Question

Write the quadratic function with the following transformations: vertical shrink by 3, left 4, down 9.

Answer 1

`\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}\\\\ --------------------\\\\`


`\bf \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}}}\\ \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}\\\\`


`\bf \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}}}`


with that template in mind, let's see

so, the parent equation could be
y = x² ====> y = A(x + C)² + D

shrink by 3, A = 3
left 4, C = +4
down 9, D = 9