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

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Answer 1 · 2017-11-29T04:37:18

here is the graph:
enter image source here

To graph #y=3/(2x#:

Vertical asymptotes occur at the zeroes of the denominator.

#D=2x#
#VA=0# and there are no holes.

There is a horizontal asymptote because the degree of the numerator is less than the degree of the denominator:

#HA: deg N > deg D = 0#

There is no slant asymptote.

Find Behaviour near V.A (generally +/-0.1 away)
the value is not important, you only need to know the sign of the result:

at #x=0.1# = #+# (positive)
at #x=-0.1# = #-# (negative)

so 0.1 will start in quadrant I in the top left corner at #x=0.1# and -0.1 will start in quadrant III in the bottom corner at #x=-0.1#

There are no x intercepts.
There is no y intercept.

Draw your graph with smooth curves.