How do you graph #y=6/(x^2+3)#?

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Answer 1 · 2017-12-21T20:14:27

See below.

First find significant points, these will help in sketching the graph.

y axis intercepts occur when #x=0#:

#6/((0)^2+3)=2color(white)(88)# coordinate #( 0 , 2 )#

x axis intercepts occur when #y=0#:

#6/(x^2+3)=0#

This can only be zero when denominator is zero, which is undefined, so no x axis intercepts.

as #x->oo# #color(white)(888)6/(x^2+3)->0#

as #x->-oo# #color(white)(888)6/(x^2+3)->0#

So the x axis is a horizontal asymptote.

#y=0#

Vertical asymptotes occur where the function is undefined:

#6/(x^2+3)# is not undefined for any real #x#, so no vertical asymptotes.

#6/(x^2+3)# attains a maximum value when the denominator is a minimum value, this can be seen to be 3 when #x=0#

( this is the y axis intercept that was previously found )

Graph:

graph{y=6/(x^2+3) [-10, 10, -5, 5]}