Question

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

Answer 1

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]}