Question

How do you find the vertical, horizontal or slant asymptotes for `y=(x^2-5x+4)/ (4x^2-5x+1)`?

Answer 1

`"vertical asymptote at "x=1/4`
`"horizontal asymptote at "y=1/4`

Explanation:

`"factorise and simplify"`

`y=(cancel((x-1))(x-4))/((4x-1)cancel((x-1)))=(x-4)/(4x-1)`

`"the removal of the factor "(x-1)" from the"`
`"numerator/denominator indicates a hole at x = 1"`

The denominator of y cannot be zero as this would make y 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-1=0rArrx=1/4" is the asymptote"`

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),ytoc" ( a constant)"`

`"divide terms on numerator/denominator by x"`

`y=(x/x-4/x)/((4x)/x-1/x)=(1-4/x)/(4-1/x)`

as `xto+-oo,yto(1-0)/(4-0)`

`rArry=1/4" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there is no slant asymptote.
graph{(x-4)/(4x-1) [-10, 10, -5, 5]}