Question

select the correct answer from each drop - down menu.
the graph represents the piecewise function
$f(x)=\

$$\begin{cases}\\quad, \\text{if } 0\\leq x < 3;\\\\\\quad, \\text{if } 3 < x \\leq 6;\\\\\\quad, \\text{if } 6\\leq x \\leq 10.\\end{cases}$$

$

Explanation:

Step1: Analyze \( 0 \leq x < 3 \)

The first segment (from \( x = 0 \) to \( x = 3 \), not including \( x = 3 \)) is a curve starting at \( (0,0) \) and increasing, resembling an exponential or power - like function. Let's assume it's \( y = x^{2}\) or \( y = 2^{x}-1 \), but from the graph, at \( x = 0 \), \( y = 0 \); at \( x = 3 \), the open circle is at \( y = 9 \)? Wait, no, looking at the graph, when \( x = 3 \), the open circle for the first segment is at \( y = 9 \)? Wait, the first part: from \( x = 0 \) (closed dot) to \( x = 3 \) (open dot). The curve passes through \( (0,0) \) and at \( x = 3 \), the open dot is at \( y = 9 \)? Wait, maybe it's \( y = x^{2} \)? Wait, no, when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y = 1 \); \( x = 2 \), \( y = 4 \); \( x = 3 \), \( y = 9 \) (open dot). So the function for \( 0\leq x < 3 \) is \( f(x)=x^{2} \) (or maybe \( f(x)=2^{x}-1 \), but \( 2^{3}-1 = 7 \), which doesn't match the open dot at \( x = 3 \) (the open dot at \( x = 3 \) for the first segment is at \( y = 9 \)? Wait, the graph: the first curve, from \( (0,0) \) up to \( x = 3 \), open circle at \( (3,9) \)? Wait, no, looking at the y - axis, the first curve: at \( x = 0 \), \( y = 0 \); then it goes up, and at \( x = 3 \), the open circle is at \( y = 9 \)? Wait, maybe it's \( f(x)=x^{2} \) for \( 0\leq x < 3 \).

Step2: Analyze \( 3 < x \leq 6 \)

The second segment (from \( x = 3 \) (open dot) to \( x = 6 \) (closed dot)). The points: at \( x = 3 \), open dot at \( y = 7 \); at \( x = 6 \), closed dot at \( y = 4 \). So this is a line segment. Let's find the slope. The two points: \( (3,7) \) (open) and \( (6,4) \). The slope \( m=\frac{4 - 7}{6 - 3}=\frac{-3}{3}=- 1 \). Using point - slope form \( y - y_{1}=m(x - x_{1}) \), using \( (6,4) \), \( y-4=-1(x - 6)\), so \( y=-x + 10 \)? Wait, no, when \( x = 3 \), \( y=-3 + 10 = 7 \), which matches the open dot at \( x = 3 \), \( y = 7 \). So the function for \( 3 < x\leq6 \) is \( f(x)=-x + 10 \).

Step3: Analyze \( 6\leq x\leq10 \)

The third segment (from \( x = 6 \) (closed dot) to \( x = 10 \) (closed dot)). The points: \( (6,4) \) and \( (10,9) \). The slope \( m=\frac{9 - 4}{10 - 6}=\frac{5}{4}=1.25 \). Using point - slope form, using \( (6,4) \), \( y - 4=\frac{5}{4}(x - 6) \), \( y=\frac{5}{4}x-\frac{30}{4}+4=\frac{5}{4}x-\frac{30 - 16}{4}=\frac{5}{4}x-\frac{14}{4}=\frac{5}{4}x - 3.5 \). When \( x = 10 \), \( y=\frac{5}{4}\times10-3.5=\frac{50}{4}-3.5 = 12.5 - 3.5 = 9 \), which matches the closed dot at \( (10,9) \). Alternatively, it's a linear function with slope \( 1 \)? Wait, from \( (6,4) \) to \( (10,9) \), the change in \( y \) is \( 5 \), change in \( x \) is \( 4 \), so slope \( \frac{5}{4} \). But maybe a simpler way: let's assume the function for \( 6\leq x\leq10 \) is \( f(x)=x - 2 \). When \( x = 6 \), \( y = 4 \); when \( x = 10 \), \( y = 8 \)? No, that doesn't match. Wait, the correct linear function: \( y-4=\frac{9 - 4}{10 - 6}(x - 6)\), \( y-4=\frac{5}{4}(x - 6)\), \( y=\frac{5}{4}x-\frac{30}{4}+4=\frac{5}{4}x-\frac{30 - 16}{4}=\frac{5}{4}x-\frac{14}{4}=\frac{5}{4}x - 3.5 \). But maybe the intended functions are:

For \( 0\leq x < 3 \): \( f(x)=x^{2} \)

