Question

answer attempt 1 out of 3
$f(x) = \

$$\begin{cases} \\boxed{\\space} \\text{ for } \\boxed{\\space} \\\\ \\boxed{\\space} \\text{ for } \\boxed{\\space} \\end{cases}$$

$

Explanation:

Step1: Analyze the upper piecewise function

The upper line segment has an open circle at \( x = -5 \) (coordinates \((-5, 8)\)) and a closed circle at \( x = -2 \) (coordinates \((-2, 2)\)). First, find the slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2 - 8}{-2 - (-5)}=\frac{-6}{3}=-2 \). Using point - slope form \( y - y_1=m(x - x_1) \) with \((x_1,y_1)=(-2,2)\), we get \( y - 2=-2(x + 2) \), which simplifies to \( y=-2x-2 \). The domain for this piece is \( -5 < x\leq - 2 \) (since the circle at \( x=-5 \) is open and at \( x = - 2 \) is closed).

Step2: Analyze the lower piecewise function

The lower line segment has an open circle at \( x=-2 \) (coordinates \((-2,-2)\)) and a closed circle at \( x = 4 \) (coordinates \((4,-5)\)). The slope \( m=\frac{-5-(-2)}{4-(-2)}=\frac{-3}{6}=-\frac{1}{2} \). Using point - slope form with \((x_1,y_1)=(4,-5)\), we get \( y + 5=-\frac{1}{2}(x - 4) \), which simplifies to \( y=-\frac{1}{2}x-3 \). The domain for this piece is \( -2 < x\leq4 \) (open circle at \( x = - 2 \), closed circle at \( x = 4 \)).

Answer:

\( f(x)=

$$\begin{cases}-2x - 2& \text{for } - 5 < x\leq - 2\\-\frac{1}{2}x-3& \text{for } - 2 < x\leq4\end{cases}$$

\)