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yan [13]
4 years ago
12

Disregard my work, but how do you get the answer? There is also a graph at the bottom which is optional to use.

Mathematics
1 answer:
kakasveta [241]4 years ago
7 0
\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)=  A(  Bx+  C)+  D
\\\\
~~~~y=  A(  Bx+  C)+  D
\\\\
f(x)=  A\sqrt{  Bx+  C}+  D
\\\\
f(x)=  A(\mathbb{R})^{  Bx+  C}+  D
\\\\
f(x)=  A sin\left( B x+  C  \right)+  D
\\\\
--------------------

\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}

\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}

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).
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