### Video Transcript

Which of the following statements
is true for the function β of π₯ equals negative one over seven minus π₯ minus
five? Is it (A) β of π₯ is decreasing on
the intervals negative β to seven and seven to β? Is it (B) β of π₯ is decreasing on
the intervals negative β to negative seven and negative seven to β? (C) β of π₯ is increasing on the
intervals negative β to negative seven and negative seven to β. Or (D) β of π₯ is increasing on the
intervals negative β to seven and seven to β.

If we look carefully, we see that β
of π₯ is a reciprocal function. Itβs one over some polynomial. And so we know that there are
probably going to be asymptotes on our graph. Letβs think about how we might
sketch the graph of β of π₯. Weβll begin by starting with the
function π of π₯ is equal to one over π₯. And then weβre going to consider
the series of transformations that map the function one over π₯ onto the function β
of π₯. Here is the function one over
π₯. It has horizontal and vertical
asymptotes made up of the π₯- and π¦-axis. Now weβll consider how we map π of
π₯ onto one over negative π₯. This is represented by reflection
in the π¦-axis.

And then how do we map this onto
the function one over seven minus π₯? Well, adding seven to the inner
part of our composite function gives us a horizontal translation by negative
seven. Thatβs a translation to the left
seven units. Now, in doing this, our horizontal
asymptote stays the same; itβs still the π₯-axis. But our vertical asymptote also
shifts left seven units. And so it goes from being the
π¦-axis, which is the line π₯ equals zero, to being the line π₯ equals negative
seven. But of course, β of π₯ is negative
one over seven minus π₯. This time, we reflect the graph in
the π¦-axis. And so our horizontal asymptote
remains unchanged, but our vertical asymptote is now at π₯ equals seven.

Our final transformation maps this
function onto β of π₯. Thatβs negative one over seven
minus π₯ minus five. And now we move the entire graph we
translated five units down. And so we now have the graph of β
of π₯ of negative one over seven minus π₯ minus five, and weβre ready to decide
whether the function is increasing or decreasing over the various intervals. Remember, if a function is
decreasing, its graph will have a negative slope, and if itβs increasing, its graph
will have a positive slope. As we move our values of π₯ from
left to right, that is, from negative β all the way up to π₯ equals seven, we see
that the graph is sloping downwards. It will approach negative β, but
never quite reach it.

Then as π₯ approaches positive β
from seven, the graph continues to slope downwards. This time, though, it approaches
negative five. And so the function is definitely
decreasing over these intervals from negative β to seven and seven to β. Since the function itself cannot
take a value of π₯ equals seven, and this is why we have the horizontal asymptote,
then we want to include open intervals. Those are the round brackets. And so the correct answer must be
(A), β of π₯ is decreasing on the open interval negative β to seven and seven to
β.