Info Request

1) y= x^2/(x^2+3)

Vertical asymptote: None since x^2+3≠0 for any x

Horizontal asymptote: y approache 1 as x approaches plus and minus infinity so y=1

Slant:None

y-intercept: when x=0, y=0 so 0

x-intercept: when y=0, x^2=0 so x=0 so 0


y= x^2/x^2+3 = (x^2+3-3)/(x^2+3) = 1 - 3/(x^2+3)

dy/dx = 6x/(x^2+3)^2

max/min at x=0

y=0 at x=0 and the graph approaches 1 at plus and minus infinity hence this must be a minimum

	View Full Image

--------------------------------------------------------------------------------------

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

Vertical asymptote: x^2-1=0 when x= 1 and x=-1

Horizontal asymptote: y approaches zero as x approaches plus or minus infinity so y=0

Slant:None

y-intercept: when x=0, y=0 so 0

x-intercept: when y=0, x^2=0 so x=0 so 0


dy/dx = 2/(x^2-1) -4x^2/(x^2-1)^2

= ( 2(x^2-1) -4x^2 ) / (x^2-1)^2

=-2(x^2+1)/ (x^2-1)^2

Hence no min/max since no value of x makes x^2+1=0

	View Full Image

-----------------------------------------------------------------

3) G(x)= x + 4/(x^2+1)

Vertical asymptote: None since x^2+1 is never zero

Horizontal asymptote: None

Slant: y approaches x as x approaches plus or minus infinity so y=x

y-intercept: when x=0, y=4 so 4

x-intercept: when y=0, g(x) =0

g(x) = x + 4/(x^2+1)

=( x(x^2+1)+4 )/(x^2+1)

= (x^3+x+4)/ (x^2+1)

so g(0) = 0 when x^3+x+4=0 (assuming x^2+1≠0 for the given x)

An x intercept lies between x=-1 and x=-2


I've either misunderstoof the function you intended or this question has got nasty. I think at this point I'll give you what I've done (no need to accept this answer as it is in complete). All the best,

Ben


Edited by Ben Brown on 2/5/2009 at 10:00 AM