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

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]}

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]}