{"question":{"question":{"content":"The graph of $y=f(x)$ is shown below. ![](web+graphie://ka-perseus-graphie.s3.amazonaws.com/21910e1ef2eec3e0ed3cd2e4ee8d711c804cdbed) **Which of the following could be function $f$?** [[☃ radio 1]] ","images":{"web+graphie://ka-perseus-graphie.s3.amazonaws.com/21910e1ef2eec3e0ed3cd2e4ee8d711c804cdbed":{"width":425,"height":425}},"widgets":{"radio 1":{"type":"radio","alignment":"default","static":false,"graded":true,"options":{"choices":[{"content":"$ f(x) = \begin{cases} (x-3)^2&&\text{if }-4 \leq x <-1\\ \\ -\sqrt{1-x^2}&&\text{if }-1 \leq x < 1 \end{cases} $ $$","correct":false},{"content":"$ f(x) = \begin{cases} (x+3)^2&&\text{if }-4 < x \leq-1\\ \\ -\sqrt{1-x^2}&&\text{if }-1 < x \leq1 \end{cases}$ $$","correct":false},{"content":"$ f(x) = \begin{cases} (x+3)^2&&\text{if }-4 \leq x <-1\\ \\ \sqrt{1-x^2}&&\text{if }-1 \leq x < 1 \end{cases} $ $$","correct":true},{"content":"$ f(x) = \begin{cases} (x-3)^2&&\text{if }-4< x \leq-1\\ \\ \sqrt{1-x^2}&&\text{if }-1 < x \leq 1 \end{cases} $ $$","correct":false}],"randomize":true,"multipleSelect":false,"displayCount":null,"hasNoneOfTheAbove":false,"onePerLine":true,"deselectEnabled":false},"version":{"major":1,"minor":0}}}},"answerArea":{"type":"multiple","options":{"content":"","images":{},"widgets":{}},"calculator":false,"periodicTable":false},"itemDataVersion":{"major":0,"minor":1},"hints":[{"content":"###The process Looking at the graph of the function $y=f(x)$, we notice that it comes in two *pieces*. We can come up with an equation for the *piecewise function* if we can come up with an *equation for each piece* of the function, and an *interval* over which each piece is defined. In other words, we want to find a function $f$ so that the following is true. >$ 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/6795b56b302cd2c11182eb8ce3cfcae4e5e54f97) 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/1e4d2d5ff30f07e9df4ec65fda602edc881e9eac":{"width":310,"height":310},"web+graphie://ka-perseus-graphie.s3.amazonaws.com/6795b56b302cd2c11182eb8ce3cfcae4e5e54f97":{"width":310,"height":310}},"widgets":{}},{"content":"###The left-hand piece $\blueD{\text{The graph on the left}}$ is $y=(x+3)^2$. [[☃ explanation 1]] It is confined to the interval $-4\leq x <-1 $. ","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 $y=x^2$. ![](web+graphie://ka-perseus-graphie.s3.amazonaws.com/e18bcc32f3e3c3789bd4591a58aa7e17f237c987) However, the graph of the function that we are interested in has been moved to the left $3$. So if $g(x)=x^2$, then the function we are interested in is $y=g(x+3)$. This function is built below. >$\begin{align}g(x)&=x^2\\\\ g(x+3)&=(x+3)^2\\\\ \end{align}$ Therefore, it follows that $y=(x+3)^2$.","widgets":{}},"version":{"major":0,"minor":0}}}},{"content":"###The right-hand piece $\greenD{\text{The graph on the right}}$ is $y=\sqrt{1-x^2}$. [[☃ explanation 1]] It is confined to the interval $-1\leq x<1$. ","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 the function on the right is the top portion of a circle with center $(0,0)$ and radius $1$. Since we know that $x^2+y^2=r^2$ is an equation for *full* circle with center $(0,0)$ and radius $r$, it follows that the equation for the full circle containing the given semicircle is $x^2+y^2=1$. We can solve the equation above for $y$. $\begin{align} x^2+y^2&= 1\\ \\ y^2 &= 1-x^2\\ \\ y&= \pm\sqrt{1-x^2} \end{align}$ Since we only want the top portion of the circle, our equation is $y=\sqrt{1-x^2}$.","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+3)^2}&&\text{if }-4 \leq x <-1\\ \\ \greenD{\sqrt{1-x^2}}&&\text{if }-1 \leq x < 1 \end{cases} $ $$","images":{},"widgets":{}}]},"problemId":37638,"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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