How do you graph #y=15/(x^2+2)# using asymptotes, intercepts, end behavior?

1 Answer
Jan 15, 2018

Answer:

See below.

Explanation:

#f(x)=15/(x^2+2)#

Vertical asymptotes occur where the function is undefined.i.e.

#x^2+2=0#

There are no real solutions to this, so no vertical asymptotes.

As #x->-oo# , #color(white)(888)15/(x^2+2)->0#

As #x->oo# , #color(white)(88888)15/(x^2+2)->0#

#x^2>0# for all #x in RR#

So, x axis is a horizontal asymptote.

( This is also the end behaviour ).

Maximum value occurs when:

#x^2+2# is at its minimum.

This is when #x=0#

#y=15/((0)^2+2)=15/2#

GRAPH:

graph{15/(x^2+2) [-18.02, 18.01, -9.01, 9]}