May 5th, 2007, 02:24 PM  #1 
Newbie Joined: May 2007 Posts: 1 Thanks: 0  graphing help
how do I sketch the graph of (x4)2+(y+3)2=25 the italic 2's mean squared I'm completely retarded and don't know how to do superscript 2's. 
May 5th, 2007, 04:14 PM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
So you're trying to graph: (x4)^2+(y+3)^2=25 That's the graph of a circle of radius 5 centered at the point 4,3 If you're trying to graph it in rectangular coordinates, forget it. Go to polar coordinates. In polar coordinates, this is equal to the function r=5 . 
May 6th, 2007, 03:03 PM  #3 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
You can graph it in rectangular form, but only as two separate equations, which are somewhat messy. Following are the steps converting it into two functions of x. (x4)²+(y+3)²=25 (y+3)²=25(x4)² y+3=±√[25(x4)²] y=3±√[25(x4)²] You would input the + and  as two separate functions, i.e., y_1=3+√(25(x4)²) y_2=3√(25(x4)²) In polar coordinates, it is not quite r=5, because its center is not the origin. Unfortunately, I do not know how to perform translations in polar coordinates. A better approach might be a parametric, rather than polar, graph. We now want to express x and y in terms of a dummy variable t. To simplify this, we define x'=x4 and y'=y+3. We now have x'=5cos(t) and y'=5sin(t) as our equation. Now, because x=x'+4 and y=y'3 (by simple rearrangement), x=5cos(t)+4 and y=5sin(t)3, a much more pleasing system of equations than that required by writing y as a function of x. I have used several symbols here that cannot simply be typed (including the superscript 2's that you were commenting about earlier). To find out how I did this, see the sticky at the top of this forum.[/url] 
May 6th, 2007, 06:05 PM  #4  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
Hence, after multiplying everything out, canceling, and applying trig identities, the equation in polar coordinates is: r = 8cos(θ)  6sin(θ)  

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