2 [ 2 + 2 ( 8 ⋅ 6 − 17 ) ]
1 answer:
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We are asked to describe the variation 
,
that is
for x≠0, in which case x=y=0, we distinguish the following intervals:
I = (-infinity, -1]
II = (-1, 0)
III = (0, 1)
III = [1, infinity)
Case I, consider the cases (examples) x= -5, and x= -6
, so we can conclude that the greater the x, the smaller the y.
Case II, consider x= -1/2 < x=-1/3
thus, the greater the x, the smaller the y.
Case III, consider the cases (examples) x= 5, and x= 6
, so we can conclude that the greater the x, the greater the y.
Case III, consider x= 1/3 < x=1/2
thus, the greater the x, the greater the y.
Conclusion:
y=0 only if x is 0. Otherwise y>0 always:
for x<0, x and y are inversely proportional
for x>0, x and y are directly proportional.
Remark, an other possible consideration in cases could be x<0 and x>0 only.
Remark, another approach is considering the graph of y=5x^2, which is a parabola, with the vertex at (0, 0) opening upwards.
From the graph we can find the same conclusion. That is we can see that the farer we get away from the y-axis, in both direction, the larger the values of y become.
that is
for x≠0, in which case x=y=0, we distinguish the following intervals:
I = (-infinity, -1]
II = (-1, 0)
III = (0, 1)
III = [1, infinity)
Case I, consider the cases (examples) x= -5, and x= -6
Case II, consider x= -1/2 < x=-1/3
thus, the greater the x, the smaller the y.
Case III, consider the cases (examples) x= 5, and x= 6
Case III, consider x= 1/3 < x=1/2
thus, the greater the x, the greater the y.
Conclusion:
y=0 only if x is 0. Otherwise y>0 always:
for x<0, x and y are inversely proportional
for x>0, x and y are directly proportional.
Remark, an other possible consideration in cases could be x<0 and x>0 only.
Remark, another approach is considering the graph of y=5x^2, which is a parabola, with the vertex at (0, 0) opening upwards.
From the graph we can find the same conclusion. That is we can see that the farer we get away from the y-axis, in both direction, the larger the values of y become.