How do you graph #y=(8x+3)/(2x-6)# using asymptotes, intercepts, end behavior?

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Answer 1 · 2016-11-20T09:50:08

See the graph below

The domain of #y,##D_y=RR-{3}#

As we cannot divide by #0#, #x!=3#

A vertical asymptote is #x=3#

For the limits, we take the terms of highest coefficients in the numerator and the denominator

#lim_(x->+-oo)y=lim_(x->+-oo)(8x)/(2x)=4#

A horizontal asymptote is #y=4#

For the intercepts,

#x=0#, #=># , #y=3/-6=-1/2#

This is the y-intercept #(0,-1/2)#

When #y=0,#=>#, #x=-3/8#

This is the x-intercept, #(-3/8,0)#

#lim_(x->3^(-))y=-oo#

#lim_(x->3^(+))y=+oo#

graph{(y-(8x+3)/(2x-6))(y-4)(x-3)=0 [-22.8, 22.85, -11.4, 11.4]}