Curve-sketching problem - stuck on the asymptotes

[mmc5311]

1. Sketch the graph of y = 2x + (1/x) where x > 0, indicating extreme points, intercepts, asymptotes, etc.

I've already managed to find most of the information.

y' = -x^-2 + 2 (to get rid of the negative, I multiplied x^2 through and came up with y = 2x^2 + 1)
y'' = 2/x^3

I used the quadradic formula for 2x^2 + 1 and came up with ?(2)/2 and -?(2)/2 as my critical points (and points of inflection). And got ?(2)/2 as my relative minima and -?(2)/2 as my relative maxima.

I found no x or y intercepts.

What I'm stuck on is the asymptotes. I'm pretty sure there is one, but I am so confused on how to go about finding it.


Thanks guys!
- Megan

 

[galactus]

The is no horizontal asymptote. The vertical asymptote is found by noting where it is undefined.

There is only one obvious solution. Setting the denominator equal to 0. Where does that place the VA?.

But, since it said x>0, there are no asymptotes other than the y-axis.

 

[Deleted member 4993]

1. Sketch the graph of y = 2x + (1/x) where x > 0, indicating extreme points, intercepts, asymptotes, etc.

I've already managed to find most of the information.

y' = -x^-2 + 2 (to get rid of the negative, I multiplied x^2 through and came up with y = 2x^2 + 1) <<< that is incorrect
y'' = 2/x^3

I used the quadradic formula for 2x^2 + 1 and came up with ?(2)/2 and -?(2)/2 as my critical points (and points of inflection). And got ?(2)/2 as my relative minima and -?(2)/2 as my relative maxima.

I found no x or y intercepts.

What I'm stuck on is the asymptotes. I'm pretty sure there is one, but I am so confused on how to go about finding it.


Thanks guys!
- Megan

y = 2x + (1/x)

At what value/s of 'x', y becomes undefined?

That is your vertical asymptote.

When x becomes large the function becomes y ? 2x

That is your oblique assymptote.