### Video Transcript

In this video, we will learn how to
graph rational functions whose denominators are linear, determine the types of their
asymptotes, and describe their end behaviors.

A rational function is an algebraic
fraction where both the numerator and denominator are polynomials. For example, π of π₯ equal to five
π₯ plus seven over two π₯ minus one is a rational function. Itβs important to note that a
regular polynomial function, for example, π of π₯ equals three π₯ minus five, is
also a rational function. We could rewrite three π₯ minus
five as three π₯ minus five over one, and one is a degree zero polynomial.

However, regular polynomial
functions like π of π₯ are slightly different to other rational functions. This is because any value of π₯ is
valid in the domain of π, whereas a rational function with a degree one polynomial
on its denominator, like this one, will have one value of π₯ for which the
polynomial is equal to zero. And therefore, the function π of
π₯ is undefined. In this case, if two π₯ minus one
is equal to zero, then π₯ is equal to one-half. Therefore, the value of one-half
must be excluded from the domain of π because it is undefined at this value.

The simplest rational function with
a non-zero-degree polynomial on its denominator is the reciprocal function π of π₯
equals one over π₯. The graph of the function π¦ equals
one over π₯ forms the shape of a hyperbola, which looks something like this. The hyperbola is symmetric about
the lines π¦ equals π₯ and π¦ equals negative π₯. In the first quadrant, the curve
approaches the line π¦ equals zero as π₯ tends to β without ever actually touching
it. The line π¦ equals zero or the
π₯-axis is called an asymptote. The curve also approaches the line
π¦ equals zero in the third quadrant, this time from below as π₯ tends to negative
β. Similarly, the curve approaches β
as π₯ tends to zero from the positive side and negative β as π₯ tends to zero from
the negative side. The graph of π¦ equals one over π₯
therefore has a horizontal asymptote at π¦ equals zero and a vertical asymptote at
π₯ equals zero.

We can use this most basic function
π¦ equals one over π₯ and apply function transformations to obtain different graphs
of different rational functions. In the first example, we will
identify the graph of a rational function using a function transformation from the
graph of π¦ equals one over π₯.

Which of the following graphs
represents π of π₯ equals one over π₯ plus one?

Letβs start by recalling the graph
of the simplest rational function π¦ equals one over π₯. This graph has a horizontal
asymptote at π¦ equals zero and a vertical asymptote at π₯ equals zero. We can obtain the graph of π of π₯
equals one over π₯ plus one by using a function transformation on π¦ equals one over
π₯. To transform the function one over
π₯ into one over π₯ plus one, we simply let π₯ map to π₯ plus one. Now, recall that a function
transformation of π₯ to π₯ plus π corresponds to a horizontal shift to the left by
π units. Therefore, the graph of π¦ equals
one over π₯ plus one is the same as the graph of π¦ equals one over π₯ shifted to
the left by one unit.

Looking at the possible answers, we
can see that graph (c) is the same as the graph of π¦ equals one over π₯ shifted to
the left by one unit. Its vertical asymptote π₯ equals
zero has shifted to the left to π₯ equals negative one. And its horizontal asymptote at π¦
equals zero is unchanged.

Now, letβs look at the opposite
kind of example where we are given a graph and asked to find the function.

What function is represented in the
figure below?

To begin with, notice that the
given graph resembles the graph of π¦ equals one over π₯. We can obtain this graph from the
graph of the parent function π¦ equals one over π₯ by applying a few function
transformations. The graph of π¦ equals one over π₯
has a horizontal asymptote at π¦ equals zero and a vertical asymptote at π₯ equals
zero. The given graph also has a vertical
asymptote at π₯ equals zero, but its horizontal asymptote is at π¦ equals negative
three. This means that a downward shift of
three units is one of the function transformations used to obtain the given graph
from the graph of the parent function. Before we apply this vertical
shift, we first need to check if any other function transformations are involved,
since the order of function transformations is very important.

There are three different types of
transformations to consider: translation, dilation, and reflection. We have already seen that there is
a translation of three units downwards. Looking at these two points
indicated on the given graph, if we translate these two points back up three units,
then we get the points negative one, one and one, negative one. These two points clearly are not on
the graph of the parent function π¦ equals one over π₯. Therefore, there must have been at
least one other transformation involved. They do look similar to the points
negative one, negative one and one, one. The two points are the same
distance apart and have the same absolute values for their π₯- and
π¦-coordinates. Therefore, there doesnβt seem to
have been any dilation involved, which would either increase or decrease the
distance between the points.

The given graph appears to be
upside down when compared with the parent function. More specifically, the value of π¦
approaches positive β as π₯ tends to zero from the negative direction. And it approaches negative β as π₯
tends to zero from the positive direction. This is exactly the opposite
behavior of the graph of the parent function, which implies that there is a
reflection involved. Due to the symmetry of the graph of
the parent function π¦ equals one over π₯, a reflection over the π₯-axis is the same
as a reflection over the π¦-axis. So, for default, we will say this
is a reflection over the π₯-axis.

