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

1 Answer
Binayaka C.
Aug 7, 2018

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]

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