# Translating Cubic with Algebra?

I'm having a little trouble figuring out how to translate $ax^3+bx^2+cx+d=y$ by vector $(1,1)$ using only algebra. If possible could someone give me a hand?

An example:

Translate $f(x)=2x^3–3x^2+x+1$ by vector $(1,-1)$

(also, it's a vector, not a coordinate! sorry, I suck at MathJaX :(..)

Thank you! Any help is greatly appreciated :)

-
Perhaps you're supposed to translate the cubic curve $y=2x^3–3x^2+x+1$ by the vector $(1,-1)$? – lhf Sep 21 '12 at 12:11
Ah yes, that is what I meant! Sorry for the confusion! – David North Sep 21 '12 at 12:28
(1,-1), Thank you! – David North Sep 21 '12 at 12:29
Thanks for clarifying. I think I've gotten the question the way you intended it. Check out the markup I used by clicking the timestamp next to my "edited" thing above. – rschwieb Sep 21 '12 at 12:34

If you are trying to translate the graph of the function in the plane with the vector $(1,-1)$, then I can offer a hint in that direction. That would suggest you are moving the graph down one unit and right one unit.
From basic study of functions, we know that $f(x)+c$ moves the graph up by $c$ if $c>0$ and down by $c$ if $c<0$. Secondly, $f(x+c)$ moves the graph left by $c$ if $c>0$ and right by $c$ if $c<0$.
So, it looks like you need to compute $f(x-1)-1$, for your function $f(x)$.