### Video Transcript

In this video, we will learn how to
identify transformations involving horizontal and vertical shifts or
translations. Letβs say that weβre given this
function π of π₯ equals two π₯ squared plus three. Itβs important to remember that a
function is just a series of inputs and outputs. In this case, the π₯ is the
input. And when we substitute a specific
input value into the expression on the right-hand side and evaluate it, then we get
a particular output. Weβll now take a look at the
different ways in which we can change the input or the output values of a
function. And weβll see how those changes
affect the way the function looks.

When we first started learning
about functions, we may have considered function machines. In a function machine, when π₯ is
the input, then π of π₯ or the function of π₯ represents the output. We can think of any of the
transformations in one of two ways. The first way is by changing the
output values. For example, we might add on three
to every one of the outputs. This means that the input values
stay the same, but every π¦-value would have three added on. Alternatively, we might change the
input values, for example, by adding on three. That means that instead of just
having π₯ as the input, we would add on three to every one of the original input
values and see what their corresponding output values would be.

It might be tempting to think that
both of these options would produce the same result. However, weβll start by looking at
how changing the output changes a function. And then weβll compare that with
how changing the input changes a function.

Letβs say that weβre given this
graph of π¦ equals π of π₯. We can transform this function by
adding four onto the output or π of π₯ values. Letβs start by taking this
coordinate zero, negative two on the original function. And we can even consider this in
terms of a function machine. The coordinate zero, negative two
means that for an input of zero, the output would be negative two. On the transformed function,
however, we need to add four to each of the outputs. So an input of zero would give an
output of negative two plus four. As negative two plus four is two,
then that means that our transformed function has the coordinate zero, two.

We can take a different point on
the original function. This coordinate of negative two,
two means that for an input of negative two, the output is two. Transforming this by adding four
means that weβll have a new π¦-value of six. So we have the coordinate negative
two, six. Taking a third coordinate, this
time the point two, two, we can see how the transformed function would have the
coordinate two, six. We can draw the transformed
function π¦ equals π of π₯ plus four by joining these three points with a
curve.

So what exactly do we notice with
this transformed function? Firstly, we notice that every
coordinate has been shifted upwards by four units. We should also note that the
transformed function is exactly the same shape and size as the original
function. Itβs just being translated. We could make a general note then
that if we transform a function by adding π units to it for any real value of π,
then the new function can be written as π¦ equals π of π₯ plus π and the graph
shifts upwards by π units.

Letβs see how this function would
change if instead of adding four to the output, we subtract four. We can start once again by looking
at this coordinate zero, negative two, remembering that that means for an input of
zero, thereβs a corresponding output of negative two. And so we can write the coordinate
on our transformed function as zero, negative six. Looking at a different point then,
the coordinate negative two, two, when thatβs transformed, we subtract four from the
output. Two minus four is negative two. So we know that the coordinate
negative two, negative two will lie on the transformed function.

We might notice the pattern here
that this time the function is being translated four units downwards. Verifying this with another
coordinate, this time the point three, seven, we can see how this new coordinate of
three, three will appear on the transformed function. We can then see what this
transformed function would look like by drawing a smooth curve through the
points. And a note about drawing functions,
donβt worry if you canβt get your curve wonderfully smooth. What examiners are really looking
for are the key points marked and the correct general idea of the shape of the
function.

We can now add to the previous note
if weβre transforming a function by subtracting π for any real value of π, then
the new equation can be written as π¦ equals π of π₯ minus π. And the graph will shift downwards
by π units. In general then, if weβre
transforming a function by adding or subtracting values to the output, then the
graph will translate in a vertical direction. Now that we have looked at changing
the outputs, letβs see what happens when we change the inputs of a function. We can keep the same representation
of π¦ equals π of π₯. And this time, weβre going to
change the input values by adding on three.

We can take this coordinate of
zero, negative two remembering that that means for this function an input of zero
gives a corresponding output of negative two. When the input was π₯, this output
was nice and simple to figure out. However, letβs see what happens
when the input is actually π₯ plus three. When the input of zero is actually
equal to π₯ plus three, then that means that the value of π₯ must be negative three
since negative three plus three gives us zero. This means that we can plot one of
the coordinates of the transformed function at negative three, negative two.

