Question

How do you graph `y=(8-5x)/(x-5)` using asymptotes, intercepts, end behavior?

Answer 1

see explanation.

Explanation:

`color(blue)"Asymptotes"`

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 : `x-5=0rArrx=5" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`y=(8/x-(5x)/x)/(x/x-5/x)=(8/x-5)/(1-5/x)`

as `xto+-oo,yto(0-5)/(1-0)`

`rArry=-5" is the asymptote"`

Oblique 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 oblique asymptotes.

`color(blue)"Intercepts"`

`x=0toy=8/(-5)rArr(0,-8/5)`

`y=0rArr8-5x=0rArrx=8/5rArr(8/5,0)`
graph{(8-5x)/(x-5) [-20, 20, -10, 10]}