# What is the line of symmetry of the graph of ##y=1/(x-1)##?

The graph is a hyperbola, so there are two lines of symmetry: ##y=x-1## and ##y=-x+1##

The graph of ##y = 1/(x-1)## is a hyperbola.

Hyperbolas have two lines of symmetry. both lines of symmetry pass through the center of the hyperbola. One goes through the vertices (and through the foci) and the other is perpendicular to the first.

The graph of ##y=1/(x-1)## is a translation of the graph of ##y=1/x##.

##y = 1/x## has center ##(0,0)## and two of symmetry: ##y = x## and ##y = -x##

For ##y = 1/(x-1)## we have replaced ##x## by ##x-1## (and we have not replaced ##y##. This translates the center to the point ##(1,0)##. Everything moves ##1## to the right, the graph, the asymptotes and the lines of symmetry.

##y = 1/(x-1)## has center ##(1,0)## and two of symmetry: ##y = (x-1)## and ##y = -(x-1)##

One way of describing this is that we translate the lines of symmetry just as we did the hyperbola: we replace ##x## with ##x-1##

The two lines are, therefore, ##y=x-1## and ##y = -x+1##

Bonus example

What are the lines of symmetry of the graph of: ##y = 1/(x+3) +5##?

Try to work it out yourself, before reading the solution below.

Did you get: ##y = x+8## and ##y = -x+2##?

If so, you are correct.

We can rewrite the equation to make the translations more clear:

##y = 1/(x+3) +5## can be written

##y-5 = 1/(x+3)## or, perhaps better yet,

##(y-5) = 1/((x+3))##

It is clear that starting with ##y=1/x##, I have replaced ##x## by ##x+3## and replaced ##y## with ##y-5##

That moves the center to ##(-3, 5)##. (Yes it is like finding the center of a circle.)

The lines of symmetry get translated as well:

Instead of ##y=x##, we have: ##(y-5) = (x+3)## and

instead of ##y = -x##, we have ##(y-5) = – (x+3)##.

Now put the lines in slope intercept form to get the answers I gave.

By the way: the asymptotes of ##y=1/x## are ##y=0## and ##x=0##, so the asymptotes of ##y = 1/(x+3) +5## are:

##(y-5) = 0##, usually written: ##y = 5##, and

##(x+3) =0##, usually written: ##x = -3##.