answer attempt 1 out of 2 $f(x) = \begin{cases} \square \text{ for } \square \\ \square \text{ for } \square \end{cases}$ (there are buttons for , ≤, ≥, =, ≠, add rule, remove rule, and a submit answer button)

Question

answer attempt 1 out of 2
$f(x) = \begin{cases} \square \text{ for } \square \\ \square \text{ for } \square \end{cases}$
(there are buttons for <, >, ≤, ≥, =, ≠, add rule, remove rule, and a submit answer button)

Accepted Answer

# Explanation:
## Step1: Analyze the left piece
The left piece has a hollow circle at $ x = -5 $ (point $(-5, 6)$) and a solid circle at $ x = 0 $ (point $(0, -4)$) and intersects the x - axis at $ x=-2 $. Let's find the slope $ m_1 $ between $(-5, 6)$ and $(0, -4)$. The slope formula is $ m=\frac{y_2 - y_1}{x_2 - x_1} $, so $ m_1=\frac{-4 - 6}{0-(-5)}=\frac{-10}{5}=- 2 $. Using the point - slope form $ y - y_1=m(x - x_1) $ with $(x_1,y_1)=(-5,6)$, we get $ y-6=-2(x + 5) $, which simplifies to $ y=-2x-10 + 6=-2x - 4 $. The domain of this piece: since the hollow circle is at $ x=-5 $ and the solid circle is at $ x = 0 $, the domain is $ -5<x\leq0 $.

## Step2: Analyze the right piece
The right piece has a hollow circle at $ x = 0 $ (point $(0,-8)$) and a solid circle at $ x = 6 $ (point $(6,7)$) and intersects the x - axis at $ x = 3 $. Let's find the slope $ m_2 $ between $(0,-8)$ and $(6,7)$. Using the slope formula $ m=\frac{y_2 - y_1}{x_2 - x_1} $, we have $ m_2=\frac{7-(-8)}{6 - 0}=\frac{15}{6}=\frac{5}{2} $. Using the point - slope form with $(x_1,y_1)=(0,-8)$, we get $ y+8=\frac{5}{2}(x - 0) $, which simplifies to $ y=\frac{5}{2}x-8 $. The domain of this piece: since the hollow circle is at $ x = 0 $ and the solid circle is at $ x = 6 $, the domain is $ 0<x\leq6 $.

# Answer:
$ f(x)=\begin{cases}-2x - 4& \text{for }-5<x\leq0\\frac{5}{2}x-8& \text{for }0<x\leq6\end{cases} $

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the left piece

Step2: Analyze the right piece

Answer: