Respuesta :
We are asked to describe the variation [tex]\displaystyle{ \frac{y}{x^2}=5[/tex],
that is [tex]y=5x^2[/tex]
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
[tex]5(-6)^2\ \textgreater \ 5(-5)^2[/tex], so we can conclude that the greater the x, the smaller the y.
Case II, consider x= -1/2 < x=-1/3
[tex]5(-\frac{1}{2})^2= \frac{5}{4}\\\\ 5(-\frac{1}{3})^2= \frac{5}{9}\\\\\frac{5}{9}\ \textless \ \frac{5}{4} [/tex]
thus, the greater the x, the smaller the y.
Case III, consider the cases (examples) x= 5, and x= 6
[tex]5(6)^2\ \textgreater \ 5(5)^2[/tex], so we can conclude that the greater the x, the greater the y.
Case III, consider x= 1/3 < x=1/2
[tex]5(\frac{1}{2})^2= \frac{5}{4}\\\\ 5(\frac{1}{3})^2= \frac{5}{9}\\\\\frac{5}{9}\ \textless \ \frac{5}{4} [/tex]
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 [tex]y=5x^2[/tex]
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
[tex]5(-6)^2\ \textgreater \ 5(-5)^2[/tex], so we can conclude that the greater the x, the smaller the y.
Case II, consider x= -1/2 < x=-1/3
[tex]5(-\frac{1}{2})^2= \frac{5}{4}\\\\ 5(-\frac{1}{3})^2= \frac{5}{9}\\\\\frac{5}{9}\ \textless \ \frac{5}{4} [/tex]
thus, the greater the x, the smaller the y.
Case III, consider the cases (examples) x= 5, and x= 6
[tex]5(6)^2\ \textgreater \ 5(5)^2[/tex], so we can conclude that the greater the x, the greater the y.
Case III, consider x= 1/3 < x=1/2
[tex]5(\frac{1}{2})^2= \frac{5}{4}\\\\ 5(\frac{1}{3})^2= \frac{5}{9}\\\\\frac{5}{9}\ \textless \ \frac{5}{4} [/tex]
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.
