Question 198906
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Given: 
{{{x-(1/8)y^2=0}}} ---> {{{x=(1/8)y^2}}} ---> {{{8x=y^2}}}
{{{y=sqrt(8x)}}}

This should be a half parabola that opens to the right.
*note: {{{x>=0}}}


Let us get the points.


When x = 0 
{{{y=0}}}


{{{x=0.50}}}
{{{y=sqrt(8*0.50)=sqrt(4)=2}}}


{{{x=1}}}
{{{y=sqrt(8*1)=sqrt(8)=2.828}}}


{{{x=2}}}
{{{y=sqrt(8*2)=sqrt(16)=4}}}


{{{x=3}}}
{{{y=sqrt(8*3)=sqrt(24)=4.9}}}


{{{x=4}}}
{{{y=sqrt(8*4)=sqrt(32)=5.66}}}


{{{x=5}}}
{{{y=sqrt(8*5)=sqrt(40)=6.32}}}


We see the points:

{{{drawing(300,300,-3,8,-3,8,grid(1),graph(300,300,-3,8,-3,8),blue(circle(0,0,.10)),blue(circle(0.50,2,.10)),blue(circle(1,2.828,.10)),blue(circle(2,4,.10)),blue(circle(3,4.9,.10)),blue(circle(4,5.66,.10)),blue(circle(5,6.32,.10)))}}}---->{{{drawing(300,300,-3,8,-3,8,grid(1),graph(300,300,-3,8,-3,8,sqrt(8x)),blue(circle(0.50,2,.10)),blue(circle(1,2.828,.10)),blue(circle(2,4,.10)),blue(circle(3,4.9,.10)),blue(circle(4,5.66,.10)),blue(circle(5,6.32,.10)),blue(circle(0,0,.10)))}}}



Thank you,
Jojo</font>