[tex]\bf ~~~~~~~~~~~~\textit{function transformations}
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% templates
f(x)= A( Bx+ C)+ D
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~~~~y= A( Bx+ C)+ D
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f(x)= A\sqrt{ Bx+ C}+ D
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f(x)= A(\mathbb{R})^{ Bx+ C}+ D
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f(x)= A sin\left( B x+ C \right)+ D
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--------------------[/tex]
[tex]\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\
\bullet \textit{ flips it upside-down if } A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if } B\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\
~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by } D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{ B}[/tex]
with that template in mind.
since the original or "parent" function is y = x²−4x+3, if we change the "x" argument to "x-2", we end up with a horizontal shift.
the x-2 part would be in the template the Bx+C part, with B = 1 and C = -2, or a horizontal shift to the right of 2/1 or 2 units.
since the parent function has a point at (2, -1), if we move that horizontally only to the right, we'd end up at (4, -1).