Answer to Question #112746 in Algebra for Moza

Question

In Writing explain how to find the asymptote of y=- 3/x-2 +11

Answer 1

The equations of slant asymptotes are usually sought in the form y = kx + b. By definition, asymptotes:

$\lim\limits_{x\rarr \infty}(kx+b- f(x))\\$

We find the coefficient k:

$k=\lim\limits_{x\rarr \infty}\frac{f(x)}{x}\\ k=\lim\limits_{x\rarr \infty}\frac{\frac{-3}{x-2}+11}{x}=\lim\limits_{x\rarr \infty}\frac{11x-25}{x^2-2x}=0\\$

We find the coefficient b:

$b=\lim\limits_{x\rarr \infty}{f(x)}-{kx}\\ b=\lim\limits_{x\rarr \infty}\frac{-3}{x-2}+11-0x=\lim\limits_{x\rarr \infty}\frac{11x-25}{x-2}=11\\$

We obtain the equation of horizontal asymptote:

$y=11\\$

Find the vertical asymptotes. To do this, define the points of discontinuity:

x₁=2

Find the asymptotic behaviour at x = 2:

$\lim\limits_{x\rarr2+0 }\frac{-3}{x-2}+11=-\infty\\ \lim\limits_{x\rarr2-0 }\frac{-3}{x-2}+11=+\infty$

x₁ = 2 is a discontinuity point of the second kind and x=2 is a vertical asymptote.