SOLUTION: Please please help!!!! h(x)=((x^2)(e^x))/x The function h is defined above. Which of the following are true about the graph of y=h(x)? -The graph has a vertical asymptote

Question 973447: Please please help!!!!
h(x)=((x^2)(e^x))/x
The function h is defined above. Which of the following are true about the graph of y=h(x)?
-The graph has a vertical asymptote at x=0
-The graph has a horizontal asymptote at y=0
-The graph has a minimum point
I know the minimum point is correct but the booklet is telling me that graph has a horizontal asymptote y=0 but I believe that is not true because for it to y=0 the denominator leading coefficient exponent value must be greater then the numerator value which is not the case unless I'm missing something. Thank you!!!

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
h(x)=((x^2)(e^x))/x
The function h is defined above. Which of the following are true about the graph of y=h(x)?
-The graph has a vertical asymptote at x=0
-The graph has a horizontal asymptote at y=0
-The graph has a minimum point
I know the minimum point is correct but the booklet is telling me that graph has a horizontal asymptote y=0 but I believe that is not true because for it to y=0 the denominator leading coefficient exponent value must be greater then the numerator value which is not the case unless I'm missing something.
---------------
No vertical asymptote.
It has a minimum @ x = -1
It has a horizontal asymptote y = 0.
As x increases in the negative direction, e^x approaches zero faster than x goes negative.
-------------
x = 0 --> y = 0
x = -5 --> y = -0.0337
x = -10 --> y = -0.0005
x = -20 --> y = -4.1e-8
x = -30 --> y = -2.8e-12
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