Question

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 - hand line

The left - hand line passes through the points \((0,2)\) and \((-2,0)\). The slope \(m\) of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For the points \((0,2)\) and \((-2,0)\), \(m=\frac{0 - 2}{-2-0}=\frac{-2}{-2} = 1\). Using the slope - intercept form \(y=mx + b\) (where \(b\) is the \(y\) - intercept), and since \(b = 2\) (the line crosses the \(y\) - axis at \((0,2)\)) and the line is defined for \(x\leq4\) (because the right - hand line starts at \(x = 4\) with an open circle), the equation of the left - hand line is \(y=x + 2\) for \(x\leq4\).

Step2: Analyze the right - hand line

The right - hand line has an open circle at \((4,4)\) and let's find another point. Let's assume it has a slope. Let's take two points on the right - hand line. We can see that as \(x\) increases, \(y\) decreases. Let's find the slope between \((4,4)\) and \((10,1)\) (we can estimate the slope). The slope \(m=\frac{1 - 4}{10 - 4}=\frac{-3}{6}=-\frac{1}{2}\). Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(4,4)\) and \(m =-\frac{1}{2}\), we get \(y-4=-\frac{1}{2}(x - 4)\), which simplifies to \(y=-\frac{1}{2}x+2 + 4=-\frac{1}{2}x + 6\)? Wait, no, let's re - calculate. Wait, if we take the two points: when \(x = 4\), \(y = 4\) (open circle) and when \(x=10\), \(y = 1\). The slope \(m=\frac{1 - 4}{10 - 4}=\frac{-3}{6}=-\frac{1}{2}\). So \(y-4=-\frac{1}{2}(x - 4)\), \(y=-\frac{1}{2}x+2 + 4\)? No, \(y-4=-\frac{1}{2}x + 2\), so \(y=-\frac{1}{2}x+6\)? Wait, no, let's check with \(x = 4\): \(y=-\frac{1}{2}(4)+6=-2 + 6 = 4\), which is correct. And for \(x = 10\), \(y=-\frac{1}{2}(10)+6=-5 + 6 = 1\), which matches our estimate. But actually, from the graph, the right - hand line is defined for \(x>4\). Wait, no, the open circle is at \(x = 4\), so the domain of the right - hand line is \(x>4\).

Wait, maybe a better way: The left - hand line: passes through \((0,2)\) and \((-2,0)\), slope \(m = 1\), equation \(y=x + 2\) for \(x\leq4\). The right - hand line: open circle at \((4,4)\), let's find the slope. Let's take two points on the right - hand line. Let's say when \(x = 4\), \(y = 4\) (open) and when \(x = 8\), \(y = 2\). Then the slope \(m=\frac{2 - 4}{8 - 4}=\frac{-2}{4}=-\frac{1}{2}\). So the equation using point - slope form with \((x_1,y_1)=(4,4)\) is \(y-4=-\frac{1}{2}(x - 4)\), so \(y=-\frac{1}{2}x+2 + 4\)? No, \(y-4=-\frac{1}{2}x + 2\), so \(y=-\frac{1}{2}x+6\)? Wait, when \(x = 4\), \(y=-\frac{1}{2}(4)+6=-2 + 6 = 4\), correct. When \(x = 8\), \(y=-\frac{1}{2}(8)+6=-4 + 6 = 2\), correct. So the right - hand line has the equation \(y=-\frac{1}{2}x + 6\) for \(x>4\). Wait, but maybe I made a mistake. Alternatively, looking at the graph, the left - hand line is \(y=x + 2\) for \(x\leq4\) (since it passes through \((0,2)\) and \((-2,0)\) and is a straight line with slope 1) and the right - hand line is \(y=-\frac{1}{2}x + 6\) for \(x>4\)? Wait, no, let's re - check the left - hand line. The line passes through \((0,2)\) and \((-2,0)\), so the slope is \(\frac{0 - 2}{-2-0}=1\), so the equation is \(y=x + 2\). And the domain of the left - hand line is \(x\leq4\) (because the right - hand line starts at \(x = 4\) with an open circle). The right - hand line: the open circle is at \((4,4)\), and let's see the slope. If we take two points on the right - hand line, say \((4,4)\) and \((10,1)\), the slope is \(\frac{1 - 4}{10 - 4}=-\frac{1}{2}\), so the equation is \(y-4=-\frac{1}{2}(x - 4)\), which simplifies to \(y=-\frac{1}{2}x+2 + 4\)? No, \(y-4=-\frac{1}{2}x + 2\), so \(y=-\frac{1}{2}x+6\). Wait, but maybe the correct equations are:

Left part: \(f(x)=x + 2\) for \(x\leq4\)

Right part: \(f(x)=-\frac{1}{2}x + 6\) for \(x>4\)

Wait, but maybe I made a mistake in the right - hand line. Alternatively, let's look at the graph again. The left - hand line goes through \((0,2)\) and \((-2,0)\), so it's \(y=x + 2\) for \(x\leq4\). The right - hand line has an open circle at \((4,4)\) and is a line with slope \(-\frac{1}{2}\), so \(y=-\frac{1}{2}x + 6\) for \(x>4\).

Answer:

The piece - wise function is \(f(x)=

$$\begin{cases}x + 2, & x\leq4\\-\frac{1}{2}x+6, & x>4\end{cases}$$

\)