Wykres funkcji $y=f\left(x-p\right)+q$  otrzymujemy przez przesunięcie wykresu funkcji $y=f\left(x\right)$  o wektor $\left[p,q\right]$ .

$\text{a)}$  

$g\left(x\right)=f\left(x+3\right)=f\left(x-\left(-3\right)\right)+0$  

$\overrightarrow{u}=\left[-3,0\right]$  

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$\text{b)}$  

$g\left(x\right)=f\left(x-1\right)+0$  

$\overrightarrow{u}=\left[1,0\right]$  

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$\text{c)}$  

$g\left(x\right)=f\left(x\right)+6=f\left(x-0\right)+6$  

$\overrightarrow{u}=\left[0,6\right]$  

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$\text{d)}$  

$g\left(x\right)=f\left(x\right)-4=f\left(x-0\right)-4$  

$\overrightarrow{u}=\left[0,-4\right]$  

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$\text{e)}$  

$g\left(x\right)=f\left(x+1\right)+2=f\left(x-\left(-1\right)\right)+2$  

$\overrightarrow{u}=\left[-1,2\right]$  

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$\text{f)}$  

$g\left(x\right)=f\left(x-7\right)-5$  

$\overrightarrow{u}=\left[7,-5\right]$  

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$\text{g)}$  

$g\left(x\right)=f\left(x+5\right)-3=f\left(x-\left(-5\right)\right)-3$  

$\overrightarrow{u}=\left[-5,-3\right]$  

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$\text{h)}$  

$g\left(x\right)=f\left(x-p\right)+q$  

$\overrightarrow{u}=\left[p,q\right]$