express the function graphed on the axes below as a piecewise function.
answer attempt 1 out of 2
$f(x)= \begin{cases}
\boxed{\space} \text{for} \boxed{\space}\\
\boxed{\space} \text{for} \boxed{\space}
\end{cases}$

Question

Accepted Answer

# Explanation:
## Step1: Analyze the left segment
The left segment has endpoints at $(-4, -7)$ (open circle) and $(-2, 2)$ (open circle). Let's find the slope $m=\frac{2 - (-7)}{-2 - (-4)}=\frac{9}{2}=4.5$. Using point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(-4,-7)$, we get $y+7 = \frac{9}{2}(x + 4)$, which simplifies to $y=\frac{9}{2}x+18 - 7=\frac{9}{2}x + 11$. The domain for this segment is $-4<x<-2$.

## Step2: Analyze the right segment
The right segment has endpoints at $(-2, 4)$ (open circle) and $(4, 9)$ (open circle). The slope $m=\frac{9 - 4}{4-(-2)}=\frac{5}{6}$? Wait, no, let's recalculate. Wait, from $(-2,4)$ to $(4,9)$, the slope is $\frac{9 - 4}{4-(-2)}=\frac{5}{6}$? Wait, no, looking at the graph, when $x = - 2$, the left segment ends at $y = 2$ (open circle) and the right segment starts at $y = 4$ (open circle). Wait, maybe I misread the points. Let's re - examine the graph:

For the right - hand line: It passes through $(-1,5)$ (wait, the y - intercept seems to be 5) and $(4,9)$. The slope $m=\frac{9 - 5}{4-(-1)}=\frac{4}{5}$? No, wait, let's take two points on the right line. Let's see, when $x=-2$, the point is $(-2,4)$ (open circle) and when $x = 4$, the point is $(4,9)$. So the slope $m=\frac{9 - 4}{4-(-2)}=\frac{5}{6}$? No, $\frac{9 - 4}{4+2}=\frac{5}{6}\approx0.833$. Wait, but when $x = 0$, $y = 5$, so the equation is $y=x + 5$? Let's check: when $x=-2$, $y=-2 + 5=3$? No, that's not right. Wait, the point at $x=-2$ for the right line is $(-2,4)$, so if $y=mx + b$, when $x=-2$, $y = 4$, and when $x = 4$, $y = 9$. Then $4=-2m + b$ and $9 = 4m + b$. Subtract the first equation from the second: $9-4=(4m + b)-(-2m + b)$, $5 = 6m$, so $m=\frac{5}{6}$, and then $b=4 + 2\times\frac{5}{6}=4+\frac{5}{3}=\frac{12 + 5}{3}=\frac{17}{3}\approx5.666$. Wait, this is getting confusing. Wait, maybe the right line has a slope of 1? Let's check the points: from $(-2,4)$ to $(4,9)$, the difference in $y$ is $9 - 4 = 5$, difference in $x$ is $4-(-2)=6$, no. Wait, maybe I made a mistake in the left - hand segment.

Wait, let's start over. The graph has two pieces:

1. The lower piece (left piece): connects $(-4,-7)$ (open) to $(-2,2)$ (open).
   - Slope $m=\frac{2-(-7)}{-2 - (-4)}=\frac{9}{2}=4.5$.
   - Equation: Using point - slope form with $(x_1,y_1)=(-4,-7)$: $y+7=\frac{9}{2}(x + 4)$
   - $y=\frac{9}{2}x+18 - 7=\frac{9}{2}x + 11$. Domain: $-4<x<-2$.

2. The upper piece (right piece): connects $(-2,4)$ (open) to $(4,9)$ (open).
   - Slope $m=\frac{9 - 4}{4-(-2)}=\frac{5}{6}$? No, wait, when $x=-2$, $y = 4$; when $x = 4$, $y = 9$. Let's use the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{9 - 4}{4-(-2)}=\frac{5}{6}$. Then using point - slope form with $(x_1,y_1)=(-2,4)$: $y - 4=\frac{5}{6}(x + 2)$
   - $y=\frac{5}{6}x+\frac{10}{6}+4=\frac{5}{6}x+\frac{5}{3}+4=\frac{5}{6}x+\frac{17}{3}$. But this seems complicated. Wait, maybe the right line has a slope of 1. Let's check $x = 0$, $y = 5$. So $y=x + 5$. Let's check $x=-2$: $y=-2 + 5 = 3$, but the point at $x=-2$ for the right line is $(-2,4)$. So that's wrong. Wait, maybe the right line passes through $(-2,4)$ and $(4,9)$. Let's calculate the slope again: $\frac{9 - 4}{4-(-2)}=\frac{5}{6}\approx0.833$.

Wait, perhaps I misidentified the points. Let's look at the graph again. The right - hand line: when $x=-2$, the $y$ - value is 4 (open circle), and when $x = 4$, $y = 9$ (open circle). The left - hand line: when $x=-4$, $y=-7$ (open circle), and when $x=-2$, $y = 2$ (open circle).

So for the left piece (domain $-4<x<-2$):
- Let's use two points $(-4,-7)$ and $(-2,2)$. The slope $m=\frac{2-(-7)}{-2 - (-4)}=\frac{9}{2}$. The equation is $y=\frac{9}{2}x + b$. Plug in $(-4,-7)$: $-7=\frac{9}{2}(-4)+b$, $-7=-18 + b$, so $b = 11$. So $y=\frac{9}{2}x+11$ for $-4<x<-2$.

For the right piece (domain $-2<x<4$):
- Use two points $(-2,4)$ and $(4,9)$. The slope $m=\frac{9 - 4}{4-(-2)}=\frac{5}{6}$. The equation is $y=\frac{5}{6}x + b$. Plug in $(-2,4)$: $4=\frac{5}{6}(-2)+b$, $4=-\frac{5}{3}+b$, so $b=4+\frac{5}{3}=\frac{12 + 5}{3}=\frac{17}{3}$. So $y=\frac{5}{6}x+\frac{17}{3}$ for $-2<x<4$.

# Answer:
$f(x)=\begin{cases}\frac{9}{2}x + 11& \text{for }-4<x<-2\\frac{5}{6}x+\frac{17}{3}& \text{for }-2<x<4\end{cases}$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the left segment

Step2: Analyze the right segment

Answer: