# Sketch The Curve

1. Nov 19, 2005

### scorpa

Hi guys,

I know this is a dumb question, but I have to ask. We are supposed to sketch the curve of y=x/(x-1)^2 using the guidelines for curve sketching such as domain, intervals of increase/decrease, concavity....ect. For the most part on this question I think I have done this all right, it really isn't all that hard. But for some reason my asymptotes are a bit messed up.

I know there is a vertical asymptote occuring at x=1, because having an x value of one makes the denominator undefined. Then to find the horizontal asymptoes I expanded the denominator to get
y = x/(x^2 -2x+1), then I found the horizontal denominator (limit as it approaches infinity) by taking an x^2 out of the whole thing, leaving me with a horizontal asymptote at y=0, which for the right half of the curve is true. But the left half of the curve does cross the axis at (0,0). How do I account for the fact that there is not a horizontal tangent at y=0 on the left side of the graph? I hope I explained my situation clearly. Thanks for any help in advance.

2. Nov 19, 2005

### Tide

The function approaches the x axis from below as x goes to negative infinity.

3. Nov 19, 2005

### scorpa

OK, I get the statement you just made. But how would I go about showing that mathematically?

4. Nov 19, 2005

### Tide

For x < 0, the function is negative and there are no other zeros of the function.

5. Nov 20, 2005

### scorpa

Ok so because I have found that there is a horizontal asymptote at x=0 I can say that as x approaches both infinity and negative infinity the graph approaches but never reaches zero, except for the intercept at (0,0) ?

6. Nov 20, 2005

### Tide

Well, there's a horizontal asymptote at y = 0 and not at x = 0 and since the function is negative for all negative values of x so the graph approaches the asymptote from below for negative x but from above for positive x.

7. Nov 20, 2005

### scorpa

Ok I think I got it now, thanks