Question

I forgot how to do this.... can somebody help?

Answer 1

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

$\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative} \\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}$

$\bf \begin{array}{llll} \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}} \end{array}$

now, with that template in mind, let's see

3 units to the right, that means C/B = -3 so hmm C = -3 and B = 1 will do, -3/1 = -3

vertical stretch by 2, so A = 2
reflected over the x-axis, so that means is flipped upside-down, so A = -2 then

and shifted down by 3, do D  = -3

$\bf f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}\implies f(x)={{ -2}}(\mathbb{R})^{{{ 1}}x-{{ 3}}}-{{3}}\\\\\\ f(x)={{ -2}}(3)^{{{ }}x-{{ 3}}}-{{3}}$