How do you graph #y=(5x+2)/(4x)# using asymptotes, intercepts, end behavior?

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Answer 1 · 2017-02-19T09:17:36

Asymptotes : #x(y-5/4)=0#. x-intercept ( y = 0 ) : #-2/5#. No intercepts at all. See asymptotes-inclusive Socratic graph.

graph{(x(y-5/4)-1/2)(y-5/4+.001x) (x--.001y)=0 [-5, 5, -2.5, 2.5]}

This is #y = 5/4+1/(2x)#, giving

#(y-5/4)(x)=1/2#, representing the rectangular hyperbola,

with center at C( 0, 15/4) , at which the asymptotes

#x(y-5/4)=0# intersect.

x-intercept ( y = 0 ) : #-2/5#

There are no intercepts at all.