The graph of f(x)=sqrt x is shifted four units up and five units to the right. What is the equation of the resulting graph?

Accepted Solution

[tex]\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % templates f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}} \\\\ ~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}} \\\\ f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\\\ f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}} \\\\ f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}} \\\\ --------------------[/tex]

[tex]\bf \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}[/tex]

[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ ~~~~~~if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \left. \qquad \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 check,

"shifted four units up"    D = +4

"and five units to the right"   B = 1, C = -5, C/B = -5/1 or -5

[tex]\bf f(x)=\stackrel{A}{1}\sqrt{\stackrel{B}{1}x\stackrel{C}{+0}}\stackrel{D}{+0}\qquad \\\\\\ f(x)=\stackrel{A}{1}\sqrt{\stackrel{B}{1}x\stackrel{C}{-5}}\stackrel{D}{+4}\implies f(x)=\sqrt{x-5}+4[/tex]