[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\[/tex]
[tex]\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}[/tex]
so.. with that template in mind, let's do the 4units shift, by setting C = +4,
and leaving B = 1, since C/B = +4/1 = +4
[tex]\bf y=(x-2\boxed{+4})^2+3\implies y=(x+2)^2+3\impliedby \textit{shifted version}[/tex]
so.. now... where do they meet? well, is just a system of equations of two variables, and they meet when both equations equal each other
since both equations are solved by "y", let's use substitution
[tex]\bf \begin{cases}
y=(x-2)^2+3\\
y=(x+2)^2+3\\
----------\\
(x-2)^2+3=(x+2)^2+3
\end{cases}
\\\\\\
(x-2)^2=(x+2)^2\implies x^2-4x+4=x^2+4x+4
\\\\\\
-4x=4x\implies 0=8x\implies \boxed{0=x}
\\\\\\
\textit{let's plug that in the first equation, to get "y"}
\\\\\\
y=(0-2)^2+3\implies y=2^2+3\implies \boxed{y=7}[/tex]
