How do you find the vertical, horizontal or slant asymptotes for # f(x) =sqrt(x^2+1)+2-x#?

Question

Elijah Allen · Accepted Answer

No Vertical and slant asymptote. Horizontal asymptote is y=2

Christian Antunez · Answer

undefined To find the vertical, horizontal, or slant asymptotes of $ f(x) = \sqrt{x^2 + 1} + 2 - x $:

1. Vertical asymptotes: There are no vertical asymptotes because the function is defined for all real numbers.

2. Horizontal asymptotes: To find horizontal asymptotes, examine the behavior of the function as $ x $ approaches positive or negative infinity. In this case, as $ x $ approaches positive or negative infinity, $ \sqrt{x^2 + 1} $ will dominate the function. Since the leading term inside the square root is $ x^2 $, the function behaves like $ \sqrt{x^2} $ for large $ x $, which simplifies to $ |x| $. Therefore, there are no horizontal asymptotes.

3. Slant asymptotes: To find slant asymptotes, perform long division or polynomial division to divide the numerator by the denominator. If the degree of the numerator is greater than the degree of the denominator by exactly 1, then the slant asymptote exists. In this case, the degree of the numerator is 1 greater than the degree of the denominator, so we can proceed with polynomial division.

$$ f(x) = \sqrt{x^2 + 1} + 2 - x $$

$$ = \sqrt{x^2 + 1} - x + 2 $$

Performing polynomial division of $ x^2 + 1 $ by $ -x + 2 $ yields:

$$ x^2 + 1 = (-x + 2)(-x) + 2x + 1 $$

So, $ f(x) = \sqrt{x^2 + 1} + 2 - x $ can be rewritten as:

$$ f(x) = (-x + 2) + \frac{2x + 1}{\sqrt{x^2 + 1}} $$

As $ x $ approaches positive or negative infinity, the fraction $ \frac{2x + 1}{\sqrt{x^2 + 1}} $ will approach zero since the numerator's degree is less than the denominator's degree. Therefore, the slant asymptote is $ y = -x + 2 $.