### Video Transcript

In this lesson, weβll learn how to
identify function transformations which include a combination of translations,
dilations, and reflections. Weβll begin by recalling the key
function transformations that we should know. These can be grouped as
transformations that cause horizontal changes to the function, and theyβre achieved
by implementing changes to the π₯-variable, and those that cause vertical changes,
which are achieved by changing the function itself.

Given a function π of π₯, the
function π of π₯ plus π for some real number π is a translation π units to the
left or by the vector negative π, zero. And the function π of π₯ plus π
is a translation π units up or by the vector zero, π. Next, we can multiply π₯ by some
constant π giving us the function π of ππ₯. This leads to a horizontal stretch
or compression by the scale factor one over π. And multiplying the entire function
by some constant π to give us ππ of π₯ is a vertical stretch by a scale factor of
π. Now, in fact, in our previous
transformation, if π is less than zero, this also results in a reflection in the
π¦-axis. And if π is less than zero, this
results in a reflection in the π₯-axis.

In the simplest terms, we can
separate the situation where π and π are equal to negative one. The function π of negative π₯ is a
reflection in the π¦-axis, and the function negative π of π₯ is a reflection in the
π₯-axis. As an aside, we notice that for
horizontal transformations, the effects of additional multiplication are the
opposite of what we would expect. For example, π of π₯ plus three,
that is, adding three to the π₯ before inputting it to the function, results in a
geometric transformation which shifts the graph three units to the left. A multiplication by two of the
value of π₯ results in a horizontal stretch by a scale factor of one-half. On the other hand, algebraic
transformations on the function itself result in the expected geometric
transformations. So with this in mind, letβs recap
how to identify function transformations.

If the following graph is reflected
in the π₯-axis and then shifted to the left by one unit and down by three units,
what is the equation of the new graph?

Remember, given some function π of
π₯, the corresponding function negative π of π₯ represents a reflection of the
original graph in the π₯-axis. In order to represent a shift or
translation left by one unit, we need to convert π of π₯ onto the function π of π₯
plus one. Finally, given some function π of
π₯, the corresponding function π of π₯ minus three is found by translating the
original function down by three units. So weβre going to take the original
function, which is given by π¦ is equal to the absolute value of π₯, and then
complete each of these transformations in turn.

To perform a reflection in the
π₯-axis, the function π¦ equals the absolute value of π₯ needs to be changed to π¦
is equal to negative the absolute value of π₯. And the corresponding graph of this
function is shown. Then, in order to shift this to the
left one unit, we add one to the value of π₯. And we do this to the new
function. So the function we have now is π¦
is equal to the negative of the modulus of π₯ plus one, and the corresponding graph
of this function is shown.

Finally, in order to translate this
graph three units down, we subtract three from the entire function. So we take the function we have and
we subtract three from it, giving us π¦ equals negative the modulus of π₯ plus one
minus three. And the corresponding graph of this
function is shown. So after this series of
transformations, the equation of the new graph is π¦ equals the negative modulus of
π₯ plus one minus three.

In our next example, weβll apply a
combination of a vertical dilation or stretch and a reflection to a given
function.

The function π of π₯ equals π₯
plus two cubed plus three is stretched in the vertical direction by a scale factor
of two then reflected in the π¦-axis. Write the equation of the
transformed function π of π₯.

There are two transformations that
have been applied to our function π of π₯. Firstly, itβs stretched in the
vertical direction; then itβs reflected. Now, we might recall that in order
to stretch a function in the vertical direction, we need to multiply that entire
function by the scale factor. So π of π₯ will be mapped onto two
times π of π₯ to achieve this. And given some function π of π₯,
the corresponding function π of negative π₯ represents a reflection in the π¦-axis
of that original function. Now, the order in which we apply
these is important. So we will follow the order in the
question, starting with the vertical stretch.

We have π of π₯ equals π₯ plus two
cubed plus three, and weβre going to multiply the entire function by two. When we do, we find that two π of
π₯ is equal to two times π₯ plus two cubed plus three. Then we can distribute this two
across the parentheses and we have that two times π of π₯ is two times π₯ plus two
cubed plus six. Now that weβve performed the
stretch in the vertical direction, weβre going to perform a reflection in the
π¦-axis.

