How do you do this ?

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\\\ \begin{array}{rllll} % left side templates f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ y=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}} \end{array}$

$\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative} \\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}$

$\bf \begin{array}{llll} \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}} \end{array}$

now... look at g(x)
$\bf g(x)=(-x)^2+1\implies g(x)=(-x)(-x)+1\implies g(x)=x^2+1 \\\\\\ \begin{array}{llllll} g(x)=&1(&1x&+0)^2&+1\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}$

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