Question

express the function graphed on the axes below as a piecewise function.

Explanation:

Step1: Analyze the left - hand line

First, we find two points on the left - hand line. Let's assume the left - hand line passes through the points \((-5,0)\) and \((-4, - 6)\). The slope \(m\) of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
So, \(m=\frac{-6 - 0}{-4-(-5)}=\frac{-6}{1}=-6\).
Using the point - slope form \(y - y_1=m(x - x_1)\), with \((x_1,y_1)=(-5,0)\) and \(m = - 6\), we get \(y-0=-6(x + 5)\), which simplifies to \(y=-6x-30\). The domain of this part of the function: we can see from the graph that the left - hand line has an open circle at \(x=-4\) and extends to the left. So, the domain for this part is \(x\lt - 4\).

Step2: Analyze the right - hand line

Let's find two points on the right - hand line. The right - hand line has an open circle at \(x = 5\) (assuming the open circle is at \((5,-5)\)) and let's take another point, say \((10,-10)\). The slope \(m\) of the line passing through \((5,-5)\) and \((10,-10)\) is \(m=\frac{-10+5}{10 - 5}=\frac{-5}{5}=-1\).
Using the point - slope form with \((x_1,y_1)=(5,-5)\) and \(m=-1\), we have \(y+5=-1(x - 5)\), which simplifies to \(y=-x\). The domain of this part of the function: since the open circle is at \(x = 5\) and the line extends to the right, the domain is \(x\gt5\).

Answer:

\(f(x)=

$$\begin{cases}-6x - 30& \text{for }x\lt - 4\\-x& \text{for }x\gt5\end{cases}$$

\)