How do you find the horizontal and vertical asymptote of the following: f(x) = (2x-3)/(x^2+2)f(x)=2x−3x2+2?
1 Answer
Explanation:
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=-2solve x2+2=0⇒x2=−2
"this has no real solutions hence there are no vertical"this has no real solutions hence there are no vertical
"asymptotes"asymptotes
"horizontal asymptotes occur as"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]}