We can now check a different
coordinate. Letβs look at the point negative
two, two. Remember, that means that when the
input is negative two, the output is two. So when π₯ plus three is equal to
the input negative two, that means that the value of π₯ must be negative five. This time, the coordinate that
weβre plotting will be negative five, two. You might already be starting to
spot a pattern. It looks as though the transformed
graph is three units to the left of the original function. We can check another coordinate
just to be sure. We can take the coordinate two,
two. When weβre transforming the
function, weβre saying that this input of π₯ plus three is two. So that means that π₯ must be equal
to negative one.

Plotting this coordinate on our
graph, we can confirm that thereβs been a translation of three units to the
left. The transformed function π¦ equals
π of π₯ plus three would look like this. We can make a note then that if we
transform a function by adding π to the input values for any positive real value of
π, then the new function can be written as π¦ equals π of π₯ plus π and the graph
will shift left by π units.

We can now consider what happens
when we transform a function by subtracting three units from the input. We can start once more with this
coordinate of zero, negative two. Remember that once again we donβt
have this nice, simple situation where π₯ equals zero is considered as the input,
but instead we have π₯ minus three going into the function machine. So when π₯ minus three is equal to
zero, that means that π₯ must be equal to three. And thatβs our first coordinate:
three, negative two.

Taking another coordinate, this
time the coordinate three, seven, remembering that when weβre transforming it, weβre
saying that the input three is actually equal to π₯ minus three. So the π₯-value is six, and itβs
giving us the coordinate of six, seven. So this time we have a translation
of three units to the right. And thatβs true for any coordinate
we might choose. We can join the points to get the
graph of the transformed function π¦ equals π of π₯ minus three like this.

For our notes then, we can say that
when we transform a function by subtracting π from the input for any real value of
π, then the new equation can be written as π¦ equals π of π₯ minus π and the
graph shifts right by π units. This transformation can actually
feel a little bit unsettling. Weβll think that if we add π
units, then the transformation will be right. But in fact the reverse is
true. It will shift left. And when we subtract π from the
input, weβll not be translating left; weβll be translating to the right.

It can be worthwhile learning these
transformations, but itβs also very important to understand whatβs happening and
why. Weβll now look at some questions
involving the transformation of functions.

Which of the following is the graph
of π of π₯ equals the cube root of π₯ minus one?

There are several different ways in
which we can approach a question like this. One way is to take several
different input values or π₯-values and see what their corresponding output or π of
π₯ values would be. Letβs say that we take the values
of π₯ of negative one, zero, and one. When π₯ is equal to negative one,
the cube root of negative one is negative one. And then subtracting one from that
will give us negative two. When π₯ is equal to zero, the cube
root of zero is zero, and subtracting one will give us negative one. Finally, when π₯ is equal to one,
the cube root of one is one, and subtracting one will give us an output of zero.

We then know that the graph must
contain the coordinates negative one, negative two; zero, negative one; and one,
zero. The only one of these graphs which
passes through these points is that given in option (b). And so that would be the
answer. An alternative method of solving
this question would be one involving the transformation of functions. We can consider the function π of
π₯ equals the cube root of π₯. And notice that the function that
we were given is different because one has been subtracted from the output.

We can draw a quick sketch of the
graph of π of π₯ equals the cube root of π₯. We can then use the fact that if we
transform a graph by subtracting π units from the output for any real value of π,
then the graph will shift down by π units. In this case, we were subtracting
one from the output, so our graph of π of π₯ equals the cube root of π₯ will shift
downwards by one unit. The graph which demonstrates this
is the one given in option (b). And so that is the answer.

Letβs take a look at another
question.

The function π¦ equals π of π₯ is
translated eight down. Write in terms of π of π₯ the
equation of the translated graph.

In these questions, weβre given a
function π¦ equals π of π₯. We donβt know what it looks
like. For example, it could be a straight
line, a quadratic, or even something like a sine wave. However, it doesnβt matter what
this function looks like as weβll still be able to write the equation of this graph
as weβre given the information about its translation.

Because weβre told that the
transition is downwards, then we can use the fact that when we change the output of
a function by subtracting π for any real value of π, then the graph shifts π
units down. And the new equation can be written
as π¦ equals π of π₯ minus π. In this question, as weβre told
that weβre shifting eight units down, so we subtract eight from the output. And so the equation of the
translated graph is π¦ equals π of π₯ minus eight.

