### Video Transcript

In this series of videos weβve looked at what happens to the graphs of functions when you transform them
in different ways. Weβve seen vertical and horizontal translations and vertical and horizontal stretches. Now weβre gonna
take a few minutes to look up what happens to the equation of a function when we transform it.

Weβve seen that π of π₯ plus π is a transformation that translates
π of π₯ by negative π units in the π₯-direction. So if π is positive, itβs gonna translate the curve of the
function to the left. And if π was negative, itβs gonna translate the curve of the function to the right. Weβve also seen
the π of π times π₯ stretches π of π₯ by a factor of one over π in the π₯-direction. And that fixes the curve at the π¦-axis and stretches the curve away from the π¦-axis. Although depending
on the value of π, it might squash the curve towards the π¦-axis.

And we have also said that π of π₯ plus π translates function π of π₯ by π
units in the π¦-direction. If π is positive, it shifts the curve upwards. And if π is negative, it shifts the curve
downwards.

And weβve seen that π times π of π₯ stretches the function π of π₯ by a factor of
π in the π¦-direction. Now in this case itβs gonna lock off all points on the π₯-axis. And itβs either gonna stretch that curve
away from the π₯-axis or itβs going to squash it towards the π₯-axis. So thereβre various different things that can happen for different values of π.

Now letβs think about some equations and see the effect of these transformations on them. First letβs think about π of π₯ plus π. So for example,
if π of π₯ with three π₯ squared plus two π₯ plus one and π was equal to five, then as we said in the introduction that curve of π¦ equals π of π₯ is translated five places
in the positive π¦-direction.

So take any point on the π¦ equals π of π₯ curve. Add five to its π¦-coordinate. And youβre gonna
map onto the π¦ equals π of π₯ plus five curve. So thatβs the graph, but letβs think of the equation of π of π₯ plus five.

Well weβre gonna start off with the expression for π of π₯, so three π₯ squared plus two π₯ plus one. And then weβre gonna be adding five to that, which simplifies to three π₯ squared plus two π₯ plus six.

Now in this format here itβs pretty easy to say that weβve taken the π¦-coordinates that we had from the
π of π₯ function and we have just added five to it. But in this format here itβs perhaps not quite so immediately
obvious whatβs happened in terms of that transformation, even though itβs not a very dramatic change to the equation.

Now letβs think of transformations of the type π times π of π₯. For example, π of π₯, again weβre gonna take three π₯ squared plus two π₯ plus one. And π is gonna be equal to five. If we look at the curves of those two
functions: π¦ equals π of π₯ and π¦ equals five times π of π₯, all we have done is weβve taken the
π¦-coordinates from π¦ equals π of π₯ and we multiplied them by three. Looking at the curves of those functions, we can see that the π¦-coordinates
from π¦ equals π of π₯ have just been multiplied by five to give us the π¦-coordinates
of π¦ equals five times π of π₯.

So one times five gives us five. Two times five gives us ten. Four times five gives us twenty and so on.

Now looking at the equation of the function five times π of π₯ is equal to five times three π₯ squared plus two π₯ plus one. And we can distribute that five across the parentheses and simplify the expression to five of π₯ is equal to fifteen π₯ squared plus ten π₯ plus five. Now in this simplified format, itβs pretty much the same format. But all of the coefficients are five times larger. But it is not immediately obvious what that transformation is doing.
So if you look at the other format here, so we got the same π¦-coordinates we have for π of π₯, but we just multiplied them all by five.

So just like before by simplifying that expression, weβve actually put it into a format which makes it
harder to work out exactly what the transformation was doing, multiplying all of the π¦-coordinates by five.

Well now letβs have a look at transformations of the type π of π₯ plus π. So again weβre
gonna use π of π₯ is equal to three squared plus two π₯ plus one. And weβre gonna use π equals five. And hopefully remember from the introduction that a transformation like π of π₯ plus π translates
the whole curve π units to the left. So because π is equal to five, weβve translated the π of π₯ curve five
units to the left.

Look for example, this point on π¦ equals π of π₯ maps over to here five places to the left
on π¦ equals π of π₯ plus five.

And similarly this point gets transformed to here. But itβs a little bit tricky when we try to think about what the function is going to look like in
terms of the equation. In π of π₯ we took π₯ and then we squared it and multiplied that by three. And then
we multiplied π₯ by two and then we added one. So in this transformed function wherever we saw π₯ in the original
function, weβre gonna have to replace that with π₯ plus five.

So instead of three π₯ squared weβre gonna get three lots of π₯ plus five squared. And instead of two π₯, weβre gonna get two lots of π₯ plus five. But the plus one on the end isnβt affected, so thatβs just gonna stay as plus one. So now we need to
multiply out those expressions.

