Question

the graph of a piecewise - defined function is given. write a definition for the function that best describes this graph.
$f(x)=\begin{cases}square&\text{if }squareleq xleqsquare\square&\text{if }square< xleqsquareend{cases}$
(simplify your answers. use integers or fractions for any numbers in the expressions.)

Explanation:

Step1: Find the first piece (left segment)

The left segment connects \((-8, 2)\) and \((0, 0)\). The slope \(m_1=\frac{0 - 2}{0 - (-8)}=\frac{-2}{8}=-\frac{1}{4}\). Using point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(0,0)\), the equation is \(y =-\frac{1}{4}x\). The domain for this segment: from \(x=-8\) (inclusive) to \(x = 0\) (inclusive), so \(-8\leq x\leq0\).

Step2: Find the second piece (right segment)

The right segment connects \((0, 0)\) and \((3,-3)\). The slope \(m_2=\frac{-3 - 0}{3 - 0}=\frac{-3}{3}=- 1\). Using point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(0,0)\), the equation is \(y=-x\). The domain for this segment: from \(x = 0\) (exclusive, since the first piece includes \(x = 0\)) to \(x=3\) (inclusive), so \(0\lt x\leq3\).

Answer:

\(f(x)=

$$\begin{cases}-\frac{1}{4}x & \text{if } - 8\leq x\leq0\\-x & \text{if } 0\lt x\leq3\end{cases}$$

\)