How do you graph $y=3/(x-3)+1$ using asymptotes, intercepts, end behavior?

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Answer 1 · 2016-12-29T13:56:04

Vertical asymptote : $uarr x = 3 darr$ .
Horizontal asymptote: $larr y = 1 rarr$ .
x-intercept ( y = 0 ) : 0. y-intercept ( x = 0 ): 0. See graph.

graph{(y-1)((y-1)(x-3)-3)=0 [-20, 20, -10, 10]}

The given equation has another form

$(y-1)(x-3)=3$

This is an example to show that the indeterminate from $0 X oo$

can take a finite limit, including 0.

As $x to +-oo$ ,

the other factor $(y-1)$ ought to $to 0$ , giving $y to 1$ .

Likewise, as $y to +-oo$ ,

the other factor $(x-3) to 0$ , giving $x to 3$ .

So, the asymptotes are given by x = 3 and y = 1.

If the limit of the product is 0, we directly get the pair of asymptotes

$(y-1)(x-3)=0$ .

This is the logic behind the structure

#(y-ax-b)((y-a'x-b'x-c')=k

for the equation of a hyperbola that has the asymptotes given by

setting k = 0, in this form.

How do you graph y=3/(x-3)+1y=3/(x-3)+1 using asymptotes, intercepts, end behavior?