When combining transformations,
dilations and reflections must be done before translations. A reflection over the π₯-axis means
we swapped the sign of the π¦-value of all of the points on the graph. So, π of π₯ goes to negative π of
π₯. This implies that we multiply the
original function by negative one. So, so far, we are taking the
parent function one over π₯ and multiplying it by negative one to give negative one
over π₯. Letβs clear a little space, and
letβs look at the graph of what we have so far: π¦ equals negative one over π₯. The graph has been reflected in the
π₯-axis. And the point negative one,
negative one has become negative one, one, and the point one, one has become one,
negative one. These are the two points that we
obtained on the given graph when shifting it up by three units.

So, now, all we need to do is apply
the translation of three units downwards. A vertical translation of π units
means adding π to the π¦-values of all of the points on the graph. So, π of π₯ goes to π of π₯ plus
π. And the sign of π will indicate
the direction the graph moves in. If π is positive, it will move
upwards, and if negative, it will move downwards. Therefore, to translate this graph
three units downwards, we subtract three from π of π₯. So, negative one over π₯ goes to
negative one over π₯ minus three. Therefore, the given graph
represents the function π of π₯ equals negative one over π₯ minus three. And we can substitute in the
original two points negative one, negative two and one, negative four to show that
they do indeed satisfy this equation.

In the next example, we will
determine missing parameters in a rational function from the given graph.

The graph shows π¦ equals π over
π₯ minus π plus π. A single point is marked on the
graph. What are the values of the
constants π, π, and π?

We begin by noticing that this
graph resembles the graph of π¦ equals one over π₯. We can obtain the given graph from
the graph of the parent function π¦ equals one over π₯ by applying some function
transformations. The graph of the parent function π¦
equals one over π₯ has a horizontal asymptote at π¦ equals zero and a vertical
asymptote at π₯ equals zero. The given graph has a horizontal
asymptote at π¦ equals negative two and a vertical asymptote at π₯ equals three. This means that a downward shift of
two units and a rightward shift of three units is one of the function
transformations used to obtain this graph from the graph of π¦ equals one over
π₯.

Before applying this translation,
however, we first need to check if any other transformations are involved, since the
order of transformations is very important. There are three different types of
transformations to consider: translation, dilation, and reflection. The given graph is oriented the
same way as the parent function, so we can rule out reflection. Dilation is a possibility, but itβs
difficult to judge by eye. Since we are given the point six,
negative one on the graph, we can use this point to determine the dilation factor,
if there is one. Recall that when combining
transformations, dilations and reflections must be done before translations.

Recall that a horizontal dilation
by a scale factor of π one means that we map π₯ to π₯ over π one. In other words, the π₯-values of
all the points on the graph are reduced by a factor of π one. A vertical dilation by a scale
factor of π two means mapping π of π₯ to π two times π of π₯. In other words, we multiply all of
the π¦-values on the graph by a factor of π two. Beginning with the parent function
π of π₯ equals one over π₯ and performing a horizontal dilation, we get one over π₯
over π one.

Next, performing a vertical
dilation, we multiply this new π of π₯ by π two, giving π two over π₯ over π
one. This simplifies to π one π two
over π₯. π one times π two is just another
constant in and of itself. So, we can call this constant
π. We do not yet know the value of
π. First, we need to move on to
translations. The positions of asymptotes are
unaffected by dilations. Therefore, these could only have
been moved by the translation of two units downwards and three units to the
right. A vertical translation of π units
means mapping π of π₯ to π of π₯ plus π. So, we change all of the π¦-values
of the points on the graph from π¦ to π¦ plus π. π being positive means the graph
moves upwards and negative means it moves downwards.

In our case, we need to perform the
dilation first. So, the parent function one over π₯
goes to π over π₯. Then, to translate this by two
units downwards, this goes to π over π₯ minus two. Now, recall that for a horizontal
translation by π units, we map all of the π₯-values to π₯ minus π. Again, the value of π will
determine the direction, with positive π giving a rightward shift and negative π
giving a leftward shift. But of course, the sign will be
reversed since we are subtracting π from π₯. So, starting from π over π₯ minus
two, to perform a rightward shift of three units, we need to change the π₯-values to
π₯ minus three. So, this gives us π over π₯ minus
three minus two. So, this graph represents the
function π of π₯ equals π over π₯ minus three minus two. So, we have found two of the
parameters in the question. π is equal to three and π is
equal to negative two.

Now we just need to find the value
of π. This equation has three unknowns in
it. We have π of π₯, π₯, and π. We are given a point on the graph:
six, negative one. And this corresponds to values of
π of π₯ and π₯, respectively, which we can then substitute into the equation and
rearrange to find π. So, we have π of π₯ equals
negative one and π₯ equals six. Simplifying and rearranging gives
π equals three. So, we now have the values of all
three constants. π is equal to three, π is equal
to negative two, and π is equal to three.

