Assume frequency response of x(t) is X(f). Find frequency response of x(-2t+4)? Frequency response of $x(t)$ is $X(f)$. Find frequency response of $x(-2t+4)?$
Here is my work using the fourier transform property:$
x(-2t+4) = x( -2(t-2) )$,
+ First, I apply the property,  $x(at)$  &lt;---> $1/a \cdot X(f/a)$, by setting $a=2$, I have:
      $x(2t) &lt;---&gt; 1/2\cdot X(f/2)$
+ Second, I use the time-delay property,  $x(t-t0)$  &lt;----> $X(f) \cdot e^{(-j \cdot 2\pi \cdot f \cdot t_0)}$, with $t_0=2$, I have:
      $x( 2(t-2) ) &lt;---&gt; 1/2 \cdot X(f/2) \cdot e^{(-j \cdot 2 \cdot \pi \cdot f \cdot 2)}$
+ Third, I use the rotation property, $x(-t)$   &lt;-----> $X(-f)$, then I have,
     $x( -2(t-2) ) $&lt;---> $1/2 \cdot X(-f/2)*e^{(-j \cdot 2 \cdot \pi \cdot f \cdot 2)}$
+ Lastly, because spectrum of FT is symmetric, means $X(-f)=X(f)$, so the final result is:
                        $1/2 \cdot X(f/2) \cdot e^{(-j \cdot 2 \cdot \pi \cdot f \cdot 2)}$
I am not sure what I did is wrong or not. Please help me point out and explain my wrong concepts or way?
The key is to apply 'shift and scale'. So for $x(-2t + 4)$, you would first do $x_1(t) = x(t+4)$, then $x_2(t) = x_1(-2t) = x(-2t + 4)$. 

So $X_1(f) = e^{-j2\pi f(-4)}X(f)$ = $e^{j8\pi f}X(f)$. 

Then $X_2(f) = \frac{1}{|-2|}X_1(f/-2)$ = $\frac{1}{2}e^{j-4\pi f}X(f/-2)$

In your answer how did you assume it is symmetric? Also, in your third step, for $x(-t)$, you should be getting $x(2(-t-2)$ instead of $x(-2(t-2))$. Your first 2 steps were correct but failed to use scaling factor -2 instead of 2 in the first step itself.