Symmetric Graphing
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What is the graph of $x^2+y^2=|x|+|y|$.
I tried solving this but I have don't understand how we know that the graph is symmetric to the axes. I read the solutions to this Area enclosed by the curve $x^2+y^2=|x|+|y|$. btw
contest-math
asked Nov 18 at 3:53
user501887
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up vote
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favorite
What is the graph of $x^2+y^2=|x|+|y|$.
I tried solving this but I have don't understand how we know that the graph is symmetric to the axes. I read the solutions to this Area enclosed by the curve $x^2+y^2=|x|+|y|$. btw
contest-math
asked Nov 18 at 3:53
user501887
113
2
Write it as $(|x| - 1/2)^2 + (|y| - 1/2)^2 = 1/2$ (completing the square twice; this is also what you would do if the absolute value signs weren't there and then you'd just have a circle).
– Qiaochu Yuan
Nov 18 at 3:57
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the graph of $x^2+y^2=|x|+|y|$.
I tried solving this but I have don't understand how we know that the graph is symmetric to the axes. I read the solutions to this Area enclosed by the curve $x^2+y^2=|x|+|y|$. btw
contest-math
asked Nov 18 at 3:53
user501887
113
What is the graph of $x^2+y^2=|x|+|y|$.
I tried solving this but I have don't understand how we know that the graph is symmetric to the axes. I read the solutions to this Area enclosed by the curve $x^2+y^2=|x|+|y|$. btw
contest-math
contest-math
asked Nov 18 at 3:53
user501887
113
asked Nov 18 at 3:53
user501887
113
asked Nov 18 at 3:53
user501887
113
asked Nov 18 at 3:53
user501887
113
asked Nov 18 at 3:53
user501887
113
113
2
Write it as $(|x| - 1/2)^2 + (|y| - 1/2)^2 = 1/2$ (completing the square twice; this is also what you would do if the absolute value signs weren't there and then you'd just have a circle).
– Qiaochu Yuan
Nov 18 at 3:57
add a comment |
2
Write it as $(|x| - 1/2)^2 + (|y| - 1/2)^2 = 1/2$ (completing the square twice; this is also what you would do if the absolute value signs weren't there and then you'd just have a circle).
– Qiaochu Yuan
Nov 18 at 3:57
2
2
Write it as $(|x| - 1/2)^2 + (|y| - 1/2)^2 = 1/2$ (completing the square twice; this is also what you would do if the absolute value signs weren't there and then you'd just have a circle).
– Qiaochu Yuan
Nov 18 at 3:57
Write it as $(|x| - 1/2)^2 + (|y| - 1/2)^2 = 1/2$ (completing the square twice; this is also what you would do if the absolute value signs weren't there and then you'd just have a circle).
– Qiaochu Yuan
Nov 18 at 3:57
add a comment |
1 Answer
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Because of the squares and the absolute value signs, changing the sign on $x$ or $y$ will not change either side of the equation. That means that if you find one solution $(x,y)$ you know that $(-x,y), (x,-y)$ and $(-x,-y)$ are also solutions. You can find the solution in the first quadrant (for example) and reflect it in both axes to get the solution in the other quadrants.
answered Nov 18 at 4:07
Ross Millikan
288k23195365
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Because of the squares and the absolute value signs, changing the sign on $x$ or $y$ will not change either side of the equation. That means that if you find one solution $(x,y)$ you know that $(-x,y), (x,-y)$ and $(-x,-y)$ are also solutions. You can find the solution in the first quadrant (for example) and reflect it in both axes to get the solution in the other quadrants.
answered Nov 18 at 4:07
Ross Millikan
288k23195365
add a comment |
up vote
1
down vote
Because of the squares and the absolute value signs, changing the sign on $x$ or $y$ will not change either side of the equation. That means that if you find one solution $(x,y)$ you know that $(-x,y), (x,-y)$ and $(-x,-y)$ are also solutions. You can find the solution in the first quadrant (for example) and reflect it in both axes to get the solution in the other quadrants.
answered Nov 18 at 4:07
Ross Millikan
288k23195365
add a comment |
up vote
1
down vote
up vote
1
down vote
Because of the squares and the absolute value signs, changing the sign on $x$ or $y$ will not change either side of the equation. That means that if you find one solution $(x,y)$ you know that $(-x,y), (x,-y)$ and $(-x,-y)$ are also solutions. You can find the solution in the first quadrant (for example) and reflect it in both axes to get the solution in the other quadrants.
answered Nov 18 at 4:07
Ross Millikan
288k23195365
Because of the squares and the absolute value signs, changing the sign on $x$ or $y$ will not change either side of the equation. That means that if you find one solution $(x,y)$ you know that $(-x,y), (x,-y)$ and $(-x,-y)$ are also solutions. You can find the solution in the first quadrant (for example) and reflect it in both axes to get the solution in the other quadrants.
answered Nov 18 at 4:07
Ross Millikan
288k23195365
answered Nov 18 at 4:07
Ross Millikan
288k23195365
answered Nov 18 at 4:07
Ross Millikan
288k23195365
answered Nov 18 at 4:07
Ross Millikan
288k23195365
288k23195365
add a comment |
add a comment |
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Write it as $(|x| - 1/2)^2 + (|y| - 1/2)^2 = 1/2$ (completing the square twice; this is also what you would do if the absolute value signs weren't there and then you'd just have a circle).
– Qiaochu Yuan
Nov 18 at 3:57