Calculus - help needed (1 Viewer)

[Petinga]

the question is sketch the graph of the following (y=x + 1 / x squared +2x) locating any staionary pts and asymptopes.

i found that there were no stationary pts and could not undersatnd how to grpah the solution in between the asymtopes as there were no stat pts.

Help needed ASAP if anyone is good at these type of questions

 

[acmilan]

do you mean y = x + (1/x² )+ 2x ?

 

[Slidey]

Think he means y=(x+1)/(x^2+2x), since y=3x+1/x^2 has a stat point but y=(x+1)/(x^2+2x) has none.

y=(x+1)/(x^2+2x)
Vertical asymptotes when x^2+2x=0, thus at x=0, -2

Stat pts when y'=0.
y'=[(x+1)(x^2+2x)^-1]'
y'=-2(x+1)^2/(x^2+2x)^2 + 1/(x^2+2x)
y'=(x^2+2x-2x^2-4x-2)/(x^2+2x)
y'=-2(x^2+x+1)/(x^2+2x)
Bottom never zero.
Discriminant of the top equals 1-4=-3, thus the top never equals zero.
Since y' never equals zero, no stationary points exist.

You say you had trouble graphing on the domain -2<=x<=0. This is because you have not accounted for the point of inflexion at (-1,0) (can be found by the second derivative).

Also, you should examine the behaviour of the curve as x approaches 0 from the negative side and 2 from the positive side. This will give you further clues.