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

##### 2 Answers

#### Answer:

Asymptotes:

#### Explanation:

Cross multiplying and reorganizing,

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

asymptotes given by

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

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

x-intercept ( y = 0 ):

asymptotes

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

#### Answer:

See explanation.

#### Explanation:

A **vertical asymptote** occurs at

V.A. when

#2x-4=0#

#<=>2x=4#

#<=>x=2#

So the equation for the V.A. is

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 *will be* a horizontal asymptote.

When the degrees are the same (like in this case), the horizontal asymptote is found at **the ratio of the leading coefficients**. Here, that happens to be

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

The ** #x#-intercept** is found by letting

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

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

#x="-5"#

So our

Similarly, the ** #y#-intercept** is found by letting

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

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

So our

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]}