To achieve this, we change the π₯
to a negative. In other words, we essentially
multiply the value of π₯ by negative one. Since weβre starting off with our
already transformed function, we need to change two π of π₯ to two π of negative
π₯. And all we do here is we multiply
the value of π₯ by negative one. So two π of negative π₯ is two
times negative π₯ plus two cubed plus six. And so we have the equation of the
transformed function, which we can now write as π of π₯. π of π₯ is two times negative π₯
plus two cubed plus six.

In our previous examples, we found
algebraic expressions of functions after a combination of transformations were
applied in a given order. Weβre now going to consider
examples in the reverse. So given an algebraic expression of
a function which is a result of a series of transformations, how do we retrace the
steps to identify all the transformations and obtain the given expression? Well, we need to know the order in
which we complete a series of several transformations if weβre given the algebraic
expression. And just as thereβs an order of
operations for calculations, for instance, we can apply an order for our combination
of transformations.

Firstly, we look for any horizontal
translations and we apply those first. Then we look for any stretches or,
of course, compressions. Remember, we can have stretches in
both the horizontal and vertical direction, and it really doesnβt matter which order
we apply those. Then, we complete any reflections
in either the π₯-, π¦-axis, or both. And then a vertical translation
will be the last thing that we do. Now, of course, this is hugely
important when applying a combination of transformations. But there will be occasions where
different combinations and orders of transformations may result in fact the same
effect on the graph of a function.

At this stage, though, we donβt
have any way to determine whether that will be the case. So weβll follow the order as
shown. Letβs demonstrate this in an
example.

This is the graph of the
exponential function π¦ equals π of π₯. Which of the following is the graph
of π¦ equals four minus π of two π₯?

Letβs compare the original equation
to the equation of our transformed function. Three distinct things have happened
here. Weβve added four, weβve multiplied
the entire function by negative one, and weβve multiplied the π₯ by two. But we need to be careful to ensure
that weβve done these in the correct order. The order is as follows. We begin by looking for any
horizontal translations. Is the graph going to move left or
right? Then we stretch it, and we can do
that in both directions if necessary. We then perform any reflections
either across the π₯- or π¦-axis or both. And finally, we deal with any
vertical translations. Does the graph move up or down?

So a horizontal translation occurs
when we add some constant to the value of π₯. So π of π₯ maps onto π of π₯ plus
π by a translation π units left. Now, if we look at our function,
nothing has actually been added to the value of π₯. And so we arenβt going to be
applying any horizontal translations. But we are going to be applying a
stretch. If we have π of π₯, π of π times
π₯ is a stretch in the horizontal direction by a scale factor of one over π. We have an expression of this
form. In fact, since the coefficient of
π₯ is two, since weβre multiplying our value of π₯ by two, we have a horizontal
stretch by a scale factor of one-half. We can add this interim graph to
our diagram.

We know it will still pass through
the π¦-axis at two. Then the π₯-value of the point with
coordinates one, approximately 0.7 will half, and we get the coordinate 0.5,
0.7. Similarly, the curve passes through
the point two and approximately 0.2. The π₯-value will halve at this
coordinate. And we get one, 0.2. Up here, we have a point which has
coordinates negative 0.6 roughly and four. And halving the π₯-coordinate, that
becomes negative 0.3, four. And so the graph of π¦ equals π of
two π₯ looks as shown, and we can see that thatβs got that horizontal
compression. Itβs a stretch by a scale factor of
one-half. And what about reflections? Well, these occur when we either
change the coefficient of π₯ to be negative or we multiply the whole function by
negative one.

Well, we do indeed have negative π
of two π₯. This is going to result then in a
reflection in the π¦-axis. And so whilst we canβt quite fit
the whole graph on, we know itβs going to pass through negative two on the π¦-axis
as shown. So weβve performed a stretch. Weβve performed a reflection. What about a vertical
translation? This would be of the form π of π₯
plus π. So given an original function π of
π₯, we map it onto π of π₯ plus π by moving it π units up.

If we rewrite our function as π¦
equals negative π of two π₯, which are the transformations weβve performed, plus
four, we should now be able to see that we need to move our graph four units up. And so it will now pass through the
π¦-axis at two. And out of our options for (a),
(b), and (c), this is graph (b). So graph (b) is the graph of π¦
equals four minus π of two π₯.

