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mckenzieandjesus
 one year ago
Can you please check my answer?
Describe how to translate the graph of y= sqrt x to obtain the graph of y=sqrt x9
shift down 9 units
shift left 9 units ****
shift right 9 units
shift up 9 units
mckenzieandjesus
 one year ago
Can you please check my answer? Describe how to translate the graph of y= sqrt x to obtain the graph of y=sqrt x9 shift down 9 units shift left 9 units **** shift right 9 units shift up 9 units

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mckenzieandjesus
 one year ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910

mckenzieandjesus
 one year ago
Best ResponseYou've already chosen the best response.1shift left 9 units

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh. Yes. It's shift left 9 unites. dw:1444435135720:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It means its moving across the "x" axis. The X is horizontal.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, you're correct :D.

mckenzieandjesus
 one year ago
Best ResponseYou've already chosen the best response.1Okay thanks :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\textit{function transformations} \\ \quad \\ \begin{array}{llll} \begin{array}{llll} shrink\ or\\ expand\\ by\ {\color{purple}{ A}}\cdot {\color{blue}{ B}}\end{array} \qquad \begin{array}{llll} vertical\\ shift\\ by \ {\color{green}{ D}} \end{array} \begin{array}{llll}{\color{green}{ D}} > 0& Upwards \\ \quad \\ {\color{green}{ D}} < 0 & Downwards\end{array} \\ \qquad \downarrow\qquad\qquad\quad\ \downarrow\\ % template start y = \sqrt{{\color{purple}{ A}} ( {\color{blue}{ B}}x + {\color{red}{ C}} )} + {\color{green}{ D}}\\ % template ends \qquad\qquad\quad\ \uparrow \\ \qquad\begin{array}{llll} horizontal\\ shift\\ by \ \frac{{\color{red}{ C}}}{{\color{blue}{ B}}}\end{array} \begin{array}{llll}\frac{{\color{red}{ C}}}{{\color{blue}{ B}}} > 0 & to\ the\ left \\ \quad \\ \frac{{\color{red}{ C}}}{{\color{blue}{ B}}} < 0& to\ the\ right\end{array} \end{array}\)
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