And π₯ plus five all squared is π₯ squared plus ten π₯ plus twenty-five. And
then two times π₯ plus five, distributing the two across the parentheses, that gives us two π₯ plus ten. Then distributing that three through these parentheses here gives
us three π₯ squared plus thirty π₯ plus seventy-five and then just tidying up at the end, ten plus one is eleven. So the fully tidied up and simplified
version π of π₯ plus five is equal to three π₯ squared plus thirty-two π₯ plus eighty-six is still a quadratic. But it doesnβt look like an
obvious simple change to transform it from the original function three π₯ squared plus two π₯ plus one.

So again if you left your transformed function in this unsimplified format, you might have a bit of a clue
that what we were doing is translating the function five places to the left where weβve basically added five to all of
the π₯-input values to the function. But if you simplified into this format, then itβs not really an easy step to work out
exactly what that transformation is going to be if you donβt know that we have called it π of π₯ plus five.

So lastly then letβs consider function transformations of this format π of π times π₯. Again weβll use the same π of π₯ and π will be equal to five. And remember that this type of transformation π¦ equals π of five π₯ is gonna be a stretch times one over five in
the π₯-direction about the π¦-axis. So weβre looking after π¦-axis. And all of the distances β all of the π₯-coordinates β are gonna be
multiplied by a fifth.

If we started off with the π₯-coordinate of one, weβd multiply that by a fifth to make nought point two. And if our π₯-coordinate on π of π₯ started off at negative two, we multiply that by a fifth or divide by five to
get negative nought point four, whereas any point which already had an π₯-coordinate of zero would map onto itself. So we can see the
whole thing is being squashed towards the π¦-axis.

So whatβs the equation of π of five π₯ going to look like? Well again π of π₯ meant that we took the π₯-value, squared it, multiplied it by three. We
took the π₯-value and doubled it, added that to the previous answer, and then added one. So wherever we see π₯ in that function in that expression there, weβre gonna replace it with five π₯. And this means that instead of getting three π₯ squared, weβre gonna get three times five π₯ all squared. And instead of just two times π₯, weβre gonna do two times five π₯.

But of course the plus one on the end is unaffected by all of this. So now we got to multiply this out. And five π₯ times five π₯ is twenty-five π₯ squared. So weβve got three times twenty-five π₯ squared. Two lots of five π₯ is ten π₯ plus one. So we can now tidy that up.

So in its simplified form π of five π₯ is seventy-five π₯ squared plus ten π₯ plus one. So again itβs still a quadratic, but itβs not an easy and obvious simple change to transform that. So again in the simplified format, not obvious to work out what the transformation was. But if we left it in this unsimplified format over here, we can see that weβve replaced the π₯βs up here with five π₯βs down here. And we can sort of work out what the transformation must have been.

So I have shown you a method for transforming a function and putting it into a simplifying format which makes it
difficult to work out what the transformation was. So whatβs the point of that? Well youβre most likely to encounter in questions
like this: show that π of π₯ is equal to seventy-five π₯ squared plus ten π₯ plus one is a transformation on the function
π of π₯ is equal to three π₯ squared plus two π₯ plus one such that π of π₯ is equal to π of five of π₯.

So what weβre gonna do is weβre gonna show that this expression here is this transformation of this expression
here. So letβs start off by saying π of π₯ is equal to three π₯ squared plus two π₯ plus one. Then weβre gonna form an expression for π of five π₯ because thatβs the transformation that the question
says itβs happened.

So weβre gonna replace π₯ in the function with five π₯ and then we can simplify it step by step. And this gives us the expression that they ask for in the question. And they told us that was equal to π of π₯. So this was just an exercise in rearranging the formula and picking out bits from the question.

Letβs look at some more questions that involve manipulating the equations of functions. π of π₯ is equal to π₯ squared minus four π₯ plus two. The function is translated π units in the π¦-direction to create function π of π₯ is equal to π₯ squared minus four π₯ plus nine. Find the value of π. Well the question told us that the function is translated π, so positive π units in the π¦-direction. Well that means and itβs creating function π of π₯ β so that means π of π₯ is π of π₯ plus π.

Now we were told that π of π₯ is equal to π₯ squared minus four π₯ plus nine. So we can replace that in our equation. And we were told that π of π₯ is equal to π₯ squared minus four π₯ plus two. So we can replace that in our equation. And then lastly weβve just got to add the π to the end. So now we can rearrange and solve this equation.

