How do you find the vertical, horizontal or slant asymptotes for #f(x)= (-10x+3)/(8x+2)#?

1 Answer
Jun 2, 2016

Answer:

vertical asymptote #x=-1/4#
horizontal asymptote #y=-5/4#

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : 8x + 2 = 0 → 8x = -2 #rArrx=(-2)/8=-1/4#

#rArrx=-1/4" is the asymptote"#

Horizontal asymptotes occur as #lim_(xto+-oo),f(x)to0#

divide terms on numerator/denominator by x

#((-10x)/x+3/x)/((8x)/x+2/x)=(-10+3/x)/(8+2/x)#

as #xto+-oo,f(x)to(-10+0)/(8+0)#

#rArry=-10/8=-5/4" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1). Hence there are no slant asymptotes.
graph{(-10x+3)/(8x+2) [-10, 10, -5, 5]}