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Gabriel Mesquita · Answer 1 · 2018-01-25T03:26:01+0000

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

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

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

now, with that template above

let's take a peek at yours

$\bf \begin{array}{llccll} y=|&1x&+0|&+0\\ &\uparrow &\uparrow &\uparrow \\ &B&C&D \end{array}$

so, give it a horizontal shift of 10 to the right, for C/B or just C since B is just 1 anyway, and give it a vertical shift of 3 down to D

Erpheus · Answer 2 · 2018-01-30T08:07:48+0000

y=|x-10|-3

Inside, is opposite. So therefore, to move it right 10 units, you subtract ten. However, outside of the absolute value, it is normal, so to move it down 3 units, you subtract 3.

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