# 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? And 3rd, how to test the validity of the range like whether it is a<y<b or y < a , y> a?

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

2. Nov 1, 2008

### Staff: Mentor

What do you mean "multiply the denominator with y"? I have no idea what you mean by this. If you multiply the denominator by anything other than 1, you will change it.

How did you get this? Also, you shouldn't use => when you mean =.
(y-2)^2 - 4(y - 2)(-y-1) => 0

I don't see any connection between it and what you started with.
Use = for expressions that are equal, not =>, which is used when one equation implies another.
It's not possible to substitute any real number into this function and get a complex number. f(-2/5) = -13/31, which is not the most convenient number to work with, but it's not a complex number.
Your function f is a rational function. The things you should be looking for are:
• x intercepts: values of x for which f(x) = 0. For this function, there are no real values for which f(x) = 0. This means that the numerator can't change sign for this function. By inspection I can see if x = 0, the value of the numerator is 1, which means that it (numerator) must always be positive.
• y intercept: the y value for f(0).
• vertical asymptotes: values of x that make the denominator zero. For this function, there are two real values that make the denominator zero. This means that the denominator will be positive or negative in three regions. The behavior of the graph is determined by the sign of the denominator for this function, since the numerator is always positive.
• horizontal asymptote(s): Since the degrees of the numerator and denominator are both two, there is a horizontal asymptote. As x gets very large or very negative the graph approaches the horizontal line y = 2.

3. Nov 1, 2008

### takercena

It's not x = -2/5 but it's y = -2/5. And => is more 'than or equal' and otherwise. Let me show you my work

First I rearrange this into like this
from y = (2x^2 + 2x + 1)/(x^2 + x - 1 ) into y(x^2 + x - 1 ) - (2x^2 + 2x + 1) = 0. Then i use b^2 - 4ac => 0 and compare the coefficient from (y-2)x^2 + (y-2)x + (-y-1) = 0 which gives (y-2)^2 - 4(y-2)(-y-1) => 0. After factoring it, i get y = 2 and y = -2/5.

You can try use the http://graph.seriesmathstudy.com/ and see that there is at least 2 horizontal asymptote there. My question is if i substitute the value of y to the original equation (to y not x) I will get complex solution and 2nd the answer to the question are 2 and 2/5.

And the more i think about this question the more confuse i am. Can some explain to me how to determine the validity of range like whether the solution is a < fx < b or fx > b, fx < b?

Last edited: Nov 1, 2008
4. Nov 2, 2008

### Staff: Mentor

First off, you shouldn't post the same problem in two different forums. Unless I'm mistaken, you have also posted this problem in the Precalc forum. Second, if you mean "greater than or equal" you should write it as >=, otherwise it looks like the arrow used for implication.

Now that I've seen your work I am able to follow your reasoning. In the discriminant (b^2 - 4ac), what you have found are two values of y that make the discriminant 0, namely y = 2 and y = -2/5.

If the discriminant is zero, that means you'll have only one solution for x, namely (-y + 2) / (2(y -2)). If y = 2, x is undefined, since both numerator and denominator are zero. If y = -2/5, x = -1/2.

What this means is that you have done a whole lot of work just to find one point on the graph of this function.

For this function, there is only one horizontal asymptote -- the line y = 2.

There are some fairly simple things you can do to find out what the graph of this function looks like, as I've already stated and will put here again.
• x intercepts: values of x for which f(x) = 0. For this function, there are no real values for which f(x) = 0. This means that the numerator can't change sign for this function. By inspection I can see if x = 0, the value of the numerator is 1, which means that it (numerator) must always be positive.
• y intercept: the y value for f(0).
• vertical asymptotes: values of x that make the denominator zero. For this function, there are two real values that make the denominator zero. This means that the denominator will be positive or negative in three regions. The behavior of the graph is determined by the sign of the denominator for this function, since the numerator is always positive. For this function, the vertical asymptotes are x $$\approx$$ -1.62 and x $$\approx$$ .62. The exact values are -1/2 - sqrt(5)/2 and -1/2 + sqrt(5)/2.
• horizontal asymptote(s): Since the degrees of the numerator and denominator are both two, there is one horizontal asymptote. As x gets very large or very negative the graph approaches the horizontal line y = 2.

I have drawn a rough graph of this function. To the left of the vertical line x $$\approx$$ -1.62, the graph is always above the line y = 2, approaching it from above as x gets more and more negative. As x approaches -1.62 from the left, the y values get larger and larger. Between approximately -1.62 and .62, the y values are all negative and approach neg. infinity as x approaches -1.62 from the right and as x approaches .62 from the left. If x is larger than .62, the y values are all larger than 2, with y approaching infinity as x approaches .62 from the right. As x gets larger and larger, y approaches 2 from above.

To find the range for this function, I think you would need calculus to find the high point of the middle section of the graph. For the two outer sections, y > 2.

5. Nov 2, 2008

### takercena

Thanks for your reply. I really appreciate that. But I still don't get it as if you said there is a curve between -1.62 and .62, what is the turning point? It's can be 0 isn't it?

As i mentioned before, the inner range is probably -2/5 because you can just factorize the y from (y-2)x^2 + (y-2)x + (-y-1) >= 0 where you will get 2 and -2/5 as you already prove the 2. Calculus will only make the function harder.

My question is if i substitute -2/5 into the original equation, i will get complex solution.

1st
Let y = 2; 2(x^2 + x - 1 ) - (2x^2 + 2x + 1) = 0, -3 = 0 which is not true. Therefore the fx will never pass through 2.

2nd
Let y = -2/5, you will get the complex solution. Now I don't know if this just an error from book or not but, the answer is fx>2 and fx < 2/5. not -2/5

Last edited: Nov 2, 2008
6. Nov 2, 2008

### Staff: Mentor

All this shows is that -2/5 is not in the range of this function. For x in the interval (-1.62, .62) --approximate values--the graph of f looks sort of lik an upside-down U. Short of using calculus I don't know any way to find the exact local maximum point on this part of the graph. However, you could get a rough idea of where it is by evaluating the function for various values of x between -1 and 0.

If you do know some calculus, you could find f'(x) and see where it is zero. That would give you the x-value at the top of the middle section of the graph.