Graphing y^2=f(x) help (2 Viewers)

[blackops23]

Hi, this is a question from Terry Lee.

There's five parts, I couldn't do the last part.
Here's the question:

Given the curve y=(x^2)/(1-x^2)
a) Determine whether the function is odd or even.
IT'S EVEN

b) Discuss the behaviour of the curve for large values of x.
APPROACHES y=-1, y=-1 is a horizontal asymptote

c) Find y' and coordinates of any turning points.
y' = (2x)/(1-x^2), turning point (0,0) --> MINIMA

d) Hence sketch the curve,
ASYMPTOTES, y=-1, x=-1, x=1
-1<(x)<1, y is >0, parabolic shape with minima at (0,0)
x>1, graph begins in negative infinity, and then curves to y=-1
x<1, same as above but other side of y-axis.

e) Without doing any further calculation, sketch the curve y^2 = (x^2)/(1-x^2)

It was part (e) that I could not do at all
The main trouble was the "without any further calculation".
I just don't see how to sketch y^2=f(x), when you have y=f(x) given to you, and you can't do any calculations. Any help would be absolutely fantastic.

Thanks guys.</x>

 

[deterministic]

To sketch y^2 = f(x) based on your sketch of f(x) consider:
- there will be 2 branches: y=sqrt(f(x)) and y=-sqrt(f(x)). Consider y=sqrt(f(x)) for now as the negative is merely a reflection about the x axis
- if f(x)-> 0 then sqrt(f(x))-> 0 and thus y=0 so zeros are the same
- vertical asymptotes of y=f(x) are vertical asymptotes of y^2
- we note that as x-> infinity, f(x)-> a where a is the horizontal asymptote, then y=sqrt(f(x))-> sqrt(a) (so square root horizontal asymptope for new asymptote)
- domain of x is such that f(x) >= 0 (obviously), so only include parts of graph where f(x) is positive
- if 0=<f(x)<1, then sqrt(f(x))>f(x) so y>f(x), vice versa for case where f(x)>1 (equality at f(x)=1)
- consider y=sqrt(f(x)), so y'= f'(x)/(2sqrt(f(x))), so when f'(x)=0, then y'=0, so f(x) and y share the same x value of turning point (y values obviously differ)

REMEMBER TO REFLECT POSITIVE CASE ABOUT X-AXIS FOR NEGATIVE CASE

 

[jyu]

Also consider the gradient at x=0, no longer a turning point.

 

[deterministic]

Just a correction on my last dot point:

- consider y=sqrt(f(x)), so y'= f'(x)/(2sqrt(f(x))), so when f'(x)=0 AND f(x)=/= 0 , then y'=0, so f(x) and y share the same x value of turning point (y values obviously differ)

 

[blackops23]

you know when it says without further calculation, exactly how much calculation can you do? - with relation to y' of course. Also i noticed, at x=0, y' becomes indeterminable, because y'=0/0 which is practically meaning less....

Also regarding:

To sketch y^2 = f(x) based on your sketch of f(x) consider:
- if f(x)-> 0 then sqrt(f(x))-> 0 and thus y=0 so zeros are the same
- we note that as x-> infinity, f(x)-> a where a is the horizontal asymptote, then y=sqrt(f(x))-> sqrt(a) (so square root horizontal asymptope for new asymptote)
- consider y=sqrt(f(x)), so y'= f'(x)/(2sqrt(f(x))), so when f'(x)=0 AND f(x)=/= 0 , then y'=0, so f(x) and y share the same x value of turning point (y values obviously differ)

Are these hints just for graphs in general or for this specific graph, because i don't get the first point.

-2nd point: domain is -1<(x)<1 so I guess the horizontal asymptote of y=-1 goes away.

-3rd point: LOL sorry don't understand... please explain more... f'(x)=0 at x=0, which makes f(x)=0, therefore y'=0/0 which is indeterminable....
-------------------------------------------------
So this is all I got for y^2=f(x)
1.Asymptotes: x=1, x=-1
2.passes through (0,0) probably intersection point of lines because gradient is indeterminable at that point
3.Domain: -1<(x)<1
4. sqrt[f(x)] > f(x) in the range 0<y<1

Anything else I could do for this graph,

thanks

 

[deterministic]

These are tips for graph of y^2=f(x) for any f(x)
-If f(x)=0, then y^2=f(x)=0 then hence y=0. So the zeros of f(x) are the zeros of the new graph.
- 3rd point refer to my correction. y'=f'(x)/(2sqrt(f(x)) (what you get when you differentiate y = sqrt(f(x))) so if f'(x)=0 and f(x)=0 y'= 0/ something not 0, then y'=0 indicating a turning point. However if f(x)=0 as well, the y'=0/0 which is indeterminable.

Everything you have so far is correct, just remember that the graphs f(x) and y^2=f(x) must cross each other somewhere above y axis, and that y=sqrt(f(x)) goes to infinity slower than f(x) (so sqrt(f(x)) should be below f(x) close to the vertical asymptotes. The graph should turn out to be an hourglass with heads cut off or something like that.
Just remember all you are doing is taking each y value in the old graph and applying a sqrt function to it and also a negative square root function as well.