Weβll now look at a similar
question. Only, this time, the translation
will be in a horizontal direction.

The function π¦ equals π₯ minus one
times two π₯ minus three times four minus π₯ is translated two units in the
direction of the positive π₯-axis. What is the equation of the
resulting function?

In this question, weβre given this
function and the translation. But donβt worry; we donβt need to
actually graph this function in order to be able to answer the question. We can instead apply some logic to
this transformation given that weβre told that itβs a translation in the direction
of the positive π₯-axis. The direction of the positive
π₯-axis is just another way of saying that this translation occurs to the right.

When a graph has shifted right by
π units, that means that the input has changed to π₯ minus π for any real value of
π. So this graph, which has shifted
right by two units, means that the input becomes π₯ minus two. To find the equation of the
transformed function, then we take every value of π₯ and replace it with π₯ minus
two. So thatβs going to be relatively
simple. For the first set of parentheses,
weβll have π₯ minus two minus one. For the second set of parentheses,
weβll need to be a little more careful. The original input of π₯ was
multiplied by two. So that means that all of π₯ minus
two will also need to be multiplied by two.

For the final expression, because
we subtracted π₯ from four, weβll also need to subtract all of π₯ minus two from
four. We can then simplify our
equation. The first set of parentheses will
simplify to π₯ minus three. Next, when we distribute two across
the parentheses containing π₯ minus two, weβll have two π₯ minus four. And subtracting three from that
will give us two π₯ minus seven.

In the final expression,
distributing the negative across the parentheses gives us four minus π₯ plus two,
which simplifies to six minus π₯. We can then give the answer for the
equation as π¦ equals π₯ minus three times two π₯ minus seven times six minus
π₯.

We can now look at a question
involving more than one transformation.

The following graph is a
transformation of the graph of π¦ equals the absolute value of π₯. What is the function it
represents? Write your answer in a form related
to the transformation.

In this question, weβre given the
graph of a function which weβre told is the transformation of the graph of π¦ equals
the absolute value of π₯. So it might be worth recalling what
the graph of π¦ equals the absolute value of π₯ actually looks like. We can consider a similar range of
π₯-values. When π₯ is a positive value, then
its absolute value will be equivalent. However, when π₯ is a negative
value, its absolute value will simply be its magnitude.

And so the graph of π¦ equals the
absolute value of π₯ will look like this. So how is this graph different to
the one that we were given? The first thing we might notice is
that this graph looks upside down. In other words, thereβs been a
reflection in the π₯-axis. However, if we just had a
reflection, then the graph would look like this in orange. So some other transformations must
also have occurred.

Letβs consider how the top of each
of these graphs is different. We can compare the coordinates
zero, zero and negative one, four. This means that there must have
been a shift of one left and four up, remembering that weβre transforming this graph
of π¦ equals the absolute value of π₯ to the unknown function and not the other way
round. We now need to write these three
transformations in terms of an actual equation. A reflection in the π₯-axis occurs
when the output π of π₯ is mapped to negative π of π₯. This is because every π¦-value on a
transformed function is simply the negative of its original value.

A shift of one left means that the
input value has changed to π₯ plus one. And the function can be written as
π¦ equals π of π₯ plus one. A shift of four up means that the
output has increased by four. And the new function can be written
as π¦ equals π of π₯ plus four. We can now apply these three
changes to the graph of π¦ equals the absolute value of π₯. Letβs start by applying the
reflection. This gives us π¦ equals negative
the absolute value of π₯. Then instead of having π₯ as the
input, weβll have π₯ plus one. Once weβve done that, we can add
four to the output. We can then give our answer in the
slightly nicer form as π¦ equals four minus the absolute value of π₯ plus one. But either of these two function
forms would be valid.

We can now summarize some of the
key points of this video. We saw how we can transform a
function by changing the input π₯ or the output π of π₯. We can summarize the changes to the
input or the output in this table. We can remember that the changes to
the output are the ones that perform in the way we expect them to. By that we mean that when we add π
to the output, the graph shifts upwards by π. When we subtract π, the graph
shifts downwards by π.

The changes to the input, however,
are the ones that are more tricky. For example, when we add π to the
input, weβll think that the graph will shift right in a positive direction. But actually it shifts left. Subtracting π means that the graph
will shift right. So we need to be careful when weβre
translating functions.