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Video: Transforming Functions: What Happens to the Equations

Tim Burnham

Consider the effect of function transformations on their equations. We will learn how to present equations of transformed functions in a form that helps us identify the nature of the transformation and use equations to explore transformations.

17:26

Video Transcript

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

We’ve seen that 𝑓 of π‘₯ plus π‘Ž is a transformation that translates 𝑓 of π‘₯ by negative π‘Ž units in the π‘₯-direction. So if π‘Ž is positive, it’s gonna translate the curve of the function to the left. And if π‘Ž was negative, it’s gonna translate the curve of the function to the right. We’ve also seen the 𝑓 of 𝑏 times π‘₯ stretches 𝑓 of π‘₯ by a factor of one over 𝑏 in the π‘₯-direction. And that fixes the curve at the 𝑦-axis and stretches the curve away from the 𝑦-axis. Although depending on the value of 𝑏, it might squash the curve towards the 𝑦-axis.

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

And we’ve seen that 𝑏 times 𝑓 of π‘₯ stretches the function 𝑓 of π‘₯ by a factor of 𝑏 in the 𝑦-direction. Now in this case it’s gonna lock off all points on the π‘₯-axis. And it’s either gonna stretch that curve away from the π‘₯-axis or it’s going to squash it towards the π‘₯-axis. So there’re various different things that can happen for different values of 𝑏.

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

So take any point on the 𝑦 equals 𝑓 of π‘₯ curve. Add five to its 𝑦-coordinate. And you’re gonna map onto the 𝑦 equals 𝑓 of π‘₯ plus five curve. So that’s the graph, but let’s think of the equation of 𝑓 of π‘₯ plus five.

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

Now in this format here it’s pretty easy to say that we’ve taken the 𝑦-coordinates that we had from the 𝑓 of π‘₯ function and we have just added five to it. But in this format here it’s perhaps not quite so immediately obvious what’s happened in terms of that transformation, even though it’s not a very dramatic change to the equation.

Now let’s think of transformations of the type 𝑏 times 𝑓 of π‘₯. For example, 𝑓 of π‘₯, again we’re gonna take three π‘₯ squared plus two π‘₯ plus one. And 𝑏 is gonna be equal to five. If we look at the curves of those two functions: 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals five times 𝑓 of π‘₯, all we have done is we’ve taken the 𝑦-coordinates from 𝑦 equals 𝑓 of π‘₯ and we multiplied them by three. Looking at the curves of those functions, we can see that the 𝑦-coordinates from 𝑦 equals 𝑓 of π‘₯ have just been multiplied by five to give us the 𝑦-coordinates of 𝑦 equals five times 𝑓 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 𝑓 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, it’s 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 𝑦-coordinates we have for 𝑓 of π‘₯, but we just multiplied them all by five.

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

Well now let’s have a look at transformations of the type 𝑓 of π‘₯ plus π‘Ž. So again we’re gonna use 𝑓 of π‘₯ is equal to three squared plus two π‘₯ plus one. And we’re gonna use π‘Ž equals five. And hopefully remember from the introduction that a transformation like 𝑓 of π‘₯ plus π‘Ž translates the whole curve π‘Ž units to the left. So because π‘Ž is equal to five, we’ve translated the 𝑓 of π‘₯ curve five units to the left.

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

And similarly this point gets transformed to here. But it’s a little bit tricky when we try to think about what the function is going to look like in terms of the equation. In 𝑓 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, we’re gonna have to replace that with π‘₯ plus five.

So instead of three π‘₯ squared we’re gonna get three lots of π‘₯ plus five squared. And instead of two π‘₯, we’re gonna get two lots of π‘₯ plus five. But the plus one on the end isn’t affected, so that’s 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 𝑓 of π‘₯ plus five is equal to three π‘₯ squared plus thirty-two π‘₯ plus eighty-six is still a quadratic. But it doesn’t 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 we’ve basically added five to all of the π‘₯-input values to the function. But if you simplified into this format, then it’s not really an easy step to work out exactly what that transformation is going to be if you don’t know that we have called it 𝑓 of π‘₯ plus five.

So lastly then let’s consider function transformations of this format 𝑓 of 𝑏 times π‘₯. Again we’ll use the same 𝑓 of π‘₯ and 𝑏 will be equal to five. And remember that this type of transformation 𝑦 equals 𝑓 of five π‘₯ is gonna be a stretch times one over five in the π‘₯-direction about the 𝑦-axis. So we’re looking after 𝑦-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, we’d multiply that by a fifth to make nought point two. And if our π‘₯-coordinate on 𝑓 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 𝑦-axis.

