Question

The graph below is comprised of transformation of tool kit functions.-Complete a piecewise defined function that describes the graph.

Answer 1

Step-by-step explanation

We need to find the equations for each of the functions that compose the piecewise function. First it's important to identify the type of each function. The graph at the left seems to be a parabola which implies that its equation is a quadratic one:

`ax^2+bx+c`

The graph on the right looks like the graph of a function with an absolute value of a linear term since it's v-shaped. We can start with this one.

As I stated, the graph of a function with an absolute value of a linear term has the shape of a v. The vertex of the v is located at the x-value for which the term inside the module is equal to 0. In this case the vertex is located at (5,-2) so the term with an absolute value could be |x-5|. Then a generic form for this equation would be:

`a|x-5|+b`

We can use two of the points that are part of this graph to find a and b. As you can see this graph passes through the points (3,-1) and (5,-2). Then we have the following:

`\begin{gathered} (3,-1)=(x,a\lvert x-5\rvert+b)=(3,a\lvert3-5\rvert+b)=(3,a\lvert-2\rvert+b)=(3,2a+b) \\ (3,-1)=(3,2a+b) \\ -1=2a+b \end{gathered}`

We can substract 2a from both sides:

`\begin{gathered} -1-2a=2a+b-2a \\ b=-1-2a \end{gathered}`

Then with (5,-2) we get:

`\begin{gathered} (5,-2)=(x,a\lvert x-5\rvert+b)=(5,a\lvert5-5\rvert+b)=(3,a\lvert0\rvert+b)=(3,b) \\ (5,-2)=(3,b) \\ b=-2 \end{gathered}`

So we know that b=-2. Then we take the equation we found before:

`b=-2=-1-2a`

We can add 2a+2 to both sides:

`\begin{gathered} -2+2a+2=-1-2a+2a+2 \\ 2a=1 \end{gathered}`

We divide both sides by 2:

`\begin{gathered} (2a)/(2)=(1)/(2) \\ a=(1)/(2) \end{gathered}`

So the function with the graph at x≥3 is:

`(1)/(2)|x-5|-2`

Now we must find the quadratic function. It passes through (0,0) so if we take x=0 the result of the quadratic expression must be 0:

`\begin{gathered} a\cdot0^2+b\cdot0+c=0 \\ c=0 \end{gathered}`

So c=0 and for now the quadratic function is:

`ax^2+bx`

Just like before we can use two points of the parabola to find a and b. As you can see the parabola passes through (2,4) and (3,3) so we have:

So we have these two equations:

`\begin{gathered} 4a+2b=4 \\ 9a+3b=3 \end{gathered}`

We can take the first equation and substract 4a from both sides:

`\begin{gathered} 4a+2b-4a=4-4a \\ 2b=4-4a \end{gathered}`

And we divide both sides by 2:

`\begin{gathered} (2b)/(2)=(4-4a)/(2) \\ b=2-2a \end{gathered}`

We use this in the second equation:

`\begin{gathered} 9a+3b=9a+3\cdot(2-2a)=9a+6-6a=3a+6=3 \\ \begin{equation*} 3a+6=3 \end{equation*} \end{gathered}`

We substract 6 from both sides and then we divide by 3:

`\begin{gathered} 3a+6-6=3-6 \\ 3a=-3 \\ (3a)/(3)=-(3)/(3) \\ a=-1 \end{gathered}`

And we use this to find b:

`b=2-2a=2-2\cdot(-1)=2+2=4`

So a=-1 and b=4 which implies that the quadratic function is:

`-x^2+4x`Answer

Then the piecewise function is:

`f(x)=\begin{cases}-x^2+4x\text{ if }x<3 \\ (1)/(2)|x-5|-2\text{ if }x\ge3\end{cases}`