If #f(x)=x^2-x#, how do you find #f(x-6)#?

Question

Elias Ackerman · Accepted Answer

For every #x# in the equation, we replace it with #x-6# An easy way to think of #f(x)# is #y#. So #f# is the transformations we apply and whatever comes in the brackets is what we apply the transformations to. In this case, #f(x)# means #x^2-x#, so we square our #x# value then minus the original #x# from #x^2#. Now, we could put anything in the brackets, and the transformations would be the same - square it then minus the original number from the result. So the same applies to #f(x-6)#. We simply square #x-6# and then minus #x-6# from the result. #f(x-6) = (x-6)^2-(x-6) = x^2-12x+36-x+6=x^2-13x+42#

Serenity Johnson · Answer

#f(x-6)=x^2-13x+42#

Another way we can do this is to first factor the function #f(x)#.

#f(color(red)x)=color(red)x(color(red)x-1)#

Therefore, if we consider the function composition:

#f(color(blue)(x-6))=(color(blue)(x-6))(color(blue)(x-6)-1)=(x-6)(x-7)=x^2-13x+42#

Serenity Johnson · Answer

undefined To find $ f(x-6) $ when $ f(x) = x^2 - x $, you simply substitute $ x-6 $ for $ x $ in the expression for $ f(x) $.

$$ f(x-6) = (x-6)^2 - (x-6) $$

$$ = (x-6)(x-6) - (x-6) $$

$$ = x^2 - 12x + 36 - x + 6 $$

$$ = x^2 - 13x + 42 $$

So, $ f(x-6) = x^2 - 13x + 42 $.