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Quick question: when in the formula:(x-b), if it was (x-2) moves to the right but when it is (x-b)+c, you have (x-2)+3 then it moves up 3. I do not understand why it is a minus sign instead of a plus sign so instead be: (x+b)+c then (x+2)+3 would move to the right 2 and up 3

Hadi Khan
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TheDragonorian
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    replacing $x$ with $x-3$ moves the graph right (direction of increasing $x$) by 3. Replacing $y$ with $y-3$ moves it up (direction of increasing $y$) by 3. So it works the same in both dimensions if you do the replacements the same way. It goes the opposite direction from what you might think because in the new equation, the $x$ that works must 3 bigger the it used to be to make the new formula with $x-3$ be like the old formula with $x$. – Ned Dec 17 '18 at 20:19
  • See https://math.stackexchange.com/a/3029278/265466. – amd Dec 17 '18 at 21:29
  • @Ned You should write that up as an answer. – amd Dec 18 '18 at 01:16
  • A duplicate of https://math.stackexchange.com/questions/2634587/explain-why-the-equation-is-y-fx-p-q-for-moving-the-original-fx-by/2634882#2634882. – amd Dec 23 '18 at 00:50

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Comment made into an answer:

Replacing $x$ with $x-3$ moves the graph $3$ to the right (direction of increasing $x$).

Replacing $y$ with $y-3$ moves the graph $3$ up (direction of increasing $y$). So it works the same in both dimensions if you do the replacements the same way.

The movement is in the opposite direction to what you might think because when you replace $x$ by $x-3$, the value of $x$ that works in the new equation with $x-3$ must be $3$ $bigger$ than it used to be to make the new formula with $x-3$ come out the same as the old formula with $x$.

Ned
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