Question

Given its parent function g(x) = (1/2)^x, what is the equation of the function shown?

Answer 1

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

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

so... notice, the graph is just g(x), flipped upside-down and shifted up by 1 unit

Answer 2

Answer:  The correct equation is (B) $f(x)=-2\left(\dfrac{1}{2}\right)^x+1.$

Step-by-step explanation:  We are given to select the correct equation of the function f(x) shown in the figure.

From the graph given in the figure, we have

$f(-1)=-3,~~f(0)=-1,~~f(1)=0.$

(A) First option is

$f(x)=-3\left(\dfrac{1}{2}\right)^x+1.$

From here, we get

$f(-1)=-3\left(\dfrac{1}{2}\right)^{-1}+1=-3\times 2+1=-5\neq -3.$

So, this option is not correct.

(B) First option is

$f(x)=--2\left(\dfrac{1}{2}\right)^x+1.$

From here, we get

$f(-1)=-2\left(\dfrac{1}{2}\right)^{-1}+1=-2\times 2+1=-3\\\\f(0)=-2\left\dfrac{1}{2}\right)^0+1=-2+1=-1,\\\\f(1)=-2\left(\dfrac{1}{2}\right)^1+1=-1+1=0.$

So, this option is correct.

(C) Third option is

$f(x)=-\left(\dfrac{1}{2}\right)^x-3.$

From here, we get

$f(-1)=-\left(\dfrac{1}{2}\right)^{-1}-3=-1\times 2-3=-5\neq -3.$

So, this option is not correct.

(D) Fourth option is

$f(x)=-\left(\dfrac{1}{2}\right)^x-1.$

From here, we get

$f(-1)=-\left(\dfrac{1}{2}\right)^{-1}-1=--\times 2-1=-3,\\\\f(0)=-\left\dfrac{1}{2}\right)^0-1=-1-1=-2\neq -1.$

So, this option is not correct.

Thus, the correct function is

(B) $f(x)=-2\left(\dfrac{1}{2}\right)^x+1.$