I have a question about HA, VA and minimum points

[Roni1985]

Homework Statement

For the function h(x)=$$\frac{x^2*e^x}{x}$$ , which of the following are true about the graph of y=h(x)?

I. The graph has a vertical asymptote at x=0
II. The graph has a horizontal asymptote at y=0
III. The graph has a minimum point

Homework Equations

h(x)=$$\frac{x^2*e^x}{x}$$

The Attempt at a Solution

'I' is wrong because we can cancel X in the denominator and it becomes a hole (or I am wrong?), so its not an asymptote.

For 'II', I tried different ways.
My first way was putting 'lan' on each side, and trying to find the limit of that function when x goes to infinite.

For the third one, when I take the derivative, I get X=-1 when the derivative is equal to 0 and a critical point at X=0...
But I am getting a max point and not a min point.

Plus, according to my answers, 'II' and 'III' are correct.

The thing is that I haven't took cal I since the 10th grade and now I am tutoring cal I at my college, so, I need to refresh the very basic things (even though it doesn't look very basic)
Also, this question allows using a calculator, but I want to know how to solve it without.

P.S: I posted it in precal, but I think its more cal than precal.

Thanks,
Roni

 

[rock.freak667]

You are correct for I being wrong. I also believe II is wrong as xe^x=0 can be solved to give a real value for x.EDIT: plotted the wrong graph ignore my post please. :)

 

[defunc]

2 and 3 are indeed correct. X=-1 is a minimum. You can prove 2 by taking the limit as x goes to -infinity, and using lhospital's rule

 

[n!kofeyn]

This problem has already been discussed and solve https://www.physicsforums.com/showthread.php?t=342328", as the original poster made a double post.

I'm also not really for sure what rock.freak667 meant by solving xe^x=0 for number II, because it's asking for a horizontal asymptote.