Question

Let f(x) = Square root x. Find g(x), the function that is f(x) shifted up 7 units and left 1 units.

Answer 1

$~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill$

$\bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}$

$\bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

keeping in mind that template, let's take a looksie

$f(x) = \sqrt{x}\implies f(x) = A\sqrt{Bx + C}+D\implies f(x) = 1\sqrt{1x+0}+0 \\\\\\ \begin{cases} D = \stackrel{\textit{shifted up}}{+7}\\ C = \stackrel{\textit{left shift}}{+1} \end{cases}~\hspace{4em}g(x) = 1\sqrt{1x+1}+7\implies g(x) = \sqrt{x+1}+7$