Question

How can you use transformations to graph this function? y=3xxFormula1" title=" 7^{-x} " alt=" 7^{-x} " align="absmiddle" class="latex-formula">+2

Answer 1

$\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\\\ % left side templates \begin{array}{llll} f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ y=&{{ A}}({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}} \\ \quad \\ f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}} \end{array}\\\\ --------------------$

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

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

with that template in mind, let's check

$\bf y=\stackrel{A}{3}(7^{\stackrel{B}{-1}x})\stackrel{D}{+2}$    <---- so the parent function will then be y = 7⁻ˣ

A = 3    it shrinks, in relation the y-axis, by a factor of 3 times, so is narrower.

B=-1     well, it flips it sideways, reflection over the y-axis.

D= +2  well, that's just a vertical shift of 2 units.