How do you find the vertical, horizontal or slant asymptotes for #(2x+sqrt(4-x^2)) / (3x^2-x-2)#?

1 Answer
Dec 6, 2016

Answer:

#-2<=x<=2#. y-intercept: x-intercept: #-2/sqrt5#. Dead ends #(-2, -1/6) and (2, 1)#. .Vertical asymptote: x = 1. See the explanation and graphs, for clarity.

Explanation:

To make y real, #-2<=x<=2#.

Accordingly, we reach dead ens at #(-2, -1/6) and (2, 1)#.

y-intercept ( x = 0 ) is 1 and x-intercept ( y = 0 ) is #-2/sqrt5#.

Note that the curve does not reach the x-axis for positive intercept.

Also, #y = (1/x)(2+sqrt(1-2/x^2))/(3-1/x+2/x^2) to 0# as # x to +-oo#

does not happen, as #|x|<=2#,

Now, #y =(2x+sqrt(4-x^2))/((x-1)(3x+2)) to +-oo# as x to 1 and x to -2/3#

The first is OK. but, the second is negated by #x >=-2#.

Important note:

The convention #sqrt(4-x^2)>=0 ( unlike (4-x^2)^(1/2)=+-sqrt(4-x^2))#

troubled me in making my data compatible with that in the graph. I

had to take negative root, breaking the convention.

graph{y(x-1)(3x-2)-2x-sqrt(4-x^2)=0 [-10.17, 10.17, -5.105, 5.065]}

graph{y(x-1)(3x-2)-2x-sqrt(4-x^2)=0 [-20.34, 20.34, -10.19, 10.15]}