Posted by Vickie 23) x^2+y^2-2x+4y-4=0 needs to be put in standard form: (x-h)^2+(y-k)^2=r^2, we do that by completeing the square:
on 10/17/2006, 10:10 pm, in reply to "MTH 163 04H"
68.96.0.104
I appologize for not getting back to you in time for your test. I still don't get the allerts when people post questions. Please email me when you have a question.
21)x^2+(y-1)^2=4
This is in stndard form: (x-h)^2+(y-k)^2=r^2
There is no h, so h=0, k=1, and r=sqrt(4)=2
The center is (0,1) and radius is 2. I can't show you the graph here, but plot the center (0,1), count 2 units up, down, right, and left plot those points and connect the dots. If you have a TI-83/84 calculator hit (zoom)(5)zoom square (2nd)(draw)(9)circle(0,1,2)(enter) and you'll see the graph of the circle. Notice the order:circle(h,k,r).
(x^2-2x+___)+(y^2+4y+___)=4+___+____
The first blank on both sides we fill with (-2/2)^2=(1)^2=1
(x^2-2x+1)+(y^2+4y+____)=4+1+____
The second blank we fill with (4/2)^2=(2)^2=4
(x^2-2x+1)+(y^2+4y+4)=4+1+4
(x-1)^2+(y+2)^2=9
h=1, k=-2, r=sqrt(9)=3
The center is (1,-2) and radius =3
Plot the center (1,-2) count 3 units up, down, right, and left and plot those points and connect the dots to form your circle.
Follow the TI-83/84 direction above with the exception: circle(1,-2,3)
Happy Calculating!!!
I'm sorry I didn't get back to you in time, hopefully it didn't hamper your test. I will not be depending on notification by this site. I will check it whenever I'm at work from now on.
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