Question 82935
I did 
  x^2-1=-1  <=== Good. You did this by setting Y1 equal to Y2
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x^2-1=1=0 <=== A little confusing here. What you need to do is to get the x^2 term by itself 
on the left side. To do that you eliminate the -1 on the left side by adding +1 to both sides.
When you do that you get the next equation that you got. 
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x^2=0  <=== This is correct.
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x=0   <=== This is also correct. It tells you that when x equals zero, the two equations
you originally were given have equal values of y.  Return to the original equations 
and substitute zero for x and you get:
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Y1 = x^2 - 1
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Then substituting zero for x results in:
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Y1 = 0 -1 = -1
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The second equation that you were given was:
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Y2 = -1
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Since this equation does not contain an x term, Y2 is -1 no matter what value is assigned
to x.
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From the first equation you know that (0, -1) is the point that has a Y value of -1. And
from the second equation you know that (0, -1) is also a point that is common with the
solution of the first equation.
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What this all means is that (0, -1) is the point that these two equations have in common.
And what is going on is that the graph of the first equation is a parabola that as you move
to the right drops down to its minimum value of -1 [at the point (0, -1)] and then as
you continue to move to the right rises. The Y axis is the line about which this parabola
is symmetrically centered.
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In the meantime the line Y2 = -1 is a horizontal line that is tangent to the bottom of
the parabola of the first equation at the point (0, -1). So there is only one point of
intersection ... and that is the point (0, -1).  The coordinate system below shows the
graphs you should have:
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{{{graph(300,300, -10,10,-10,10, x^2-1,-1)}}}
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Hope this helps to clarify the points you were looking for.
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