So what’s the equation of 𝑓 of five π‘₯ going to look like? Well again 𝑓 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, we’re gonna replace it with five π‘₯. And this means that instead of getting three π‘₯ squared, we’re gonna get three times five π‘₯ all squared. And instead of just two times π‘₯, we’re 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 we’ve 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 𝑓 of five π‘₯ is seventy-five π‘₯ squared plus ten π‘₯ plus one. So again it’s still a quadratic, but it’s 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 we’ve replaced the π‘₯’s up here with five π‘₯’s 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 what’s the point of that? Well you’re most likely to encounter in questions like this: show that 𝑔 of π‘₯ is equal to seventy-five π‘₯ squared plus ten π‘₯ plus one is a transformation on the function 𝑓 of π‘₯ is equal to three π‘₯ squared plus two π‘₯ plus one such that 𝑔 of π‘₯ is equal to 𝑓 of five of π‘₯.

So what we’re gonna do is we’re gonna show that this expression here is this transformation of this expression here. So let’s start off by saying 𝑓 of π‘₯ is equal to three π‘₯ squared plus two π‘₯ plus one. Then we’re gonna form an expression for 𝑓 of five π‘₯ because that’s the transformation that the question says it’s happened.

So we’re 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 𝑔 of π‘₯. So this was just an exercise in rearranging the formula and picking out bits from the question.

Let’s look at some more questions that involve manipulating the equations of functions. 𝑓 of π‘₯ is equal to π‘₯ squared minus four π‘₯ plus two. The function is translated π‘Ž units in the 𝑦-direction to create function 𝑔 of π‘₯ is equal to π‘₯ squared minus four π‘₯ plus nine. Find the value of π‘Ž. Well the question told us that the function is translated π‘Ž, so positive π‘Ž units in the 𝑦-direction. Well that means and it’s creating function 𝑔 of π‘₯ β€” so that means 𝑔 of π‘₯ is 𝑓 of π‘₯ plus π‘Ž.

Now we were told that 𝑔 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 𝑓 of π‘₯ is equal to π‘₯ squared minus four π‘₯ plus two. So we can replace that in our equation. And then lastly we’ve just got to add the π‘Ž 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‘Ž is equal to seven. And if I’ve got enough time like at the end of an exam, I can check that answer. Translating a function by π‘Ž units in the 𝑦-direction is like doing 𝑓 of π‘₯ plus π‘Ž. So if we look at the answer π‘Ž is equal to seven, we can work out 𝑓 of π‘₯ plus seven. So we got 𝑓 of π‘₯, our original 𝑓 of π‘₯ function. And we just add seven to the 𝑦-coordinates like we do there. And that gives us π‘₯ squared minus four π‘₯ plus nine, which indeed is the same as 𝑔 of π‘₯. So we know we’ve got the right answer.

Our next example then, 𝑓 of π‘₯ is equal to three π‘₯ plus nine. This function is translated positive two units in the 𝑦-direction and 𝑐 or positive 𝑐 units in the π‘₯-direction to form function 𝑔 of π‘₯ is equal to three π‘₯ plus two. Find the value of 𝑐. Now to translate a function two units in the 𝑦-direction, we’re gonna map 𝑓 of π‘₯ onto 𝑓 of π‘₯ plus two. We’re adding two to all of the 𝑦-coordinates. And to translate positive 𝑐 units in the π‘₯-direction, we would map a function like this: 𝑓 of π‘₯ is gonna map to 𝑓 of π‘₯ minus 𝑐.

So combining those two different transformations, 𝑓 of π‘₯ is gonna map onto 𝑓 of π‘₯ minus 𝑐 plus two like this. Now we’re told in the question that this forms function 𝑔 of π‘₯. So this is equal to 𝑔 of π‘₯.

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

So simplifying that 𝑔 of π‘₯ is equal to three π‘₯ minus three 𝑐 plus eleven. But remember we know that 𝑔 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 𝑐 plus eleven.

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

The function 𝑓 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 𝑓 of π‘₯ onto 𝑓 of π‘₯ minus five. So in our function up here 𝑓 of π‘₯, we need to replace π‘₯ with π‘₯ minus five. So 𝑓 of π‘₯ minus five is equal to?

Well instead of π‘₯ in our first parentheses there π‘₯ minus five, we’re gonna use π‘₯ minus five, so that’s π‘₯ minus five minus five. And again in the next parentheses, we’re replacing π‘₯ with π‘₯ minus five. So that becomes π‘₯ minus five minus two. And in the last parentheses again we’re 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 that’s our answer. The question only said find an equation; it didn’t tell us to multiply out the parentheses and simplify it down. We’ve said find an equation, so technically we have got away with this line here. But I think that’s a little bit cheeky. I think to tidy up a little bit to this answer is probably preferable.

Lastly then the function 𝑔 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, that’s the same as translating negative two units in the π‘₯-direction. And to achieve that we need to map 𝑔 of π‘₯ onto 𝑔 of π‘₯ plus two. And to calculate 𝑔 of π‘₯ plus two, we’re 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, I’m 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 I’m taking away π‘₯, but I’m also taking away two. So I’m 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 I’ve got negative π‘₯, that’s two take away π‘₯. So my answer is 𝑔 of π‘₯ plus two is equal to π‘₯ minus one times three π‘₯ plus eight times two minus π‘₯.