F(x) symmetric about the line x=2

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

 

## f(x)=a(x-1)(x-2)(x-3) \rightarrow f(y)=a(y+1)y(y-1) ##
## 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).

 

Its the same trick. Just do the tranformation ## x\rightarrow x+2 ## and check whether the resulting function is even. If its even, then the original function is symmetric around x=2.

 

You're doing something wrong. You should be able to reduce f(x+2) to ## \frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1 ## which is even.

 

Another way to look at it is as follows:

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.

 

## f(x+2)=a[ \frac 1 4 (x+2)^4-2(x+2)^3+\frac{11}{2} (x+2)^2-6(x+2)]+1=\\ a(x+2)[ \frac 1 4 (x+2)^3-2(x+2)^2+\frac{11}{2} (x+2)-6]+1=\\ a(x+2)(\frac 1 4 x^3+\frac 3 2 x^2+3x+2-2x^2-8x-8+\frac{11}{2}x+5)+1=\\ \frac 1 2 a(x+2)(\frac 1 2 x^3-x^2+x-2)+1=\frac 1 2 a (\frac 1 2 x^4+x^3-x^3-2x^2+x^2+2x-2x-4)+1=\\ \frac 1 2 a (\frac 1 2 x^4-x^2-4)+1##

 

Simple. When we say a function is even, we mean its symmetric around x=0. So if a function is symmetric around ## x=a \neq 0 ##, it means if we move the origin to x=a, the resulting function would be even.

 

Check here!

 

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.

 

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