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How Many Points?

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symmetry
 
Reply Mon 26 Sep, 2005 03:24 pm
When drawn on the same set of axes, the graphs of x^2 + y^2 = 16 and y = x^2 - 4 intersect at

a) 1 point

b) 2 points

c) 3 points

d) 4 points

and WHY???
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Type: Discussion • Score: 1 • Views: 1,513 • Replies: 27
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Milfmaster9
 
  1  
Reply Mon 26 Sep, 2005 04:40 pm
Well an x^2 + y^2 graph would show that its not a graph but a circle u have there....
0 Replies
 
patiodog
 
  1  
Reply Mon 26 Sep, 2005 05:26 pm
Nothing wrong with graphing a circle.

The easy way would be to graph the two equations and count the occasions of overlap. If you need to know where, change one of the equations so that it can be set equal to the other (setting both equal to y^2 would seem to be the easiest thing to do), solve for the remaining unknown (x), plug x back into the two equations, and figure out the coordinates of overlap.
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spendius
 
  1  
Reply Mon 26 Sep, 2005 05:29 pm
I'm sorry Madame.I don't do overlaps anymore.
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patiodog
 
  1  
Reply Mon 26 Sep, 2005 05:31 pm
Over dewlaps?
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Milfmaster9
 
  1  
Reply Mon 26 Sep, 2005 05:39 pm
well if they are both circles they can only overlap twice, expect if they only touch at a certain point...
I'll do it the long way despite the fact that there is many ways of doing this... x^2 + y^2 = 16 and y = x^2 - 4...

so if i slot in x^2 - 4 for y, i get...

x^2 + (x^2 - 4)^2 = 16
therefore x^2 + x^4 - 8x^2 + 16 = 16
cancelation leaves x^4 - 7x^2 = 0
so (x^2)(x^2 - 7) = 0
they intersect where x = 0 and x = (square root) of 7
therefore they only intersect twice...

u seem to need alot of help msn me... [email protected]
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patiodog
 
  1  
Reply Mon 26 Sep, 2005 05:45 pm
They're not both circles. One is a concave parabola. And what if x=0?
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Milfmaster9
 
  1  
Reply Mon 26 Sep, 2005 05:53 pm
well i never solved for y because if there is two x solutions then they must cross there.. well i believe they intersect at the y axis and where x is root 7...
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patiodog
 
  1  
Reply Mon 26 Sep, 2005 06:15 pm
7 has positive and negative square roots.
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Milfmaster9
 
  1  
Reply Mon 26 Sep, 2005 06:21 pm
well we arent dealing with imaginary numbers... **** your right, i blame the lack of mathematical keys on the keyboard, well that makes 3 a possiblity... well im most likely wrong anyways... my graph plotter refuses to plot this graph...
0 Replies
 
patiodog
 
  1  
Reply Mon 26 Sep, 2005 06:24 pm
I just drew it by hand. Waste of pen and paper, I suppose, and certainly of computing power...
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Milfmaster9
 
  1  
Reply Mon 26 Sep, 2005 06:27 pm
what it look like, well the x^2 + y^2 = 16 is definatly a circle... how many intersections and was i wrong bout the x co-ordinates... i did it on toilet paper and ive used it since!
0 Replies
 
patiodog
 
  1  
Reply Mon 26 Sep, 2005 06:32 pm
Better to use toilet paper as scratch paper than vice versa...
0 Replies
 
Milfmaster9
 
  1  
Reply Mon 26 Sep, 2005 06:36 pm
well im concerned was i right??? drawing graphs usually solves all, literally..
0 Replies
 
symmetry
 
  1  
Reply Mon 26 Sep, 2005 07:14 pm
ok
WOW! My question started an argument here. LOL. I want to thank all the great replies.
0 Replies
 
patiodog
 
  1  
Reply Mon 26 Sep, 2005 07:41 pm
you've got it right.

unless i'm wrong.

in which case, we're wrong together, and the right can take a flying hike, or somesuch...
0 Replies
 
g day
 
  1  
Reply Tue 27 Sep, 2005 03:54 am
So its (0, -4), (-root 7, 3) and (+ root 7,3) - the why part gets me though, why - cause it does! A second and fourth order function can overlap in at most 4 spots I guess.
0 Replies
 
Milfmaster9
 
  1  
Reply Tue 27 Sep, 2005 12:48 pm
yes id agree there, it means that the function enters the circle once and must touch the edge once.. x^2 + y^2 = R^2 is the formula of a circle (a special case), therefore one function is a circle...
0 Replies
 
el pohl
 
  1  
Reply Sat 1 Oct, 2005 09:03 am
Haha, thank god for TI graph calculators.

Yes, the first equation is a circle of radius = 4 if I remember my math correctly. The other one has a displacement of 4... they might not even overlap... but since 0 its not one of the options, the answer is 2! Tables work.
0 Replies
 
markr
 
  1  
Reply Sat 1 Oct, 2005 09:40 am
A: x^2 + y^2 = 16
B: y = x^2 - 4
C: x^2 = y + 4 (from B)
Sub C into A to get:
y^2 + y - 12 = 0
(y+4)*(y-3) = 0
y = -4, y = 3
Plug -4 into A or B to get x^2 = 0
Therefore, (0,-4) is a solution.
Plug 3 into A or B to get x^2 = 7
Therefore (+/-sqrt(7),3) are solutions.
0 Replies
 
 

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