I asked a similar question earlier. The intercepts both are the origin (0, 0). The graph is a 'V' shape where the lines are in the first and second quadrants, meeting at origin which is the vertex of the 'V'. The lines are mirror images of each other. One line represents y = -x and the other represents y = x
2007-06-07 01:56:07
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answer #1
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answered by Akilesh - Internet Undertaker 7
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That's not a determinant symbol. That's absolute value, where for any value of x, y is the positive form of x. To determine the x and y intercepts, just substitute 0 in for one, which will give you 0 for the other. So the x and y intercepts are both 0 b/c the point where this graph crosses either axis is at the origin (0,0).
2007-06-07 01:38:54
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answer #2
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answered by whatcanmaxdo4u?everythingupscant 3
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determinatns? I assume | | is the absolute value operator, so that |x| = x if x >= 0, = -x if x<0.
so the x-intercept and the y-intercept are both 0.
2007-06-07 01:39:11
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answer #3
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answered by holdm 7
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The origin, (0, 0), is the intercept, for both axes.
For positive x --> y = x, a ray from the origin with slope 1
For negative x -- > y = -x, a ray to the origin with slope -1.
2007-06-07 01:36:42
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answer #4
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answered by jcsuperstar714 4
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We've y = |x|:
== For x intercept(s) put y = 0 above ==> |x| = 0 ==> the only x-intercept is (0,0)
== For y-intercept put x = 0 above ==> y = 0 ==> we get the same point as above (0,0).
Regards
Tonio
2007-06-07 01:34:43
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answer #5
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answered by Bertrando 4
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???
'Determinants'?? That looks like y equals absolute value of x to me ☺ And the x and y intercepts are both at the origin.
Doug
2007-06-07 01:38:22
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answer #6
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answered by doug_donaghue 7
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x ...-2 -1 0 1 2...
y ...2 1 0 1 2...
the graph is a symmetrical V-shape, found in the positive y axis whose vertex is at the origin
2007-06-07 01:40:09
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answer #7
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answered by regreg 3
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y=x or y=-x
2007-06-07 04:08:35
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answer #8
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answered by giuggiolina90 2
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