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

1 Answer
Dec 29, 2016

Answer:

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

Explanation:

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.