{"question":{"question":{"content":"The graph of $y=f(x)$ is shown below. ![](web+graphie://ka-perseus-graphie.s3.amazonaws.com/2e2493b7a8586948084d178a6848d0370cda1012) **Which of the following could be function $f$?** [[☃ radio 1]] ","images":{"web+graphie://ka-perseus-graphie.s3.amazonaws.com/2e2493b7a8586948084d178a6848d0370cda1012":{"width":425,"height":425}},"widgets":{"radio 1":{"type":"radio","alignment":"default","static":false,"graded":true,"options":{"choices":[{"content":"$ f(x) =\begin{cases} |x-2|-2&&\text{if } -8$ f(x) = \left\{ \begin{array}{lr} \blueD{\text{Equation of graph on left}} &\text{ if left interval} \\ \greenD{\text{Equation of graph on right}} &\text{ if right interval} \end{array} \right.$ ![](web+graphie://ka-perseus-graphie.s3.amazonaws.com/0a41a49b6e7d8dca8029c942fdf3f77066448e9f) An open circle at $(c,d)$ indicates that $c$ *is not included* in the interval associated with that piece, while a closed circle at $(c,d)$ indicates that $c$ *is included* in the interval associated with that piece.","images":{"web+graphie://ka-perseus-graphie.s3.amazonaws.com/0a41a49b6e7d8dca8029c942fdf3f77066448e9f":{"width":310,"height":310}},"widgets":{}},{"content":"###The left-hand piece $\blueD{\text{The graph on the left}}$ is $y=|x+2|-2$. [[☃ explanation 1]] $\blueD{\text{The function on the left}}$ is confined to the interval $-8 $\begin{align}g(x)&=|x|\\\\ g(x+2)&=|x+2|\\\\ g(x+2)-2&=|x+2|-2 \end{align}$ Therefore, it follows that $y=|x+2|-2$. ","widgets":{}},"version":{"major":0,"minor":0}},"explanation 2":{"type":"explanation","alignment":"default","static":false,"graded":true,"options":{"static":false,"showPrompt":"Why are there two options?","hidePrompt":"Got it, thanks!","explanation":"Because the point $(2,2)$ is on the graph of $f$, and it is the *joining point* between the two pieces, the value $2$ can be included in the interval of either piece, so long as it is included in at least one of the intervals. We can use the answer options to help us make a decision here. ","widgets":{}},"version":{"major":0,"minor":0}}}},{"content":"###The right-hand piece $\greenD{\text{The graph on the right}}$ is $y=\sqrt{x-2}+2$. [[☃ explanation 1]] $\greenD{\text{The function on the right}}$ is confined to the interval $2\leq x\leq 9$ or $2< x\leq9$. As mentioned above, $x=2$ doesn't necessarily have to be included in this interval, but it has to be included in one of the intervals of the two pieces. ","images":{},"widgets":{"explanation 1":{"type":"explanation","alignment":"default","static":false,"graded":true,"options":{"static":false,"showPrompt":"Explain","hidePrompt":"Got it, thanks!","explanation":"Notice that the graph of this function has the basic shape of $y=\sqrt{x}$. ![](web+graphie://ka-perseus-graphie.s3.amazonaws.com/5220e4c031e1a75fe02982dc43ddbdc79bd25d97) However, the graph of the function that we are interested in has been shifted right $2$ units and up $2$ units. So if $h(x)=\sqrt{x}$, then the function we are interested in is $y=h(x-2)+2$. This function is built below. >$\begin{align}h(x)&=\sqrt{x}\\\\ h(x-2)&=\sqrt{x-2}\\\\ h(x-2)+2&=\sqrt{x-2}+2\\\\ \end{align}$ Therefore, it follows that $y=\sqrt{x-2}+2$.","widgets":{}},"version":{"major":0,"minor":0}}}},{"content":"###Assessing the options We found the equations of both pieces and their intervals. The only thing left undetermined is whether $x=2$ is included in the interval of the first piece... $ f(x) =\begin{cases} |x+2|-2&&\text{if } -8< x \leq 2 \\ \sqrt{x-2}+2 &&\text{if }2< x \leq 9 \end{cases} $ $$ ... or in the interval of the second piece. $ f(x) =\begin{cases} |x+2|-2&&\text{if } -8< x < 2 \\ \sqrt{x-2}+2 &&\text{if }2\leq x \leq 9 \end{cases} $ $$ Of the possible options given above, only the first function is listed as an answer choice. Also, it is important to note that the following choice is ***incorrect***, since for this option, $f(2)$ is undefined. $ f(x) =\begin{cases} |x+2|-2&&\text{if } -8< x< 2 \\ \sqrt{x-2}+2 &&\text{if }2< x \leq 9 \end{cases} $ $$ [[☃ explanation 1]]","images":{},"widgets":{"explanation 1":{"type":"explanation","alignment":"default","static":false,"graded":true,"options":{"static":false,"showPrompt":"Could x=2 be included in both intervals?","hidePrompt":"Got it, thanks!","explanation":"In this case, $x=2$ could be included in both intervals since the value of each *piece* at $x=2$ is the same ensuring that $f$ is indeed a function. >$|\maroonD 2+2|-2=2\qquad$ and $\qquad \sqrt{{\maroonD{2}-2}}+2=2$ However, it is not customary to see an $x$-value repeated in the intervals, and so it was not listed above or as an answer choice. ","widgets":{}},"version":{"major":0,"minor":0}}}},{"content":"###Conclusion Putting all of this together we get the following piecewise function. $ f(x) =\begin{cases} \blueD{ |x+2|-2}&&\text{if } -8< x \leq 2 \\ \\ \greenD{\sqrt{x-2}+2} &&\text{if }2< x \leq 9 \end{cases} $ $$","images":{},"widgets":{}}]},"problemId":37641,"exerciseId":3162,"exerciseSessionId":null,"hintsEnabled":false,"readOnly":false,"submitterId":null,"checkAnswerButtonText":"Check answer","correctButtonText":"🌟 Yes! You got it!","tryAgainButtonText":"Try again","problemAiConversationId":null}

Graphs of nonlinear piecewise functions

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