Subtracting π₯ squared from both sides gives me this, then adding four π₯ to both sides gives me this, and finally subtracting two from both sides gives me this; π is equal to seven. And if Iβve got enough time like at the end of an exam, I can check that answer. Translating a function by π units in the π¦-direction is like doing π of π₯ plus π. So if we look at the answer π is equal to seven, we can work out π of π₯ plus seven. So we got π of π₯, our original π of π₯ function. And we just add seven to the π¦-coordinates like we do there. And that gives us π₯ squared minus four π₯ plus nine, which indeed is the same as π of π₯. So we know weβve got the right answer.

Our next example then, π of π₯ is equal to three π₯ plus nine. This function is translated positive two units in the π¦-direction and π or positive π units in the π₯-direction to form function π of π₯ is equal to three π₯ plus two. Find the value of π. Now to translate a function two units in the π¦-direction, weβre gonna map π of π₯ onto π of π₯ plus two. Weβre adding two to all of the π¦-coordinates. And to translate positive π units in the π₯-direction, we would map a function like this: π of π₯ is gonna map to π of π₯ minus π.

So combining those two different transformations, π of π₯ is gonna map onto π of π₯ minus π plus two like this. Now weβre told in the question that this forms function π of π₯. So this is equal to π of π₯.

So although theyβve given us an expression for π of π₯ here, weβre gonna work out an alternative expression. And then the difference between the two would tell us the value of π. Now we know that π of π₯ is equal to three π₯ plus nine. So weβve got to try and work out what π of π₯ minus π is. Well weβre gonna have to replace the π₯ in a π of π₯ function with π₯ minus π. So instead of three times π₯, we can have three times π₯ minus π. And we still gonna have the plus nine on the end of it, but then remember we got to add two.

So simplifying that π of π₯ is equal to three π₯ minus three π plus eleven. But remember we know that π of π₯ is equal to three π₯ plus two, so we can equate those two things. So three π₯ plus two is equal to three π₯ minus three π plus eleven.

Well subtracting three π₯ from both sides gives me two is equal to negative three π plus eleven. Then I think Iβd add three π to both sides and then subtract two. And then I could divide both sides by three, giving me π is equal to three.

The function π of π₯ equals π₯ minus five times π₯ minus two times π₯ plus seven is translated positive five units in the direction of the positive π₯-axis. Find an equation for the transformed function. Well to translate positive five units in the π₯-direction, we need to map
π of π₯ onto π of π₯ minus five. So in our function up here π of π₯, we need to
replace π₯ with π₯ minus five. So π of π₯ minus five is equal to?

Well instead of π₯ in our first parentheses there π₯ minus five, weβre gonna use π₯ minus five, so thatβs π₯ minus five minus five. And again in the next parentheses, weβre replacing π₯ with π₯ minus five. So that becomes π₯ minus five minus two. And in the last parentheses again weβre gonna replace π₯ with π₯ minus five. So now all we have to do is tidy up those parentheses.

Well the first one π₯ minus five minus another five is π₯ minus ten. And in the second one π₯ minus five minus another two is π₯ minus seven. And in the last one π₯ minus five plus seven is π₯ plus two. So in fact thatβs our answer. The question only said find an equation; it didnβt tell us to multiply out the parentheses and simplify it down. Weβve said find an equation, so technically we have got away with this line here. But I think thatβs a little bit cheeky. I think to tidy up a little bit to this answer is probably preferable.

Lastly then the function π of π₯ equals π₯ minus three times three π₯ plus two times four minus π₯ is translated two units in the direction of the negative π₯-axis. Find an equation for the transformed function. Now to translate two units in the direction of the negative π₯-axis, thatβs the same as translating negative two units in the π₯-direction. And to achieve that we need to map π of π₯ onto π of π₯ plus two. And to calculate π of π₯ plus two, weβre gonna replace π₯ with π₯ plus two in each case in the function.

So the first parenthesis of π₯ minus three becomes π₯ plus two minus three, the second one becomes instead of three π₯ plus two three times π₯ plus two plus two, and the last parenthesis instead of four minus π₯ is four minus the whole of π₯ plus two.

So we need to be a little bit careful about how we evaluate these. The first one is probably the simplest: π₯ plus two minus three is just π₯ minus one. Now the second one, Iβm just going to multiply out the parentheses: so distribute that three across the π₯ plus two before I evaluate this, so that makes it three π₯ plus six plus two. And the last parentheses here Iβm taking away π₯, but Iβm also taking away two. So Iβm just gonna write that out in full.

So that becomes four minus π₯ minus two. So now simplifying those two last sets of parentheses, well three π₯ plus six plus two is three π₯ plus eight. And in the last parentheses I got four take away two gives me two. And then Iβve got negative π₯, thatβs two take away π₯. So my answer is π of π₯ plus two is equal to π₯ minus one times three π₯ plus eight times two minus π₯.