How do you graph #y=(x^3-8)/(6-x^2)# using asymptotes, intercepts, end behavior?

1 Answer
Dec 21, 2016

Answer:

Slant asymptote: #y = - x#. Vertical asymptotes #uarr x = +-sqrt6 darr#. x-intercept ( y = 0 ):2. y-intercept ( x = 0 ): #-4/3#

Explanation:

By actual division.

#y = -x +(3-4/sqrt6)/(x-sqrt6)+(3+4?sqrt6)/(x+sqrt6)#, revealing

horizontal asymptote y = quotient = -x and vertical asymptotes

#x = +-sqrt6#, that makes #y +-oo at r = oo#.

The y-intercept is -4/3 and x-intercept is 2.

y' is 0 at #x = 0, +-3sqrt2#.

Local maximum ( for y'' < 0 ) at x = 3sqrt2, local minimum ( for y''>0 )

at x = -3sqrt2 and point of inflexion ( for y'' = 0 and y''' not 0 ) at x = 0.

Look at the graph that illustrates all these aspects.

graph{y(6-x^2)-x^3+8 =0 [-38.28, 38.28, -19.14, 19.14]}