Question

express the function graphed on the axes below as a piecewise function.
answer attempt 1 out of 3
$f(x)=\

$$\begin{cases}\\square\\text{ for }\\square\\\\\\square\\text{ for }\\square\\end{cases}$$

$

Explanation:

Step1: Analyze the left piece (through origin and (1,6))

The left piece passes through \((0,0)\) and \((1,6)\). The slope \(m = \frac{6 - 0}{1 - 0}=6\). So the equation is \(y = 6x\), and it's defined for \(x\leq1\) (since the point at \(x = 1\) is a closed dot here).

Step2: Analyze the right piece (starts at open circle, let's assume the open circle is at (4,8) and has slope 1 (since it's a line with rise 1 over run 1, looking at the graph trend))

Wait, actually, from the open circle (let's check coordinates: the open circle is at \(x = 4\), \(y = 8\)? Wait, no, looking at the graph, the left line goes through (0,0) and (1,6), so slope 6. The right line: the open circle is at, say, \(x = 4\), and the line has a slope. Wait, maybe the right piece: let's see, the open circle is at (4,8)? Wait, no, maybe the right line has slope 1? Wait, no, let's re - examine.

Wait, the left part: when \(x = 1\), \(y = 6\) (closed dot). The right part: open circle at, say, \(x = 4\), and the line is, for example, if we take two points on the right line, but maybe the right line is \(y=x + 4\)? Wait, no, maybe I made a mistake. Wait, the left line: passes through (0,0) and (1,6), so equation \(y = 6x\) for \(x\leq1\). The right line: the open circle is at (4,8)? Wait, no, the open circle is at \(x = 4\), \(y = 8\)? Wait, no, let's look at the graph again. The left line goes from the bottom (negative y) through (0,0) to (1,6) (closed dot). The right line starts at an open circle, let's say at \(x = 4\), \(y = 8\), and has a slope of 1 (since it's a line with a positive slope, and if we assume the line is \(y=x + 4\), when \(x = 4\), \(y=8\), and for \(x>4\), the line is \(y=x + 4\)? Wait, no, maybe the right line has slope 1, and the open circle is at (4,8), so the equation is \(y=x + 4\) for \(x>4\)? Wait, no, maybe the right line is \(y=x + 4\) when \(x>4\), and the left line is \(y = 6x\) when \(x\leq1\)? Wait, no, there is a gap between \(x = 1\) and \(x = 4\)? Wait, maybe the graph has two pieces: one for \(x\leq1\) (the line through (0,0) and (1,6)) and one for \(x>4\) (the line with open circle). Wait, perhaps the right line has slope 1, and the open circle is at (4,8), so the equation is \(y=x + 4\) for \(x>4\).

Wait, let's correct:

Left piece: passes through (0,0) and (1,6). Slope \(m=\frac{6 - 0}{1 - 0}=6\). So equation \(y = 6x\), domain \(x\leq1\) (because at \(x = 1\), it's a closed dot, so included).

Right piece: the open circle is at (4,8). Let's find the slope of the right line. If we assume the line goes through (4,8) and (5,9) (since it's a line with slope 1), then the equation is \(y-8 = 1\times(x - 4)\), so \(y=x + 4\), and the domain is \(x>4\) (because the dot at \(x = 4\) is open, so not included).

So the piecewise function is:

\(f(x)=

$$\begin{cases}6x, & x\leq1\\x + 4, & x>4\end{cases}$$

\)

Wait, but maybe the right line has a different slope. Wait, maybe the right line is \(y = x+4\) for \(x>4\), and the left line is \(y = 6x\) for \(x\leq1\).

Answer:

\(f(x)=

$$\begin{cases}6x, & x\leq1\\x + 4, & x>4\end{cases}$$

\)