In the previous example, we saw how
dilation of a graph is difficult to judge by eye. So, we used an unknown scale
factor, π, and then found this scale factor by using a point on the graph. Specifically, if we start with the
reciprocal function π¦ equals one over π₯ and perform a horizontal dilation by a
scale factor of π one, we get one over π₯ over π one. And then performing a vertical
dilation by a scale factor of π two gives π two over π₯ over π one, which
simplifies to π one π two over π₯.

Because π one and π two are
multiplied together in the end, this transformation makes no distinction between
horizontal and vertical dilations. For example, if the horizontal
dilation factor π one is equal to one and the vertical dilation factor π two is
equal to two, then we get two over π₯. Conversely, if the horizontal
dilation factor is two and the vertical dilation factor is one, we get exactly the
same result, two over π₯. This is a clear demonstration of
the symmetry of the reciprocal function, but it does actually go further than
this.

If we take the function after
performing a horizontal and a vertical dilation and reflect it over the π₯-axis, we
swap the sign of π of π₯. So, this gives negative π one π
two over π₯. Conversely, if we reflect the
function over the π¦-axis, we swap the sign of π₯, which gives π one π two over
negative π₯. Rearranging this, we get exactly
the same result as the reflection over the π₯-axis. Whatβs more, just as we did in the
previous example, this numerator here is still a constant, which we can call π.

So, after a horizontal dilation and
a vertical dilation and a reflection over either axis or both, we get just π over
π₯. So, this scalar constant, π,
accounts for all horizontal and vertical dilations and all reflections. This greatly simplifies the
function transformation since we can account for all dilations and reflections by
simply multiplying by a constant π.

Now, all that remains are
translations. We can perform a horizontal
translation by π units by mapping π₯ two π₯ minus π. And we vertically translate the
graph by π units by adding π to the π¦-values. Therefore, a hyperbola with
vertical asymptote π₯ equals π and horizontal asymptote π¦ equals π is the graph
of a rational function π of π₯ equals π over π₯ minus π plus π, for some π not
equal to zero.

In the next example, we will use
this to identify the range of values for an unknown parameter in the given graph of
a rational function.

The graph shows π¦ equals π over
π₯ minus three minus two. We can see that the intersection of
its asymptotes is at three, negative two and that the points 0.5, negative 1.5 and
1.5, negative one are below and above the graph, respectively. Determine the interval in which π
lies.

Recall first that a hyperbola with
a horizontal asymptote at π₯ equals π and a vertical asymptote π¦ equals π is the
graph of the function π of π₯ equals π over π₯ minus π plus π, for π not equal
to zero. In this case, the vertical
asymptote is at π₯ equals three. So, π is equal to three. And the horizontal asymptote is at
π¦ equals negative two. So, π is equal to negative
two. This matches the equation of the
given function π of π₯ equals π over π₯ minus three minus two.

The value of π cannot be
determined exactly unless we have any of the points on the graph, which we do
not. Instead, we have one point above
and one point below the graph, which we can use to determine an interval in which π
must lie. Consider the point above the graph:
1.5, negative one. Since this point is above the
graph, its π¦-value is greater than π of π₯ at this point. In other words, π of 1.5 is less
than the π¦-value of this point, negative one. Therefore, π over 1.5 minus three
minus two is less than negative one. Taking the two to the other side
and simplifying gives us negative π over 1.5 is less than one. And rearranging for π gives us π
is greater than negative 1.5. We now have a lower bound for π of
negative 1.5.

Now, letβs consider the other point
below the graph: 0.5, negative 1.5. Since this point is below the
graph, its π¦-value is less than the π¦-value of π of π₯ at this point. In other words, π of 0.5 is
greater than negative 1.5, which means that π over 0.5 minus three minus two is
greater than negative 1.5. Taking the two over to the
right-hand side and simplifying gives us negative π over 2.5 is greater than
0.5. Solving for π gives us the upper
bound of π. π is less than negative 1.25. This gives us our final answer, the
interval for π. π is greater than negative 1.5 and
less than negative 1.25.

Letβs now finish this video by
recapping some key points. Unlike a graph of a nonconstant
polynomial, the graph of a rational function may have vertical and horizontal
asymptotes, that is, straight lines which the graph may approach but never
touch. The graph of π¦ equals one over π₯
is a hyperbola with a horizontal asymptote π¦ equals zero and a vertical asymptote
π₯ equals zero. And finally, a hyperbola with
vertical asymptote π₯ equals π and horizontal asymptote π¦ equals π is the
transformed graph of a rational function π of π₯ equals π over π₯ minus π plus
π, for some π not equal to zero.