So weβve now seen how to apply a
series of transformations and the order in which they should be applied if
necessary. And of course, it follows that to
reverse a series of transformations, in other words, to work out the original
function, weβd need to work backwards through these steps. And if weβre given the order in
which a series of transformations is applied, we can also work out the original
function by reversing those. So letβs see what this would look
like.

The graph represents a function π
of π₯ after a vertical positive shift by three units followed by a reflection on the
π₯-axis. Which of the following represents
the original function π of π₯? Is it (A) π of π₯ equals negative
π₯ plus two cubed minus two? Is it (B) π of π₯ equals negative
three π₯ minus one cubed minus six? Is it (C) π of π₯ equals π₯ minus
one cubed plus two? (D) π of π₯ equals negative π₯
minus four cubed minus two. Or (E) π of π₯ equals negative
three π₯ minus one cubed plus two.

Weβre told that π of π₯ comes from
the original function π of π₯ after applying two transformations. We can recover the original
function π of π₯ from π of π₯ by reversing the transformations but beginning from
the second one. The process will give π of π₯ as
an expression involving π of π₯. And then we can finish this problem
by finding an equation for π of π₯.

So letβs begin with the second
transformation which is a reflection over the π₯-axis. And we might remember that given
some function β of π₯, the corresponding negative β of π₯ is a reflection of the
original graph across the π₯-axis. We reverse it by applying the same
transformation. So we take π of π₯ and we map it
onto negative π of π₯.

Next, weβll reverse the first
transformation. That was a vertical positive shift
by three units. The opposite to this is a vertical
negative shift, in other words, shifting or translating the graph three units
down. Now, we achieve that by taking the
function β of π₯ and subtracting three from it. Or, in the case of our transformed
function negative π of π₯, we subtract three from that, so we have negative π of
π₯ minus three. And since weβve reversed two
transformations, this must be equal to our original function π of π₯.

So all we need to do now is find
the equation for π of π₯. We simply look at the shape and we
can already see that it must come from the parent function π¦ equals π₯ cubed. Itβs definitely a cubic graph. But of course if the coefficient of
π₯ cubed is positive, the cubic graph has the opposite shape. In other words, it looks like this
graph but reflected across the π¦-axis.

So to achieve a reflection across
the π¦-axis, we make the entire function negative. So π¦ equals negative π₯ cubed
definitely has the correct shape and orientation. But of course, π¦ equals π₯ cubed
and π¦ equals negative π₯ cubed intersect the π¦-axis at zero. In fact, this is also the
stationary point of the function or the point of inflection. On our curve, that corresponds to
the point with coordinates one, negative five. So we take π¦ equals negative π₯
cubed and translate it one unit right and five down. Translating it one unit right
occurs when we subtract one from the value of π₯. And to move it five units down, we
simply subtract five from the whole function. So it certainly appears as if we
have the equation for our function π of π₯.

It might be sensible just to check
this by substituting a couple of the coordinates in. For instance, when π₯ is equal to
negative one, π¦ is equal to negative negative one minus one cubed minus five, which
equals three as we expected. Similarly, when π₯ is equal to
zero, π¦ is equal to negative zero minus one cubed minus four, which is negative
four also as we expected. So with some certainty, we can be
assured that we have the correct equation for this graph. So π of π₯ is negative π₯ minus
one cubed minus five, and that means we can now find an expression for π of π₯.

By replacing π of π₯ with this
expression in our equation for π of π₯, we have π of π₯ is negative negative π₯
minus one cubed minus five minus three. And then we distribute the
parentheses to get π₯ minus one cubed plus five minus three, which is π₯ minus one
cubed plus two. And we can now look at our options
and see that that corresponds to option (C) π of π₯ is equal to π₯ minus one cubed
plus two.

Weβve now demonstrated several
examples of how to apply a series of transformations. Weβve used graphs and weβve used
algebraic expressions. And weβve also seen how to reverse
such processes.

Letβs recap the key points. The order in which we apply a
combination of transformations is followed. We find horizontal translations
first, then perform any stretches before performing reflections, finally identify
any vertical translations. And whilst this order is important
because applying them in a different order can result in a different final graph,
sometimes different combinations and orders of transformations may actually result
in the same effect of the graph of a function.