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## Main Question or Discussion Point

Why is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?

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Why is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?

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ShayanJ

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Change the variable to y=x-2 and see whether the resulting function is even or odd or none!

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ShayanJ

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## f(-y)=a(-y+1)(-y)(-y-1)=-a(y-1)y(y+1)=-f(y) ##

So f(y) is odd which means f(x) is not symmetric around x=2 but is not that much asymmetric because we have f(2-x)=-f(x-2).

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And this function is said to be symmetrical about the line x = 2. But I am unable to see how?

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ShayanJ

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ShayanJ

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Another way to look at it is as follows:^{4}/4 -2x^{3}+ 11x^{2}/2 - 6x) + 1

And this function is said to be symmetrical about the line x = 2. But I am unable to see how?

Imagine starting at x = 2 and moving the same distance, d, to the right and left (d > 0). So, to the right we have 2 + d and to the left we have 2 - d.

Now, if f is symmetrical about x = 2, then f(2-d) = f(2+d) for all d. You could try that approach.

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How did you reduce it down to that? Can you show me? I am unable to get to that point.

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ShayanJ

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And what is the reason that we transformed x to x+2?

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ShayanJ

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But we aren't shifting the curve as in your link but instead we are shifting the coordinate axes. The curve stays where it was. The axes are what shift.Check here!

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This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.

If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.

But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.

In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.

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Wait. So you are saying that f(x) should change to a new variable so that it becomes f(z), right? But that would again mean f(x-2) and not f(x+2).. got me confusedThis is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.

If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.

But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.

In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.

- #19

ShayanJ

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## z=x-2 \Rightarrow x=z+2 ## so ## f(x)=f(z+2) ##!(Forget the confusing ## x\rightarrow x+2 ##!)

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For example, earlier x

- #21

ShayanJ

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Let's start from the beginning._{old}) = f(x_{new}) provided that x_{new}= x_{old}- 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?

For example, earlier x_{old}= 2 gave f(2) = c. Now x_{new}= 0 (i.e. x_{old}-2, where x_{old}= 2 ) would give the result c.

At first we have a coordinate system which we call xy. Now I define a function y=f(x). Then I move the origin to x=a and name the new coordinate system zy. But this doesn't change the function, only the coordinate system has moved. But if I insist that the function f has the same form in terms of both x and z, then this means that the function has changed which isn't right.(Imagine y=x^2. z=x-a so x=z+a. But if I say that y=z^2, this function would have its minimum at z=0 so x=a which means the function has changed!) So I should have y=g(z). But g should be related to f somehow that we actually get the same function. So let's see what's the relationship. At x=0, f gives f(0), so at z=-a, g should give f(0). Then at x=a, f gives f(a), so at z=0, g should give f(a). Now we have two relationships g(-a)=f(0) and g(0)=f(a) and so we can deduce that g(z)=f(z+a)=f(x).

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