How do you graph `f(x)=(x-2)/(x-4)` using holes, vertical and horizontal asymptotes, x and y intercepts?

`color(blue)("Hole - undefined")`

Hole is where the denominator becomes 0. The function becomes undefined at that point. So for this condition we have `x=4`

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Lets consider the behaviour close to `x=4`

`color(blue)("Vertical Asymptot "->+oo)`

If `x=4+delta4` where `delta4>0` and very small then we have

`(4+delta4-2)/(cancel(4)+delta4-cancel(4)) ->2/(delta4)+1`

`lim_(delta4->color(white)()^+ 0)2/(delta4)+1 color(white)("dd")->color(white)("dd")kcolor(white)("dd")->color(white)("dd")+oo+1=+oo`

`color(blue)("Vertical Asymptot "->-oo)`

If `x=4-delta4` where `delta4>0` and very small then we have

`(4-delta4-2)/(cancel(4)-delta4-cancel(4)) ->(2-delta4)/(-delta4)=-2/(delta4)+1`

`lim_(delta4->color(white)()^+ 0)-2/(delta4)+1color(white)("dd") ->color(white)("dd")kcolor(white)("dd")->color(white)("dd")-oo+1=-oo`
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`color(blue)("Horizontal Asymptot "->+oo)`

`(x-2)/(x-4)`

As `x>0` becomes increasing greater and greater the influences of the -2 and -4 become less and less significant. This continues until we have

`lim_(x->+oo) (x-2)/(x-4)color(white)("dd")->color(white)("dd")kcolor(white)("dd") ->color(white)("dd")oo/oo=+1`

As `x<0` becomes increasing less and less the influences of the -2 and -4 become less and less significant. This continues until we have

`lim_(x->-oo) (x-2)/(x-4)color(white)("dd")->color(white)("dd")kcolor(white)("dd") ->color(white)("dd")(-oo)/(-oo)=+1`