### Video Transcript

Find the horizontal and vertical
asymptotes of π of π₯ equals one over π₯ minus one minus two.

We begin by recalling the
definition of a horizontal and vertical asymptote. We say that the line π¦ equals πΏ
is a horizontal asymptote of a function π¦ equals π of π₯ if the limit as π₯
approaches either negative or positive β of π of π₯ is equal to πΏ. Similarly, π₯ equals π is a
vertical asymptote of a function π¦ equals π of π₯ if any of the six statements are
true. That is, the limit as π₯ approaches
π from the left of the function is either positive or negative β. The limit as π₯ approaches π from
the right of the function is either positive or negative β. Or the limit as π₯ approaches π of
the function is positive or negative β.

Now, our function π of π₯ is equal
to one over π₯ minus one minus two. So weβll begin by evaluating the
limit as π₯ approaches β of one over π₯ minus one minus two. And in fact, very little
manipulation is required here. As π₯ gets sufficiently large, one
over π₯ minus one approaches zero. And so the limit as π₯ approaches β
of one over π₯ minus one minus two is zero minus two, which is simply negative
two. And in fact, as π₯ approaches
negative β, one over π₯ minus one still approaches zero. So the limit as π₯ approaches
negative β of π of π₯ is equal to negative two. And we find the equation π¦ equals
negative two is a horizontal asymptote of our function.

Next, weβll consider the vertical
asymptotes. Less formally, we say that, with
rational functions, thatβs those written as the quotient of two functions, vertical
asymptotes are the lines which correspond to the zeros of the denominator of the
function. So in fact, weβre going to begin by
subtracting two from one over π₯ minus one by creating a common denominator. We multiply two over one by π₯
minus one over π₯ minus one. And then, we can subtract the
numerator. So we have one minus two times π₯
minus one over π₯ minus one. We simplify. And we find that π of π₯ can be
written as three minus two π₯ over π₯ minus one.

Weβre interested in the values of
π₯ so that our denominator π₯ minus one is equal to zero. Well, if we solve this equation for
π₯ by adding one to both sides, we find that π₯ is equal to one. And so we found the horizontal and
vertical asymptotes of π of π₯ equals one over π₯ minus one minus two. The horizontal asymptote is π¦
equals negative two. And the vertical asymptote is π₯
equals one.