How do you graph `y=(x^2-9x+20)/(2x)` using asymptotes, intercepts, end behavior?

Given: `y=(x^2-9x+20)/(2x)`

First and foremost, the equation becomes undefined when the denominator is 0

Thus `2x!=0 => x!=0 larr "Asymptote"`

The true behaviour can be determined if we carry out the division.

`y=x/2-9/2+10/x`

`color(blue)("Consider the case "lim_(x->0^("+"))`

`y=lim_(x->0^("+")) x/2-9/2+lim_(x->0^("+"))(10/x)`

`y=0-9/2+oo = oo`

`color(blue)("Consider the case "lim_(x->0^("-"))`

`y=lim_(x->0^("-")) x/2-9/2+lim_(x->0^("-"))(10/x)`

`y=0-9/2-oo = -oo`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider the case " lim_(x->+oo) )`

`y = lim_(x->+oo)x/2-9/2+lim_(x->+oo)10/x`

`y=oo-9/2+0 = oo`

`color(brown)("Observe the this is of general form ")`
`color(brown)(y=1/2x-9/2 larr" Asymptote")`

`color(blue)("Consider the case " lim_(x->-oo) )`

`y = lim_(x->-oo)x/2-9/2+lim_(x->-oo)10/x`

`y=-oo-9/2+0 = -oo`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Are there any "x" intercepts?")`

Set `y=0=1/2x +10/x-9/2`

`0=(x^2-9x+20 )/(2x)`

`0=x^2-9x+20`

`0=(x-9/2)^2+20-81/4`

`(x-9/2)^2=1/4`

`x=9/2+-1/2`

`x=4 and 5`
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