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AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Are you familiar with what sort of shape you'll get from an equation of this form?
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
Not this one
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Okay. Equations of the form \((x  h)^2 + (y  k )^2 = r^2\) are considered circles. The center of the circle is (h, k), and the radius is r. So, you can usually make comparisons between the 'standard form'and the given equation to figure out your center + radius, and graph from that. :)
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
will the center b (0,3)?
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Yes, that is correct.
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
So then this will be a circle intersecting at ±3 all round
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Hmm... well, the circle simply has a radius of three, so basically all the points that are a distance of 3 from (0, 3) are points on your graph. It makes sense if you consider r the distance between the center and all the points on the circle that are 3 units away. :)
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
dw:1352106893652:dw
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
The exact center of the circle will be at (0, 3), like this: dw:1352107269477:dw
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
Since the radius is 9 and the √9=3
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Our radius is determined as r^2 = 9 => r = 3. :) So, we can draw in all the points around (0, 3) that are a distance of 3 units away for the circle.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
dw:1352107502707:dw So, we can pick out a few simple ones that line up on the axes, and then we have to estimate from there. My graph isn't drawn particularly well, so it doesn't look very nice, but you should be able to get a circle. :P
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
So the points are (0,0) (3,3) (0,6) (3,3)?
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
Yes, those are a few points on the graph of the circle. If you start from your center, it is easiest to move horizontally or vertically by the same distance as the radius to find four points to draw the rest of the graph from.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
The process of graphing the circle basically goes like this: 1) Find your centerpoint and radius 2) Plot the centerpoint on your graph 3) Plot those four points at a distance of one radius away and draw in the circle from there as best you can. Usually teachers will be forgiving as long as you have those four points marked and your circle looks somewhere close. :P
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
You've been so much helpful
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
I have about 6 more qtns to go. I'm staying awake till class time which is 7am
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
I'm glad to be able to help! :) Although, I will have to get off. I should really be sleeping...
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
Sure. Thank a million.
 one year ago

odehye Group TitleBest ResponseYou've already chosen the best response.0
Good night and sound sleep.
 one year ago
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