Question

motion below by dragging the endpoints of the lines to the correct locations, then sele
in the graph or not.

$f(x) = \

$$\begin{cases} x^2 - 2 & \\text{if } x < 0 \\\\ x + 1 & \\text{if } 0 \\leq x \\leq 4 \\\\ -2 & \\text{if } x > 4 \\end{cases}$$

$

Explanation:

To graph the piece - wise function \(f(x)=

$$\begin{cases}x^{2}-2, & x < 0\\x + 1, & 0\leq x\leq4\\-2, & x>4\end{cases}$$

\), we analyze each piece:

1. For the piece \(y = x^{2}-2\) when \(x < 0\)

• This is a parabola of the form \(y=x^{2}+k\) (here \(k = - 2\)) with the vertex at \((0,-2)\). But since we are only considering \(x < 0\), we take the left - hand side of the parabola \(y=x^{2}-2\).
• When \(x=-1\), \(y=(-1)^{2}-2=1 - 2=-1\). When \(x = - 2\), \(y=(-2)^{2}-2=4 - 2 = 2\). We can plot the points \((-1,-1)\), \((-2,2)\) and draw the curve for \(x < 0\) (the left - half of the parabola \(y=x^{2}-2\)).

2. For the piece \(y=x + 1\) when \(0\leq x\leq4\)

• This is a linear function in the form \(y=mx + b\) where \(m = 1\) and \(b = 1\).
• When \(x = 0\), \(y=0 + 1=1\). When \(x=4\), \(y=4 + 1=5\). We draw a line segment connecting the points \((0,1)\) and \((4,5)\) (including both endpoints since \(0\leq x\leq4\)).

3. For the piece \(y=-2\) when \(x>4\)

• This is a horizontal line. For any \(x>4\), the \(y\) - value is \(-2\). We draw a horizontal line starting from the point \((4,-2)\) (but not including \((4,-2)\) since \(x>4\)) and extending to the right.

After graphing each piece, we can check the continuity of the function at the boundary points:

• At \(x = 0\):
• For \(x\to0^{-}\) (approaching \(0\) from the left), we use \(y=x^{2}-2\). \(\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}(x^{2}-2)=0^{2}-2=-2\).
• For \(x = 0\) (using the second piece \(y=x + 1\)), \(f(0)=0 + 1=1\). Since \(\lim_{x\to0^{-}}f(x)

eq f(0)\), the function is discontinuous at \(x = 0\).

• At \(x = 4\):
• For \(x\to4^{-}\) (approaching \(4\) from the left), we use \(y=x + 1\). \(\lim_{x\to4^{-}}f(x)=\lim_{x\to4^{-}}(x + 1)=4 + 1=5\).
• For \(x\to4^{+}\) (approaching \(4\) from the right), we use \(y=-2\). \(\lim_{x\to4^{+}}f(x)=-2\). And \(f(4)=4 + 1=5\). Since \(\lim_{x\to4^{+}}f(x)

eq f(4)\), the function is discontinuous at \(x = 4\).

If the question was to graph the function, the steps above help in constructing the graph. If the question was about continuity, we can conclude that the function is discontinuous at \(x = 0\) and \(x = 4\).

Since the original problem statement seems to be cut off (mentioning dragging endpoints of lines to correct locations and then selecting if something is in the graph or not), if we assume the task is to graph the function, the key is to plot each piece as described.

If we were to find, for example, \(f(-1)\):

Step 1: Determine the relevant piece

Since \(-1<0\), we use the first piece \(f(x)=x^{2}-2\).

Step 2: Substitute \(x=-1\) into the function

\(f(-1)=(-1)^{2}-2=1 - 2=-1\)

If we were to find \(f(2)\):

Step 1: Determine the relevant piece

Since \(0\leq2\leq4\), we use the second piece \(f(x)=x + 1\).

Step 2: Substitute \(x = 2\) into the function

\(f(2)=2 + 1=3\)

If we were to find \(f(5)\):

Step 1: Determine the relevant piece

Since \(5>4\), we use the third piece \(f(x)=-2\).

