What are the asymptotes and removable discontinuities, if any, of #f(x)=(x-3)/(2x+5)#?
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 "2x+5=0rArrx=-5/2" is the asymptote"#
#"Horizontal asymptotes occur as"#
#lim_(xto+-oo),f(x)toc" ( a constant)"#
#"divide terms on numerator/denominator by x"#
#f(x)=(x/x-3/x)/((2x)/x+5/x)=(1-3/x)/(2+5/x)#
#"as "xto+-oo,f(x)to(1-0)/(2+0)#
#rArry=1/2" is the asymptote"#
#"Removable discontinuities occur when common"#
#"factors are cancelled from the numerator/denominator."#
#"this is not the case here hence there are no removable"#
#"discontinuities"#
graph{(x-3)/(2x+5) [-10, 10, -5, 5]}