Question

Write a formula for the function g(x) obtained when the graph of f(x)= square root x is shifted up 1 unit and to the left 9 units .

Answer 1

$\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ f(x)= A( Bx+ C)+ D \\\\ ~~~~y= A( Bx+ C)+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \\\\ f(x)= A sin\left( B x+ C \right)+ D \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's take a peek.

$\bf f(x)=\sqrt{x}~\hspace{7em}\stackrel{\textit{up by 1}}{D=1}\qquad \stackrel{\textit{left by 9}}{C=9} \\\\[-0.35em] ~\dotfill\\\\ g(x)=\stackrel{A}{1}\sqrt{\stackrel{B}{1}x+\stackrel{C}{0}}+\stackrel{D}{0}\implies \stackrel{\textit{shifted}}{g(x)=\stackrel{A}{1}\sqrt{\stackrel{B}{1}x+\stackrel{C}{9}}+\stackrel{D}{1}}\implies \blacktriangleright g(x)=\sqrt{x+9}+1 \blacktriangleleft$