How do you graph #y = (2x^2)/x^2 +9# ?

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Answer 1 · 2017-10-17T00:50:58

The horizontal straight line through #(0,11)#

#y = (2x^2)/x^2 +9#

Assuming #x!=0#

#y = (2cancelx^2)/cancelx^2 +9 =11#

Which is the horizontal straight line through #(0,11)# shown below.

NB: #y=11 forall x in (-oo,0)uu(0,+oo)#

graph{11+0.0001x [-65.77, 65.9, -33, 32.8]}

However, if you meant #y =(2x^2)/(x^2+9)#

#y=0# at #x=0#

#lim_"x-> +-oo" y =2#

graph{(2x^2)/(x^2+9) [-4.86, 5.006, -2.57, 2.36]}