Question

Suppose f (x) = x^2. Find the graph of f(x+2)

Answer 1

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

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis}$

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

with that template in mind,

$\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}$

C = 2         B = 1        C/B = 2/1 or +2,    horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).