Question

How do you graph `f(x) = x + (5x)/(x-1)` and identify all the asymptotes?

Answer 1

Vertical asymptote: `x=1` , slant asymptote: `y=x+5`
`x` intercepts: `x=-4 , x=0, y` intercept: `y=0`,
end behavior: `y-> -oo` as `x -> -oo ,y-> +oo` as `x -> oo`

Explanation:

`f(x)= x+ (5 x)/(x-1)= (x^2-x+5 x)/(x-1)` or

`f(x)=(x^2+4 x)/(x-1)` Vertical asymptote occur when

denominator is zero.

`:. x-1=0 :. x= 1; lim(x->1^-) ; y -> -oo`

`lim (x->1^+); y -> +oo` numerator's degree is greater

(by a margin of 1), then,we have a slant asymptote which is

found by long division.

`f(x)=(x^2+4 x)/(x-1)= (x+5) + 5/(x-1)` , therefore, slant

asymptote is `y= x+5`

`y` intercept: Putting `x=0` in the equation we get,

`y=0 , x` intercepts: Putting `y=0` in the equation

we get , `x^2+4 x or x(x+4)=0 or x =0 , x=-4`

End behavior: `y-> -oo` as `x -> -oo` and

`y-> +oo` as `x -> oo`

graph{x+ 5x/(x-1) [-80, 80, -40, 40]} [Ans]