Image Set of fx = (2x^2 + 2x + 1)/(x^2 + x - 1)

[takercena]

Homework Statement

Find the image set of function
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.

 

[cse63146]

Are you trying to find the verticle or horizontal asymptote?

 

[takercena]

horizontal

 

[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.

 

[takercena]

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

 

[cse63146]

That was the horizontal asymptote, now you need to find the verticle asymptote. http://archives.math.utk.edu/visual.calculus/1/vertical.4/index.html page does a good job of explaining on how to do it.

 

[takercena]

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

 

[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?

 

[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.

 

[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 realize this sooner) you need to use the quadratic equation. You won't get integers, but you'll get real numbers.

 

[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?

 

[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.