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I am not yet sure how to paste graph here in this site. so am just using equations to explain the question I have .

The graph $f(t) = 2t+1$ has a positive slope , and the initial value starts from constant $1$. In one of text book example it shows, if $t$ is subtracted with value $-2$, then the graph moves towards right. That is , $f(t-2) = 2t-3$.

Question:
When I plot this graph I get the graph moves down not right as explained below. Can you explain me if my understanding is correct ?

I have explained as below.

If plot graph for value $t=0,1,2,$ to function $f(t-2) = 2t-3$, then the function value will be
$f(-2) = -3$
$f(-1)= -1$
$f(0)= 1$.

Widawensen
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cyne
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    Look carefully at your picture. If you forget the values of the coordinates and just look at the two lines you should see that moving the first line either down or right makes it inot the second line. – Ethan Bolker Apr 22 '17 at 14:49

1 Answers1

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Suppose $y=f(x)=ax+b$, as we have here. Then the graph of $y=f(x-k)$ is $$ y = f(x-k) = a(x-k)+b = ax + b-ak = f(x)-ak. $$ So in fact translating the graph right by $k$ looks the same as translating it down by $ak$. (This is only the case for $f$ of this form: $y=(x-k)^2$, for example, looks different from $y=x^2-\alpha k$ for any $\alpha$)

Chappers
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