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

Question

Ariana Alvarez · Accepted Answer

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

Christian Antunez · Answer

undefined To find the vertical asymptote(s) of the function $ f(x) = \frac{-10x + 3}{8x + 2} $, set the denominator equal to zero and solve for $ x $. There's one vertical asymptote at $ x = -\frac{1}{4} $.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since both have the same degree (1), divide the leading coefficients. Therefore, the horizontal asymptote is $ y = -\frac{10}{8} = -\frac{5}{4} $.

There are no slant asymptotes since the degree of the numerator is not exactly one more than the degree of the denominator.