Respuesta :
[tex]\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ f(x)= A( Bx+ C)+ D \\\\ ~~~~y= A( Bx+ C)+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \\\\ f(x)= A sin\left( B x+ C \right)+ D \\\\[-0.35em] ~\dotfill\\\\ \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}[/tex]
[tex]\bf ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the 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}[/tex]
with that template in mind, let's take a peek.
[tex]\bf f(x)=\sqrt{x}~\hspace{7em}\stackrel{\textit{up by 1}}{D=1}\qquad \stackrel{\textit{left by 9}}{C=9} \\\\[-0.35em] ~\dotfill\\\\ g(x)=\stackrel{A}{1}\sqrt{\stackrel{B}{1}x+\stackrel{C}{0}}+\stackrel{D}{0}\implies \stackrel{\textit{shifted}}{g(x)=\stackrel{A}{1}\sqrt{\stackrel{B}{1}x+\stackrel{C}{9}}+\stackrel{D}{1}}\implies \blacktriangleright g(x)=\sqrt{x+9}+1 \blacktriangleleft[/tex]
The function after the transformations is:
[tex]g(x) = \sqrt{x+9}+1[/tex]
To solve this question, transformations are applied to the original function.
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Original function:
The original function is:
[tex]f(x) = \sqrt{x}[/tex]
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Shifted up 1 unit
Shifting a function f(x) up a units means that: [tex]f(x) = f(x) + a[/tex]
Thus, the function shifted up 1 unit is: [tex]f(x) = \sqrt{x} + 1[/tex]
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Shifted left 9 units
Shifting a function f(x) left a units means that: [tex]f(x) = f(x+a)[/tex]
Thus, the function shifted left 9 units, combined with the shifting up, is:
[tex]g(x) = \sqrt{x+9}+1[/tex]
A similar question is found at https://brainly.com/question/15196246
