How do you find the horizontal and vertical asymptote of the following: #f(x) = (2x-3)/(x^2+2)#?

Question

Sophie Adams · Accepted Answer

#"horizontal asymptote at "y=0#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "x^2+2=0rArrx^2=-2#

#"this has no real solutions hence there are no vertical"# #"asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"# #"power of x that is "x^2#

#f(x)=((2x)/x^2-3/x^2)/(x^2/x^2+2/x^2)=(2/x-3/x^2)/(1+2/x^2)#

#"as "xto+-oo,f(x)to(0-0)/(1+0)#

#rArry=0" is the asymptote"# graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}

Greyson Brown · Answer

undefined To find the horizontal and vertical asymptotes of $ f(x) = \frac{2x - 3}{x^2 + 2} $, we analyze the behavior of the function as $ x $ approaches positive or negative infinity.

1. **Vertical Asymptotes**: Vertical asymptotes occur where the denominator of the rational function equals zero but the numerator does not. To find vertical asymptotes, set the denominator $ x^2 + 2 $ equal to zero and solve for $ x $.

$$ x^2 + 2 = 0 $$

This equation has no real solutions, so there are no vertical asymptotes for the function $ f(x) $.

2. **Horizontal Asymptotes**: To find horizontal asymptotes, we examine the behavior of the function as $ x $ approaches positive or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is at $ y = 0 $.

So, the horizontal asymptote is $ y = 0 $, and there are no vertical asymptotes for the function $ f(x) = \frac{2x - 3}{x^2 + 2} $.