How do you find vertical, horizontal and oblique asymptotes for (8x^2-5)/(2x^2+3)8x2−52x2+3?
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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
"solve "2x^2+3=0rArrx^2=-3/2solve 2x2+3=0⇒x2=−32
"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)=((8x^2)/x^2-5/x^2)/((2x^2)/x^2+3/x^2)=(8-5/x^2)/(2+3/x^2)
"as "xto+-oo,f(x)to(8-0)/(2+0)
rArry=4" is the asymptote"
"Oblique asymptotes occur when the degree of the "
"numerator is greater than the degree of the denominator"
"this is not the case here hence there are no oblique"
"asymptotes"
graph{(8x^2-5)/(2x^2+3 [-10, 10, -5, 5]}
