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

Aug 7, 2018

Vertical asymptote: $x = 1$ , slant asymptote: $y = x + 5$
$x$ intercepts: $x = - 4 , x = 0 , y$ intercept: $y = 0$,
end behavior: $y \to - \infty$ as $x \to - \infty , y \to + \infty$ as $x \to \infty$

#### Explanation:

$f \left(x\right) = x + \frac{5 x}{x - 1} = \frac{{x}^{2} - x + 5 x}{x - 1}$ or

$f \left(x\right) = \frac{{x}^{2} + 4 x}{x - 1}$ 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.

$f \left(x\right) = \frac{{x}^{2} + 4 x}{x - 1} = \left(x + 5\right) + \frac{5}{x - 1}$ , 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 , ${x}^{2} + 4 x \mathmr{and} x \left(x + 4\right) = 0 \mathmr{and} x = 0 , x = - 4$

End behavior: $y \to - \infty$ as $x \to - \infty$ and

$y \to + \infty$ as $x \to \infty$

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