What is the end-behavior asymptote for f(x)=(x^2+4x-5)/(x-2) ?
There is an oblique asymptote at y=x+6 and a vertical asymptote at x=2. Is one of those the end-behavior asymptote?
You are right to have those two asymptotes.
look at
http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x%5E2%2B4x-5%29%2F%28x-2%29+%2C+y+%3D+x%2B6
What happens at the far right and the far left ?
it approaches positive and negative infinity?
yes
To find the end-behavior asymptote of the function f(x) = (x^2 + 4x - 5)/(x - 2), we can analyze the behavior of the function as x approaches positive or negative infinity.
To determine the end behavior, we can divide the highest power term in the numerator (x^2) by the highest power term in the denominator (x) using long division or synthetic division.
Let's perform the long division:
x + 5
____________________
x - 2 | x^2 + 4x - 5
-(x^2 - 2x)
___________
6x - 5
-(6x - 12)
__________
7
The result of the long division is x + 5 with a remainder of 7.
Therefore, the function can be written as:
f(x) = x + 5 + 7/(x - 2)
As x approaches positive or negative infinity (i.e., x → ±∞), the term 7/(x - 2) approaches zero. So, the end-behavior asymptote is y = x + 5.
In the given function, y=x+6 is the oblique asymptote and x=2 is the vertical asymptote. These are not the end-behavior asymptotes. The end-behavior asymptote is y = x + 5.