need help with this one.

draw the curve y^2=2x-1

y=(2x-1)^0.5 x > 1/2

?

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- January 26th 2011, 08:45 AMpaulaadraw curve
need help with this one.

draw the curve y^2=2x-1

y=(2x-1)^0.5 x > 1/2

? - January 26th 2011, 09:13 AMyeKciM
- January 26th 2011, 09:16 AMpaulaa
do you mean like this:

y x

0 0.5

1 1

2 2.5

and then ? - January 26th 2011, 09:24 AMyeKciM
for x = 0.5 you have y = 0

for x = 1 you have y = (2*1-1)^(1/2) = 1 and also y = -1 :D

so you have 3 points and now just connect them :D your curve goes from x = 0.5 to infinity or less :P so it's moving to the "right" above and belove x axis :D - January 26th 2011, 09:45 AMpaulaa
how do you know that it is to the right ?

- January 26th 2011, 10:11 AMyeKciM
how in detail you need your answer ? Because i don't know for which level do I approach because this is simple curve (parabola) and some curve's like this you get to know .... y = x^2 , -y=x^2 , x = y^2 , -x = y^2 ... and all can be done by 3 points :D or more if you like and connecting them... but if you need to do , region for which function is defined , add or even or neither, sing , asymptote's , first and second derivate... we will do it different :D :D

- January 26th 2011, 10:13 AMpaulaa
thz for the answers :)

- January 26th 2011, 10:21 AMyeKciM
you should know also that y = x^2 and y = x^2 -1 or y= x^2 +1 are the same parabola just shifted up or down, left or right depending on which do you have ... y = x^2 have zero in point (x=0,y=0) , and y = x^2 +1 looks the same but have zero in (x=0, y = 1) so it's just shifted upwards by the y axis :D play little and you will have no problem with this type of curves :D

- January 26th 2011, 10:22 AMpaulaa
yes i did know that :P but thanks :)