Why is there no vertical asymptote on
F(x) = x/(x^2+1)
What i learned is that to find vertical asymptote you have to set the denominator To 0 and solve for x ?
In that case I find that x=-1,1
vertical asymptotes occur when you try to divide by zero.
x^2+1 is never zero.
You solved for x^2-1 = 0
I don't get it ?
huh? huh? How can you say that?
x^2 is always positive, right? At least it is never negative.
So, x^2+1 is always positive, and always at least 1.
So, dividing by x^2+1 will never yield a vertical asymptote.
Yesss okay thank you sir
To determine if a function has a vertical asymptote, we indeed need to check if there are any values of x that make the denominator equal to zero. However, just finding such values does not guarantee the existence of a vertical asymptote. Let's break down the steps to understand why there is no vertical asymptote in the given function F(x) = x/(x^2+1).
Step 1: Find the values of x that make the denominator equal to zero by setting x^2 + 1 = 0:
x^2 + 1 = 0
x^2 = -1
Here, we encounter a small problem. The equation x^2 = -1 has no real solutions because the square of a real number yields a positive result. Therefore, there are no values of x that make the denominator zero, so we can conclude that F(x) has no vertical asymptotes.
In simpler terms, a vertical asymptote exists when the denominator of a rational function becomes zero at a certain x-value, typically resulting in an unbounded value for the function. However, if the equation involving the denominator does not have any real solutions, like in this case, then there is no vertical asymptote.
It's important to note that while finding the values of x that make the denominator zero is a necessary step in determining vertical asymptotes, it is not the only step. The existence of a vertical asymptote depends on the behavior of the function around those points as well.