How do you graph #y=(3x^2)/(x^2-25)# using asymptotes, intercepts, end behavior?

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Answer 1 · 2017-01-25T12:21:53

No intercepts. Yet, origin is a point on the graph.
Horizontal asymptote : #larr y = 3 rarr#
Vertical asymptote : #uarr x = +-5 darr#

x-intercept ( y = 0 ) : 0, giving origin as a point on the graph.

y-intercept ( x = 0 ): 0

#y= 3/(1-25/x^2) to 3#, as x to +-3.#

By actual division,

#y=P+Q/R= 3+75/(x^2-25)#

The asymptotes are given by

#y = P=3 and R=x^2-25=0# that gives #x=+-5#

An asymptotes-inclusive Socratic graph is inserted.

graph{(3x^2/(x^2-25)-y)(y-3)(x-5+.01y)(x+5+.01y)=0 [-40, 40, -20, 20]}