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 line

The left line passes through (-8, 0) and (-4, 5). The slope \( m_1=\frac{5 - 0}{-4-(-8)}=\frac{5}{4} \). Using point - slope form \( y - y_1=m(x - x_1) \) with \((x_1,y_1)=(-8,0)\), we get \( y-0=\frac{5}{4}(x + 8) \), which simplifies to \( y=\frac{5}{4}x + 10 \). The domain for this line is \( x\lt - 4 \) (since there is an open circle at \( x=-4 \)).

Step2: Analyze the right line

The right line passes through (2, 6) and (8, 0). The slope \( m_2=\frac{0 - 6}{8 - 2}=\frac{-6}{6}=-1 \). Using point - slope form \( y - y_1=m(x - x_1) \) with \((x_1,y_1)=(2,6)\), we get \( y - 6=-1(x - 2) \), which simplifies to \( y=-x + 8 \). The domain for this line is \( x\geq2 \) (since there is an open circle at \( x = 2 \))? Wait, no, looking at the graph again, the left line is from \( x\lt - 4 \) and the right line is from \( x\gt2 \)? Wait, maybe I misread the points. Wait, the left line: let's check the x - intercept. The left line crosses the x - axis at \( x=-8 \), and has an open circle at \( x = - 4 \), \( y = 5 \). The right line has an open circle at \( x = 2 \), \( y=6 \) and crosses the x - axis at \( x = 8 \). So the left segment: when \( x\lt - 4 \), the function is \( y=x + 8 \)? Wait, if \( x=-8 \), \( y = 0 \), and \( x=-4 \), \( y=4 \)? No, earlier calculation was wrong. Let's recalculate the left line. Let's take two points on the left line: let's say when \( x=-8 \), \( y = 0 \); when \( x=-4 \), \( y = 4 \)? Wait, the open circle on the left is at \( x=-4 \), \( y = 5 \)? Wait, maybe the grid is 1 unit per square. Let's count the grid. From \( x=-8,y = 0 \) to \( x=-4,y = 4 \)? No, the open circle on the left is at \( x=-4 \), \( y = 5 \). So the slope between \( (-8,0) \) and \( (-4,5) \): \( m=\frac{5 - 0}{-4-(-8)}=\frac{5}{4} \), so \( y=\frac{5}{4}x+10 \) (since when \( x=-8 \), \( y=\frac{5}{4}(-8)+10=-10 + 10 = 0 \), correct). For the right line: open circle at \( x = 2 \), \( y = 6 \), and goes to \( x = 8 \), \( y = 0 \). The slope is \( \frac{0 - 6}{8 - 2}=-1 \), so \( y=-x + 8 \) (when \( x = 2 \), \( y=-2 + 8 = 6 \), correct; when \( x = 8 \), \( y=-8 + 8 = 0 \), correct). The domain of the left part: \( x\lt - 4 \), the domain of the right part: \( x\gt2 \)? Wait, but the problem's piecewise function has two parts. Wait, maybe the left part is for \( x\lt - 4 \) and the right part is for \( x\geq2 \)? No, the graph: the left line is from \( x\) approaching \( -\infty \) up to \( x=-4 \) (open circle), and the right line is from \( x = 2 \) (open circle) to \( x\) approaching \( +\infty \). Wait, maybe there is a gap between \( x=-4 \) and \( x = 2 \). So the piecewise function is:

\( f(x)=\begin{cases}x + 8, & x\lt - 4\\-x + 8, & x\gt2\end{cases} \)

Wait, when \( x=-8 \), \( y=-8 + 8 = 0 \), correct. When \( x=-4 \), \( y=-4 + 8 = 4 \), but the open circle is at \( y = 5 \), so my previous slope calculation was wrong. Oh! I see, I miscounted the y - value at \( x=-4 \). The open circle on the left is at \( x=-4 \), \( y = 5 \). So the two points on the left line: \( (-8,0) \) and \( (-4,5) \). So slope \( m=\frac{5 - 0}{-4-(-8)}=\frac{5}{4} \). So the equation is \( y-0=\frac{5}{4}(x + 8)\), \( y=\frac{5}{4}x+10 \). When \( x=-4 \), \( y=\frac{5}{4}(-4)+10=-5 + 10 = 5 \), correct.

For the right line: points \( (2,6) \) and \( (8,0) \). Slope \( m=\frac{0 - 6}{8 - 2}=-1 \). Equation: \( y - 6=-1(x - 2)\), \( y=-x + 8 \). When \( x = 2 \), \( y=-2 + 8 = 6 \), correct. When \( x = 8 \), \( y=-8 + 8 = 0 \), correct.

So the domain for the left function is \( x\lt - 4 \), and for the right function is \( x\gt2 \) (since there are open circles at \( x=-4 \) and \( x = 2 \), so the function is defined as two pieces: one for \( x\lt - 4 \) and one for \( x\gt2 \)).

Answer:

\( f(x)=

$$\begin{cases}\frac{5}{4}x + 10, & x\lt - 4\\-x + 8, & x\gt2\end{cases}$$

\)