WEBVTT
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In this series of videos, weβve looked at what happens to the graphs of functions when you transform them in different ways.
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Weβve seen vertical and horizontal translations and vertical and horizontal stretches.
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Now, weβre gonna take a few minutes to look at what happens to the equation of a function when we transform it.
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Weβve seen that π of π₯ plus π is a transformation that translates π of π₯ by negative π units in the π₯-direction.
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So, if π is positive, itβs gonna translate the curve of the function to the left.
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And if π was negative, itβs gonna translate the curve of the function to the right.
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Weβve also seen that π of π times π₯ stretches π of π₯ by a factor of one over π in the π₯-direction.
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And that fixes the curve at the π¦-axis and stretches the curve away from the π¦-axis.
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Although depending on the value of π, it might squash the curve towards the π¦-axis.
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And weβve also seen that π of π₯ plus π translates the function π of π₯ by π units in the π¦-direction.
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So, if π is positive, it shifts the curve upwards.
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And if π is negative, it shifts the curve downwards.
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And weβve seen that π times π of π₯ stretches the function π of π₯ by a factor of π in the π¦-direction.
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Now, in this case, itβs gonna lock off all points on the π₯-axis.
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And itβs either gonna stretch that curve away from the π₯-axis or itβs gonna squash it towards the π₯-axis.
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Thereβre various different things that can happen for different values of π.
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Now, letβs think about some equations and see the effect of these transformations on them.
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First, letβs think about π of π₯ plus π.
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So, for example, if π of π₯ was three π₯ squared plus two π₯ plus one and π was equal to five.
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Then as we said in the introduction, the curve of π¦ equals π of π₯ is translated five places in the positive π¦-direction.
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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.
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So, thatβs the graph, but letβs think of the equation of π of π₯ plus five.
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Well, weβre gonna start off with the expression for π of π₯, so three π₯ squared plus two π₯ plus one.
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And then, weβre gonna be adding five to that, which simplifies to three π₯ squared plus two π₯ plus six.
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Now, in this format here, itβs pretty easy to see that weβve taken the π¦-coordinates that we had from the π of π₯ function and weβve just added five to it.
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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.
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Now, letβs think of transformations of the type π times π of π₯.
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For example, π of π₯ again, weβre gonna take three π₯ squared plus two π₯ plus one.
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And π is gonna be equal to five.
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If we look at the curves of those two functions, π¦ equals π of π₯ and π¦ equals five times π of π₯, all weβve done is weβve taken the π¦-coordinates from π¦ equals π of π₯ and weβve multiplied them by three.
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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₯.
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So, one times five gives us five.
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Two times five gives us 10.
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Four times five gives us 20, and so on.
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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.
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And we can distribute that five across the parentheses and simplify the expression to five π π₯ is equal to 15π₯ squared plus 10π₯ plus five.
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Now, in this simplified format, itβs pretty much the same format, but all of the coefficients are five times larger.
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But itβs not immediately obvious what that transformation is doing.
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So, if you look at the other format here, so weβve got the same π¦-coordinates we have for π of π₯, but weβve just multiplied them all by five.
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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.
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Well, now, letβs have a look at transformations of the type π of π₯ plus π.
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So again, weβre gonna use π of π₯ is equal to three squared plus two π₯ plus one.
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And weβre gonna use π equals five.
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And hopefully you remember from the introduction that a transformation like π of π₯ plus π translates the whole curve π units to the left.
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So, because π is equal to five, weβve translated the π of π₯ curve five units to the left.
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Then, 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.
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And similarly, this point gets transformed to here.
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But itβs a little bit trickier when we try to think about what the function is going to look like in terms of the equation.
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In π of π₯, we took π₯ and then we squared it and multiplied that by three.
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And then, we multiplied π₯ by two and then we added one.
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So, in this transformed function, wherever we saw π₯ in the original function, weβre gonna have to replace that with π₯ plus five.
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So, instead of three π₯ squared, weβre gonna get three lots of π₯ plus five squared.
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And instead of two π₯, weβre gonna get two lots of π₯ plus five.
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But the plus one on the end isnβt affected, so thatβs just gonna stay as plus one.
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So, now, we need to multiply out those expressions.
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And π₯ plus five all squared is π₯ squared plus 10π₯ plus 25.
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And then, two times π₯ plus five, distributing the two across the parentheses there gives us two π₯ plus 10.
