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

2 Answers
Dec 10, 2016

Answer:

Asymptotes: #uarr x = 2 darr and larr y = 1/2 rarr#.y-intercept: #5/2#. x-intercept : 2. #(x, y) to (+-oo, 1/2) and (2, +-oo)#, in the opposite directions of the asymptotes #y =1/2 and x = 2#.

Explanation:

Cross multiplying and reorganizing,

(x-2)(y-1/2)=7/2# that represents a rectangular hyperbola (RH) having

asymptotes given by

#x=2 and y=1/2#.

The center of the RH is (2, 1/2).

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

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

#(x, y) to (+-oo, 1/2) and (2, +-oo)#, in the opposite directions of the

asymptotes #y = 1/2 and x = 2#, respectively.

graph{y(2x-4)-x-5=0 [-10, 10, -5, 5]}

Dec 10, 2016

Answer:

See explanation.

Explanation:

A vertical asymptote occurs at #x#-values that make the denominator 0. To find the vertical asymptote(s) (V.A.'s), set your denominator equal to zero and solve for #x:#

V.A. when #2x-4=0#
#<=>2x=4#
#<=>x=2#

So the equation for the V.A. is #x=2.#

A horizontal asymptote occurs when the degree of the numerator is less than (or equal to) the degree of the denominator. ("Degree" means the highest power of #x.#) Since both sides of the fraction have a degree of 1, there will be a horizontal asymptote.

When the degrees are the same (like in this case), the horizontal asymptote is found at #y=#the ratio of the leading coefficients. Here, that happens to be #y=1/2# (from #(color(red)1x+5)/(color(red)2x-4)#).

(In the case that the denominator has a higher degree, the asymptote is always #y=0.#)

The #x#-intercept is found by letting #y=0# and solving for #x:#

#0=(x+5)/(2x-4)#

#0=x+5# [multiply both sides by (2x-4) ]

#x="-5"#

So our #x#-intercept is at #("-5",0).#

Similarly, the #y#-intercept is found by letting #x=0# and solving for #y:#

#y=((0)+5)/(2(0)-4)#

#y=5/"-4"=-5/4#

So our #y#-intercept is at #(0,-5/4).#

With all this information, we can now draw our hyperbola:

graph{(y-(x+5)/(2x-4))(y-(x-2.0001)/(2x-4))=0 [-8.835, 11.165, -3.91, 6.09]}