for ANY kind of light you can shed, and any way you can help me with the formula so I can "graph" it.

ANSWER: Hi Lorri~
    The first thing you need to notice is that there are only two variables and they have just the first and second powers on them. What geometric shapes have you studied that have first and second degree variables and more importantly 2 different variables? Hopefully you think of conic sections:
circles, ellipses, hyperbolas, or parabolas. Next you need to simplify what you have by combining all your like terms to see what 'form' you have:

x^2 + 20 - y^2 -7x -2y + 21 -8x + 6y -2 = 2 -4y-5x-2y^2

Sometimes it is easier to re arrange things so make it easier to see things but first I am going to get everybody on one side of the equation;
x^2 + 20 - y^2 -7x -2y + 21 -8x + 6y -2 -2 + 4y + 5x + 2y^2 = 0

Now get all x's together and all y's together and all constants together. Notice I am not doing any more math until I organize what I have:

x^2 -7x -8x + 5x - y^2 -2y + 6y + 4y + 2y^2 + 20 + 21 -2 -2 = 0

Count the number of terms in the original and see if you have that many now (kind of a double check)
I count 13 in both so we are probably good to proceed:

x^2 -10x + y^2 + 8y + 37 = 0

With 2nd degree equations and two variables and conics you generally need to complete the square.

x^2 -10x + (-5)^2 - (-5)^2 + y^2 + 8y + (4)^2 - (4)^2 + 37 = 0

I've just added 25 and subtracted 25 and added and subtracted 16 on the left.

Re write your 'perfect' squares: (x-5)^2 + (y-4)^2 - 25 - 16 + 37 = 0

-> (x-5)^2 + (y-4)^2 - 4 = 0
-> (x-5)^2 + (y-4)^2 = 4
-> (x-5)^2 + (y-4)^2 = 2^2

This is an equation of a circle centered at (h,k) with a radius of r taken from the standard formula of a circle: (x-h)^2 + (y-k)^2 = r^2


Therefore the center of the circle is (5,4) and the radius is 2.

Math Prof


---------- FOLLOW-UP ----------

QUESTION: Hi again ... I have a question in the 2nd step of this ... when you say to get all the Xs together, all the Ys together and all the constants together. RE: the constants 2: I see that there is a -2 on the left side of the equation. The right side of the equation begins with 2. Why did you make that 2 on the right side of the equation a -2 when putting it together with the other constants? It can't be just because it was at the beginning, right? Or otherwise the x^2 at the beginning of the of the left side of the equation would also become a negative (and it does not appear that it ever does), and also, I thought that when it does not specificy a "+" or a "-" at the beginning, that it is a "+". I am confused about when to change something from a negative or a positive, and vice-versa. Oh, and I guess I do have another question: what is the procedure (formula?) for cancelling out a negative/positive, and making it a positive/negative (and vice-versa). I know they do that, just to make it "just like" another number. (e.g., making a -2 a +2 by making something else a negative?) Thank you SO much ... and I may have another question or 2 as I'm working my way through this problem, trying to comprehend all the steps in a way that makes sense to me!...

Answer
Any time a number (term) is moved from one side of an equation to the other it changes the sign. For example if you have x + 2 = 5 and you want to move the 5 to the left, then to make it go away from the right hand side and appear on the left hand side you need to subtract it (it's the opposite operation). Notice though that if it is 3x -2 = -4 you would need to add 4 or subtract -4 which is equivalent to -(-4) = 4. Two minus signs next to each other become a positive.

You said: Why did you make that 2 on the right side of the equation a -2 when putting it together with the other constants?

Answer: you had a positive 2 on the right hand side (rhs) so you subtract it from the rhs or merely change the sign and put it on the other side. So it's sign was negative so call it positive and put it on the lhs (left hand side). The key here is to use the opposite operation. If something is being added or subtracted subtract or add. If something is being multiplied or divided, divide or multiply. Note when working with equations and multiplying a negative number you will need to divide by a negative and vice versa.

You said: Or otherwise the x^2 at the beginning of the of the left side of the equation would also become a negative

answer: the x^2 was already on the lhs so it didn't need to be moved in the gathering of all the x's and y's .

Math Prof