takercena said:
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.
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.
takercena said:
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.
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.
takercena said:
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?
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.
