How do you graph #y=(x^2-5x-36)/(3x)# using asymptotes, intercepts, end behavior?

2 Answers
Dec 1, 2016

Answer:

See the explanation below

Explanation:

The domain of #y# is #D_y=RR-{0} #

As #x!=0#

So, #x=0# is a vertical asymptote.

The numerator is #x^2-5x-36#

#=(x+4)(x-9)#

So, the intercepts when #y=0#

are #(-4,0)# and #(9,0)#

For the limits #x->+-oo#, we take the terms of highest degree in the numerator and the denominator

#lim_(x->+-oo)y=lim_(x->+-oo)x^2/(3x)=lim_(x->+-oo)x/3=+-oo#

#lim_(x->0^(-))y=(0^(+)+5-36)/(3*0^(-))=+oo#

#lim_(x->0^(+))y=(0^(+)-5-36)/(3*0^(+))=-oo#

Now you can draw your graph

graph{(x^2-5x-36)/(3x) [-32.47, 32.47, -16.24, 16.25]}

Dec 1, 2016

Answer:

Asymptotes: #x = 0 and x-3y-5/3=0.# x-intercepts by the two branches: -4 and 9. As #x to 0, y to +-oo# and the other asymptote tends to reach the hyperbola, at infinite distance in #Q_1 and Q_3#

Explanation:

graph{x^2-3xy-5x-36=0 [-20, 20, -10, 10]} The second degree equation #ax^2+2hxy+by^2+...=0#

represents a hyperbola, when #ab-h^2<0# and the second degree

terms reveal the first degree terms in the equations of its

asymptotes.

Here, after cross multiplication,

#x(x-3y)-5x-36=0#. Reconstructing as

((x+a)(x-3y+b)+c=0, we find that

#a =0, b=-5/3 and c=-36#.

So, the asymptotes are given by

x=0 and x-3y-5/3=0, meeting at the center C(0, -5/9)#.

x-intercepts by the two branches: -4 and 9.

As #x to 0, y to +-oo# and the other asymptote tends to reach the

hyperbola, at infinite distance in #Q_1 and Q_3#