finding equation of function from given figure
According to me the right answer is option 2 as it satisfies the point (2,-1) also. But answer says it's option 1 which is the answer. Iam not convinced with the answer. Is my answer correct?
graphing-functions
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
asked Feb 4 '17 at 11:40
shadow kh
541111
add a comment |
According to me the right answer is option 2 as it satisfies the point (2,-1) also. But answer says it's option 1 which is the answer. Iam not convinced with the answer. Is my answer correct?
graphing-functions
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
asked Feb 4 '17 at 11:40
shadow kh
541111
1
Your answer is correct, and is the only correct one among the three options.
– dxiv
Feb 4 '17 at 23:36
1
@dxiv Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems to me like your comment answers the question pretty succinctly.
– Robert Howard
Nov 24 at 21:16
1
@RobertHoward Feel free to convert my comment into an answer, that would be entirely fair play here.
– dxiv
Nov 25 at 5:01
add a comment |
According to me the right answer is option 2 as it satisfies the point (2,-1) also. But answer says it's option 1 which is the answer. Iam not convinced with the answer. Is my answer correct?
graphing-functions
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
asked Feb 4 '17 at 11:40
shadow kh
541111
According to me the right answer is option 2 as it satisfies the point (2,-1) also. But answer says it's option 1 which is the answer. Iam not convinced with the answer. Is my answer correct?
graphing-functions
graphing-functions
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
asked Feb 4 '17 at 11:40
shadow kh
541111
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
asked Feb 4 '17 at 11:40
shadow kh
541111
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
edited Feb 6 '17 at 12:42
Harry Peter
5,43111439
5,43111439
asked Feb 4 '17 at 11:40
shadow kh
541111
asked Feb 4 '17 at 11:40
shadow kh
541111
asked Feb 4 '17 at 11:40
shadow kh
541111
541111
1
Your answer is correct, and is the only correct one among the three options.
– dxiv
Feb 4 '17 at 23:36
1
@dxiv Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems to me like your comment answers the question pretty succinctly.
– Robert Howard
Nov 24 at 21:16
1
@RobertHoward Feel free to convert my comment into an answer, that would be entirely fair play here.
– dxiv
Nov 25 at 5:01
add a comment |
1
Your answer is correct, and is the only correct one among the three options.
– dxiv
Feb 4 '17 at 23:36
1
@dxiv Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems to me like your comment answers the question pretty succinctly.
– Robert Howard
Nov 24 at 21:16
1
@RobertHoward Feel free to convert my comment into an answer, that would be entirely fair play here.
– dxiv
Nov 25 at 5:01
1
1
Your answer is correct, and is the only correct one among the three options.
– dxiv
Feb 4 '17 at 23:36
Your answer is correct, and is the only correct one among the three options.
– dxiv
Feb 4 '17 at 23:36
1
1
@dxiv Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems to me like your comment answers the question pretty succinctly.
– Robert Howard
Nov 24 at 21:16
@dxiv Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems to me like your comment answers the question pretty succinctly.
– Robert Howard
Nov 24 at 21:16
1
1
@RobertHoward Feel free to convert my comment into an answer, that would be entirely fair play here.
– dxiv
Nov 25 at 5:01
@RobertHoward Feel free to convert my comment into an answer, that would be entirely fair play here.
– dxiv
Nov 25 at 5:01
add a comment |
1 Answer
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As dxiv pointed out, your answer is correct. We can verify that by plugging $y=-1$ into the three equations given to see which equation yields $x=2$:
$$begin{align}
y-|y|&=-1-|-1|=-1-1=-2 :::large{times}\[1ex]
-(y-|y|)&=-(-1-1)=-(-2)=2 qquadcheckmark \[1ex]
y+|y|&=-1+1=0 hspace{3.4cm}largetimes
end{align}$$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
As dxiv pointed out, your answer is correct. We can verify that by plugging $y=-1$ into the three equations given to see which equation yields $x=2$:
$$begin{align}
y-|y|&=-1-|-1|=-1-1=-2 :::large{times}\[1ex]
-(y-|y|)&=-(-1-1)=-(-2)=2 qquadcheckmark \[1ex]
y+|y|&=-1+1=0 hspace{3.4cm}largetimes
end{align}$$
add a comment |
As dxiv pointed out, your answer is correct. We can verify that by plugging $y=-1$ into the three equations given to see which equation yields $x=2$:
$$begin{align}
y-|y|&=-1-|-1|=-1-1=-2 :::large{times}\[1ex]
-(y-|y|)&=-(-1-1)=-(-2)=2 qquadcheckmark \[1ex]
y+|y|&=-1+1=0 hspace{3.4cm}largetimes
end{align}$$
add a comment |
As dxiv pointed out, your answer is correct. We can verify that by plugging $y=-1$ into the three equations given to see which equation yields $x=2$:
$$begin{align}
y-|y|&=-1-|-1|=-1-1=-2 :::large{times}\[1ex]
-(y-|y|)&=-(-1-1)=-(-2)=2 qquadcheckmark \[1ex]
y+|y|&=-1+1=0 hspace{3.4cm}largetimes
end{align}$$
As dxiv pointed out, your answer is correct. We can verify that by plugging $y=-1$ into the three equations given to see which equation yields $x=2$:
$$begin{align}
y-|y|&=-1-|-1|=-1-1=-2 :::large{times}\[1ex]
-(y-|y|)&=-(-1-1)=-(-2)=2 qquadcheckmark \[1ex]
y+|y|&=-1+1=0 hspace{3.4cm}largetimes
end{align}$$
answered Nov 25 at 20:07
community wiki
Robert Howard
add a comment |
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1
Your answer is correct, and is the only correct one among the three options.
– dxiv
Feb 4 '17 at 23:36
1
@dxiv Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems to me like your comment answers the question pretty succinctly.
– Robert Howard
Nov 24 at 21:16
1
@RobertHoward Feel free to convert my comment into an answer, that would be entirely fair play here.
– dxiv
Nov 25 at 5:01