# Curve sketching

1. Nov 1, 2008

### takercena

1. The problem statement, all variables and given/known data
Find the image set of fucntion
fx = (2x^2 + 2x + 1)/(x^2 + x - 1 )

The attempt at a solution

1. multiply the denominator with y and then assume x is real i get
(y-2)^2 - 4(y - 2)(-y-1) => 0
5y^2 - 8y - 4 => 0
y = 2 and y = -2/5

But the answer is y => 2 and y =< 2/5 not -2/5
Then i substitute the -2/5 to the original equation and i got complex solution.
So is it mean that -2/5 is out. And 2nd, why the answer is 2/5?

I sketched using tools on http://graph.seriesmathstudy.com/ and -2/5 is valid asymptote.

2. Nov 1, 2008

### cse63146

Are you trying to find the verticle or horizontal asymptote?

3. Nov 1, 2008

horizontal

4. Nov 1, 2008

### cse63146

1.First take the limit of the function as x approaches infinity

2.You divide each term by the highest degree of x (in both the numerator and the denominator)

remeber: the $$lim_{x \rightarrow \infty} \frac{x^p}{x^n} = 0$$ where n>p

You basically just need to worry about the highest degree of x in both the numerator and the denominator since the smaller degrees' limit would be 0.

5. Nov 1, 2008

### takercena

The limit will be 2. But that's not the only answer.

6. Nov 1, 2008

### cse63146

That was the horizontal asymptote, now you need to find the verticle asymptote. This page does a good job of explaining on how to do it.

7. Nov 1, 2008

### takercena

I am sorry, but can you help me how to find the vertical asymptote? I am total confuse now :(

8. Nov 1, 2008

### cse63146

You have to take the limit of f(x) as x approaches a, where a is the value where f(x) doesn't exist

but I can't get x^2 + x - 1 to equal 0. Did you copy it right?

9. Nov 1, 2008

### takercena

yes that is the question but might be placed in wrong section. First, you try to solve the real linear factor for denominator which is not the exist here. 2nd I don't think vertical asymptote exist as first if you factorize it, it will always not equal to 0 both denominator and numerator.

10. Nov 1, 2008

### cse63146

A vertical asymptote does exist, and only the denominator's limit has to be 0, the numerator's limit can't be 0. To determine the roots of the denominator, (can't believe I didn't realise this sooner) you need to use the quadratic equation. You won't get integers, but you'll get real numbers.

11. Nov 1, 2008

### takercena

Thanks, but after I get the x value, what should i do next? Can you tell me what is the value for horizontal asymptote?

12. Nov 2, 2008

### cse63146

The roots of x^2 + x - 1 are the verticle asymptotes. Now you'll need to determine whether it approaches infinity or negative infinity at that value. Let A be a root of x^2 + x -1. Take the limit of (x -> $$A^+$$) f(x) and the limit of (x -> $$A^-$$) f(x). You'll need to know this when you're sketching the function.