How do you graph #y=(x^2-9)/(2x^2+1)# using asymptotes, intercepts, end behavior?

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Answer 1 · 2017-02-08T14:23:11

y-intercept ( x = 0 ) :#-9#. x-intercept ( y = 0 ) : #+-3#
Horizontal asymptote : #larr y = 1/2 rarr#
Vertical asymptote : #x = 0darr#

graph{((x^2-9)/(2x^2+1)-y)(y-1/2)x=0 [-10, 10, -5, 5]}

y-intercept ( x = 0 ) :#-9#

x-intercept ( y = 0 ) : #+-3#

By actual division,

#y = 1/2-(19/2)/(2x^2+1)>1/2#

So, y = 1/2 is the asymptote.

As #x to +-oo, to 1/2#

Inversely, x =+-sqrt((y=9)/(1-2y)=+-sqrt((1+9/y)/(1/y-2)) to 0, as y to -oo#.

So,# x - 0 darr# is yet another asymptote.