How do you find the asymptotes for #f(x)=(1-2x)/sqrt(1+x^2) #?

Question

Sadie Ackerman · Accepted Answer

You find the asymptotes by subbing a number into x, in order to find the value which is undefined.

#f(x)# = #(1-2x)/(sqrt(1+x^2)# #f(-1)# = #(1-2(-1))/(sqrt(1+(-1)^2)# = Underfinded #:.# #-1# is the asymptotpe

Christian Antunez · Answer

undefined To find the asymptotes for $ f(x) = \frac{1 - 2x}{\sqrt{1 + x^2}} $, we examine both the horizontal and vertical behavior of the function.

1. Vertical asymptotes:
The vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. In this case, the denominator is $ \sqrt{1 + x^2} $, which equals zero when $ x = \pm i $ (imaginary). Since these are not real values, there are no vertical asymptotes for this function.

2. Horizontal asymptotes:
To find the horizontal asymptote, we examine the behavior of the function as $ x $ approaches positive and negative infinity. We take the limit of the function as $ x $ approaches infinity and negative infinity.

$$ \lim_{x \to \infty} \frac{1 - 2x}{\sqrt{1 + x^2}} $$

$$ \lim_{x \to \infty} \frac{-2x}{\sqrt{1 + x^2}} = -2 $$

$$ \lim_{x \to -\infty} \frac{1 - 2x}{\sqrt{1 + x^2}} $$

$$ \lim_{x \to -\infty} \frac{-2x}{\sqrt{1 + x^2}} = -2 $$

Therefore, the horizontal asymptote of the function is $ y = -2 $.