Question

Describe the transformation of the following vertex equation f(x)=(x-10)^2+15

Answer 1

`\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % templates f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}} \\\\ ~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}} \\\\ f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\\\ f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ 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}\\\\ \left. \qquad \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, let's check,


`\bf \stackrel{\stackrel{parent~function}{g(x)=x^2}}{g(x)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+0})^2\stackrel{D}{+0}}\qquad \qquad \qquad \qquad f(x)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{-10})\stackrel{D}{+15}`

notice, A and B are the same, just 1.

but C is -10, and D is +15

horizontal shift by C/B or -10/1 or -10, 10 units to the right.

vertical shift of D, or +15, upwards 15 units.