Question

How do you find the vertical, horizontal and slant asymptotes of: `(-2x+7)/(4x-3)`?

Answer 1

vertical asymptote at `x=3/4`
horizontal asymptote at `y=-1/2`

Explanation:

Let f(x)`=(-2x+7)/(4x-3)`

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: `4x-3=0rArrx=3/4" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" (a constant)"`

divide terms on numerator/denominator by x

`f(x)=(-(2x)/x+7/x)/((4x)/x-3/x)=(-2+7/x)/(4-3/x)`

as `xto+-oo,f(x)to(-2+0)/(4-0)`

`rArry=-1/2" 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{(-2x+7)/(4x-3) [-10, 10, -5, 5]}