Step 2: Substitute \(x = 5\) into the function

\(f(5)=-2\)

Since the original problem is not fully stated, but based on the given function, we can perform operations like evaluating the function at a point or analyzing its graph as above. If you can provide the complete question (e.g., evaluate \(f(a)\) for a specific \(a\), graph the function, check continuity at a point), I can give a more targeted answer.

Answer:

To graph the piece - wise function \(f(x)=

$$\begin{cases}x^{2}-2, & x < 0\\x + 1, & 0\leq x\leq4\\-2, & x>4\end{cases}$$

\), we analyze each piece:

1. For the piece \(y = x^{2}-2\) when \(x < 0\)

• This is a parabola of the form \(y=x^{2}+k\) (here \(k = - 2\)) with the vertex at \((0,-2)\). But since we are only considering \(x < 0\), we take the left - hand side of the parabola \(y=x^{2}-2\).
• When \(x=-1\), \(y=(-1)^{2}-2=1 - 2=-1\). When \(x = - 2\), \(y=(-2)^{2}-2=4 - 2 = 2\). We can plot the points \((-1,-1)\), \((-2,2)\) and draw the curve for \(x < 0\) (the left - half of the parabola \(y=x^{2}-2\)).

2. For the piece \(y=x + 1\) when \(0\leq x\leq4\)

• This is a linear function in the form \(y=mx + b\) where \(m = 1\) and \(b = 1\).
• When \(x = 0\), \(y=0 + 1=1\). When \(x=4\), \(y=4 + 1=5\). We draw a line segment connecting the points \((0,1)\) and \((4,5)\) (including both endpoints since \(0\leq x\leq4\)).

3. For the piece \(y=-2\) when \(x>4\)

• This is a horizontal line. For any \(x>4\), the \(y\) - value is \(-2\). We draw a horizontal line starting from the point \((4,-2)\) (but not including \((4,-2)\) since \(x>4\)) and extending to the right.

After graphing each piece, we can check the continuity of the function at the boundary points:

• At \(x = 0\):
• For \(x\to0^{-}\) (approaching \(0\) from the left), we use \(y=x^{2}-2\). \(\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}(x^{2}-2)=0^{2}-2=-2\).
• For \(x = 0\) (using the second piece \(y=x + 1\)), \(f(0)=0 + 1=1\). Since \(\lim_{x\to0^{-}}f(x)

eq f(0)\), the function is discontinuous at \(x = 0\).

• At \(x = 4\):
• For \(x\to4^{-}\) (approaching \(4\) from the left), we use \(y=x + 1\). \(\lim_{x\to4^{-}}f(x)=\lim_{x\to4^{-}}(x + 1)=4 + 1=5\).
• For \(x\to4^{+}\) (approaching \(4\) from the right), we use \(y=-2\). \(\lim_{x\to4^{+}}f(x)=-2\). And \(f(4)=4 + 1=5\). Since \(\lim_{x\to4^{+}}f(x)

eq f(4)\), the function is discontinuous at \(x = 4\).

If the question was to graph the function, the steps above help in constructing the graph. If the question was about continuity, we can conclude that the function is discontinuous at \(x = 0\) and \(x = 4\).

Since the original problem statement seems to be cut off (mentioning dragging endpoints of lines to correct locations and then selecting if something is in the graph or not), if we assume the task is to graph the function, the key is to plot each piece as described.

If we were to find, for example, \(f(-1)\):

Step 1: Determine the relevant piece

Since \(-1<0\), we use the first piece \(f(x)=x^{2}-2\).

Step 2: Substitute \(x=-1\) into the function

\(f(-1)=(-1)^{2}-2=1 - 2=-1\)

If we were to find \(f(2)\):

Step 1: Determine the relevant piece

Since \(0\leq2\leq4\), we use the second piece \(f(x)=x + 1\).

Step 2: Substitute \(x = 2\) into the function

\(f(2)=2 + 1=3\)

If we were to find \(f(5)\):

Step 1: Determine the relevant piece

Since \(5>4\), we use the third piece \(f(x)=-2\).

Step 2: Substitute \(x = 5\) into the function

\(f(5)=-2\)

Since the original problem is not fully stated, but based on the given function, we can perform operations like evaluating the function at a point or analyzing its graph as above. If you can provide the complete question (e.g., evaluate \(f(a)\) for a specific \(a\), graph the function, check continuity at a point), I can give a more targeted answer.