Question

look at the correct answer
graph a
graph of a
graph b
graph of b
graph c
graph of c
graph d
graph of d
which is graph of the function?
$f(x) = \

$$\begin{cases} 2x + 4 & \\text{for } x \\leq 2 \\\\ 8 - x & \\text{for } x > 2 \\end{cases}$$

$
a. graph a
b. graph b
c. graph c
d. graph d

Explanation:

Step1: Analyze the piecewise function

The function is \( f(x) =

$$\begin{cases} \sqrt{x + 3} & \text{for } x \geq -3 \\ 5 - x & \text{for } x > 3 \end{cases}$$

\) (assuming the second part is for \( x > 3 \) from the graph context). First, check the domain and behavior of each part.

For \( y=\sqrt{x + 3} \), the domain is \( x \geq - 3 \), and when \( x=-3 \), \( y = 0 \); when \( x = 1 \), \( y=\sqrt{4}=2 \); when \( x = 6 \), if we consider the first part, but wait, the second part starts at \( x>3 \). Wait, maybe the second part is for \( x \geq 3 \)? Wait, let's check the graphs.

Step2: Check the first part \( y = \sqrt{x + 3} \)

• When \( x=-3 \), \( y = 0 \)? No, \( \sqrt{-3 + 3}=0 \), but wait, in the function, maybe the first part is \( \sqrt{x + 3} \) for \( x \geq - 3 \), and the second part \( 5 - x \) for \( x \geq 3 \) (since the second part is a line decreasing). Let's check the points:

For \( y=\sqrt{x + 3} \):

• At \( x = 1 \), \( y=\sqrt{4}=2 \)? No, wait \( \sqrt{1 + 3}=2 \), but in the graph, let's see Graph C: The left part (first function) starts at \( x=-3 \) (y=0? No, wait \( \sqrt{-3 + 3}=0 \), but in Graph C, the left part starts at (0, 2)? Wait, maybe I misread the function. Wait the function is \( f(x)=

$$\begin{cases} \sqrt{x + 3} & \text{for } x \geq -3 \\ 5 - x & \text{for } x > 3 \end{cases}$$

\) (assuming the first part is \( \sqrt{x + 3} \), so when \( x = 1 \), \( \sqrt{4}=2 \); when \( x = 6 \), \( \sqrt{9}=3 \)? Wait no, the second part is \( 5 - x \), so when \( x = 3 \), \( 5 - 3 = 2 \); when \( x = 5 \), \( 5 - 5 = 0 \).

Now check the graphs:

Graph A: The left part (first function) has a point at (1,4)? No.

Graph B: The left part starts at (0,2), but \( x=-3 \) should be in the domain.

Graph C: Let's check the left part (first function): when \( x=-3 \), \( y = 0 \)? No, in Graph C, the left part starts at (0,2)? Wait, maybe the function is \( \sqrt{x + 3} \), so when \( x = 1 \), \( y = 2 \); \( x = 6 \), \( y = 3 \) (but the second part is \( 5 - x \), so at \( x = 3 \), \( y = 2 \); \( x = 5 \), \( y = 0 \).

Looking at Graph C: The left curve (first function) goes from ( - 3, 0)? No, maybe the graph is shifted. Wait, the key is to check the two parts: the square root function (increasing, domain \( x \geq - 3 \)) and the line \( y = 5 - x \) (decreasing, starting at \( x = 3 \), \( y = 2 \), ending at \( x = 5 \), \( y = 0 \)).

Graph C: The left part (first function) has a curve that starts, and the right part is a line from (3,2) to (5,0), which matches \( y = 5 - x \) (when \( x = 3 \), \( y = 2 \); \( x = 5 \), \( y = 0 \)). The left part: when \( x = 1 \), \( y=\sqrt{4}=2 \); \( x = 6 \), \( y=\sqrt{9}=3 \)? Wait no, the right part is for \( x > 3 \), so at \( x = 4 \), \( y = 5 - 4 = 1 \)? Wait, maybe I made a mistake. Wait the function is \( f(x)=

$$\begin{cases} \sqrt{x + 3} & \text{if } x \leq 3 \\ 5 - x & \text{if } x > 3 \end{cases}$$

\). Then for \( x \leq 3 \), \( y=\sqrt{x + 3} \): when \( x = 1 \), \( y = 2 \); \( x = 3 \), \( y=\sqrt{6}\approx 2.45 \)? No, maybe the function is \( \sqrt{x + 3} \) for \( x \geq - 3 \) and \( 5 - x \) for \( x \geq 3 \).

Wait, looking at the graphs, Graph C has the right part (line) from (3,2) to (5,0), which is \( y = 5 - x \) (slope - 1, y - intercept 5, but when \( x = 3 \), \( y = 2 \); \( x = 5 \), \( y = 0 \)). The left part: the curve, when \( x = 1 \), \( y = 2 \) (since \( \sqrt{1 + 3}=2 \)), which matches. So Graph C is the correct one.

Answer:

C. Graph C