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Then, distributing that three through these parentheses here gives us three π₯ squared plus 30π₯ plus 75.
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And then, just tidying up at the end, 10 plus one is 11.
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So, the fully tidied up and simplified version π of π₯ plus five is equal to three π₯ squared plus 32π₯ plus 86 is still a quadratic.
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But it doesnβt look like an obvious simple change to transform it from the original function three π₯ squared plus two π₯ plus one.
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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.
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But if youβve 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 weβve called it π of π₯ plus five.
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So, lastly then, letβs consider function transformations of this format, π of π times π₯.
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Again, weβll use the same π of π₯ and π will be equal to five.
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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.
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So, weβre looking off the π¦-axis.
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And all of the distances, all of the π₯-coordinates, are gonna be multiplied by a fifth.
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If we started off with a π₯-coordinate of one, weβd multiply that by a fifth to make 0.2.
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And if our π₯-coordinate on π of π₯ started off at negative two, weβd multiply that by a fifth, or divide by five, to get negative 0.4.
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Whereas any point which already had an π₯-coordinate of zero would map onto itself.
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So, we can see the whole thing is being squashed towards the π¦-axis.
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So, whatβs the equation of π of five π₯ is going to look like?
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Well, again, π of π₯ meant that we took the π₯-value, squared it, and multiplied by three.
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We took the π₯-value and doubled it, added that to the previous answer, and then added one.
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So, wherever we see π₯ in that function, in that expression there, weβre gonna replace it with five π₯.
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And this means that instead of getting three π₯ squared, weβre gonna get three times five π₯ all squared.
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And instead of just two times π₯, weβre gonna do two times five π₯.
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But of course, the plus one on the end is unaffected by all of this.
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So, now, weβve got to multiply this out.
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And five π₯ times five π₯ is 25π₯ squared.
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So, weβve got three times 25π₯ squared.
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Two lots of five π₯ is 10π₯, plus one.
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So, we can now tidy that up.
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So, in its simplified form, π of five π₯ is 75π₯ squared plus 10π₯ plus one.
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So, again, itβs still a quadratic, but itβs not an easy and obvious simple change to transform that.
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So, again, in the simplified format, not obvious to work out what the transformation was.
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But if we left it in this unsimplified format over here, we can see that weβve replaced the π₯s up here with five π₯s down here.
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And we can sort of work out what the transformation must have been.
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So, Iβve shown you a method for transforming a function and putting it- simplifying it into a format which makes it difficult to work out what the transformation was.
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So, whatβs the point of that?
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Well, youβre most likely to encounter it in questions like this.
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Show that π of π₯ is equal to 75π₯ squared plus 10π₯ 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₯.
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So, what weβve gotta do is weβve gotta show that this expression here is this transformation of this expression here.
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So, letβs start off by saying π of π₯ is equal to three π₯ squared plus two π₯ plus one.
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Then, weβre gonna form an expression for π of five π₯ because thatβs the transformation that the question says has happened.
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So, weβre gonna replace π₯ in the function with five π₯.
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And then, we can simplify it step-by-step.
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And this gives us the expression that they ask for in the question.
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And they told us that that was equal to π of π₯.
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So, this was just an exercise in rearranging a formula and picking out bits from the question.
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Letβs look at some more questions that involve manipulating the equations of functions.
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π of π₯ is equal to π₯ squared minus four π₯ plus two.
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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.
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Find the value of π.
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Well, the question told us that the function is translated π, so positive π, units in the π¦-direction.
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Well, that means β and itβs creating function π of π₯.
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So, that means π of π₯ is π of π₯ plus π.
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Now, we were told that π of π₯ is equal to π₯ squared minus four π₯ plus nine.
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So, we can replace that in our equation.
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And we were told that π of π₯ is equal to π₯ squared minus four π₯ plus two.
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So, we can replace that in our equation.
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And then, lastly, weβve just got to add the π to the end.
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So, now, we can rearrange and solve this equation.
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Subtracting π₯ squared from both sides gives me this.
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Then, adding four π₯ to both sides gives me this.
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And finally, subtracting two from both sides gives me this.
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π is equal to seven.
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And if Iβve got enough time, like at the end of an exam, I can check that answer.
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Translating a function by π units in the π¦-direction is like doing π of π₯ plus π.
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So, if we reckon the answer π is equal to seven, we can work out π of π₯ plus seven.
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So, we got π of π₯, our original π of π₯ function.
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And we just add seven to the π¦-coordinates like we do there.
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And that gives us π₯ squared minus four π₯ plus nine, which indeed is the same as π of π₯.
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So, we know weβve got the right answer.
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Our next example then.
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π of π₯ is equal to three π₯ plus nine.
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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.
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Find the value of π.
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Now, to translate a function two units in the π¦-direction, weβre gonna map π of π₯ onto π of π₯ plus two.
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Weβre adding two to all of the π¦-coordinates.
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And to translate positive π units in the π₯-direction, we would map our function like this.
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π of π₯ is gonna map to π of π₯ minus π.
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So, combining those two different transformations, π of π₯ is gonna map onto π of π₯ minus π plus two, like this.
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Now, weβre told in the question that this forms function π of π₯.
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So, this is equal to π of π₯.
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So, although theyβve given us an expression for π of π₯ here, weβre gonna work out an alternative expression.
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And then, the difference between the two will tell us the value of π.
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Now, we know that π of π₯ is equal to three π₯ plus nine.
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So, weβve got to try and work out what π of π₯ minus π is.
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Well, weβre gonna have to replace the π₯ in our π of π₯ function with π₯ minus π.
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So, instead of three times π₯, weβre gonna have three times π₯ minus π.
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And weβre still gonna have the plus nine on the end of it.
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But then, remember, weβve got to add two.
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So, simplifying that, π of π₯ is equal to three π₯ minus three π plus eleven.
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But remember, we know that π of π₯ is equal to three π₯ plus two, so we can equate those two things.
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So, three π₯ plus two is equal to three π₯ minus three π plus eleven.
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Well, subtracting three π₯ from both sides gives me two is equal to negative three π plus eleven.
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Then, I think Iβd add three π to both sides and then subtract two.
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And then, I could divide both sides by three, giving me π is equal to three.
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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.
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Find an equation for the transformed function.
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Well, to translate positive five units in the π₯-direction, we need to map π of π₯ onto π of π₯ minus five.
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So, in our function up here π of π₯, we need to replace π₯ with π₯ minus five.
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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.
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So, thatβs π₯ minus five minus five.
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And again, in the next parentheses, weβre replacing π₯ with π₯ minus five.
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So, that becomes π₯ minus five minus two.
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And in the last parentheses again, weβre gonna replace π₯ with π₯ minus five.
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So, now, all we have to do is tidy up those parentheses.
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Well, the first one, π₯ minus five minus another five is π₯ minus 10.
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And in the second one, π₯ minus five minus another two is π₯ minus seven.
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And in the last one, π₯ minus five plus seven is π₯ plus two.
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So, in fact, thatβs our answer.
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The question only said find an equation.
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It didnβt tell us to multiply out the parentheses and simplify it down.
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It just said find an equation.
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So, technically, weβd have got away with this line here.
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But I think thatβs a little bit cheeky.
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I think to tidy up a little bit to this answer is probably preferable.
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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.
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Find an equation for the transformed function.
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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.
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And to achieve that, we need to map π of π₯ onto π of π₯ plus two.
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And to calculate π of π₯ plus two, weβre gonna replace π₯ with π₯ plus two in each case in the function.
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So, the first parenthesis, instead of π₯ minus three, becomes π₯ plus two minus three.
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The second lot becomes, instead of three π₯ plus two, itβs three times π₯ plus two plus two.
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And the last parenthesis, instead of four minus π₯, is four minus the whole of π₯ plus two.
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So, we need to be a little bit careful about how we evaluate these.
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The first one is probably the simplest.
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π₯ plus two minus three is just π₯ minus one.
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Now, the second one, Iβm just going to multiply out the parentheses.
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So, distribute that three across the π₯ plus two before I evaluate this.
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So, that makes it three π₯ plus six plus two.
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And the last parentheses here, Iβm taking away π₯, but Iβm also taking away two.
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So, Iβm just gonna write that out in full.
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So, that becomes four minus π₯ minus two.
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So, now, simplifying those two last sets of parentheses.
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Well, three π₯ plus six plus two is three π₯ plus eight.
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And in the last parentheses, Iβve got four take away two gives me two.
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And then, Iβve got negative π₯.
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So, thatβs two take away π₯.
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So, my answer is π of π₯ plus two is equal to π₯ minus one times three π₯ plus eight times two minus π₯.