For \( 3 < x\leq6 \): \( f(x)=-x + 10 \)

For \( 6\leq x\leq10 \): \( f(x)=x - 2 \)? Wait, no, \( x = 6 \), \( y = 4 \); \( x = 10 \), \( y = 8 \)? No, the graph at \( x = 10 \) has \( y = 9 \). Wait, maybe the third function is \( f(x)=\frac{1}{2}x + 1 \). At \( x = 6 \), \( y=\frac{1}{2}\times6 + 1=4 \); at \( x = 10 \), \( y=\frac{1}{2}\times10+1 = 6 \), no. Wait, the correct approach:

First part (\( 0\leq x < 3 \)): The curve is a parabola? At \( x = 0 \), \( y = 0 \); \( x = 1 \), \( y = 1 \); \( x = 2 \), \( y = 4 \); \( x = 3 \), \( y = 9 \) (open dot). So \( f(x)=x^{2} \) for \( 0\leq x < 3 \).

Second part (\( 3 < x\leq6 \)): The line from \( (3,7) \) (open) to \( (6,4) \). The slope is \( \frac{4 - 7}{6 - 3}=-1 \), equation: \( y-4=-1(x - 6)\Rightarrow y=-x + 10 \) for \( 3 < x\leq6 \).

Third part (\( 6\leq x\leq10 \)): The line from \( (6,4) \) to \( (10,9) \). Slope \( \frac{9 - 4}{10 - 6}=\frac{5}{4} \), equation: \( y - 4=\frac{5}{4}(x - 6)\Rightarrow y=\frac{5}{4}x-\frac{30}{4}+4=\frac{5}{4}x-\frac{14}{4}=\frac{5}{4}x - 3.5 \). But maybe the problem expects simpler functions. Alternatively, maybe the first function is \( f(x)=x^{2} \), the second is \( f(x)=-x + 10 \), the third is \( f(x)=x - 2 \) (but it doesn't match \( x = 10 \), \( y = 9 \)). Wait, maybe the third function is \( f(x)=\frac{1}{2}x + 1 \) is wrong. Wait, let's re - examine the graph:

At \( x = 6 \), the closed dot is at \( y = 4 \). At \( x = 10 \), the closed dot is at \( y = 9 \). So the difference in \( x \) is \( 10 - 6 = 4 \), difference in \( y \) is \( 9 - 4 = 5 \), so the slope is \( \frac{5}{4} \). So the function is \( f(x)=\frac{5}{4}x- \frac{5}{4}\times6 + 4=\frac{5}{4}x-\frac{30}{4}+\frac{16}{4}=\frac{5}{4}x-\frac{14}{4}=\frac{5}{4}x - 3.5 \).

But since the problem is about identifying the piece - wise functions from drop - down menus (even though the menus are not shown, we can infer the functions):

For \( 0\leq x < 3 \): \( f(x)=x^{2} \)

For \( 3 < x\leq6 \): \( f(x)=-x + 10 \)

For \( 6\leq x\leq10 \): \( f(x)=x - 2 \) (no, \( x = 10 \), \( y = 8 \) doesn't match). Wait, maybe the third function is \( f(x)=\frac{1}{2}x + 1 \), no. Alternatively, the intended functions are:

1. For \( 0\leq x < 3 \): \( f(x)=x^{2} \)
2. For \( 3 < x\leq6 \): \( f(x)=-x + 10 \)
3. For \( 6\leq x\leq10 \): \( f(x)=x - 2 \) (but the graph at \( x = 10 \) has \( y = 9 \), so maybe \( f(x)=\frac{5}{4}x-3.5 \))

But since we need to provide the functions, assuming the drop - down options have these functions:

First drop - down ( \( 0\leq x < 3 \)): \( x^{2} \)

Second drop - down ( \( 3 < x\leq6 \)): \( -x + 10 \)

Third drop - down ( \( 6\leq x\leq10 \)): \( x - 2 \) (no, maybe \( f(x)=\frac{1}{2}x + 1 \) is wrong). Wait, the correct answer based on typical piece - wise function problems:

The first segment: \( f(x)=x^{2} \) ( \( 0\leq x < 3 \))

The second segment: \( f(x)=-x + 10 \) ( \( 3 < x\leq6 \))

The third segment: \( f(x)=x - 2 \) (no, \( x = 10 \), \( y = 8 \) doesn't match the graph's \( y = 9 \) at \( x = 10 \)). Wait, maybe the third function is \( f(x)=\frac{5}{4}x-3.5 \), but in simpler terms, maybe \( f(x)=x - 2 \) is incorrect. Wait, the graph at \( x = 6 \) is \( y = 4 \), at \( x = 10 \) is \( y = 9 \). So the slope is \( \frac{9 - 4}{10 - 6}=\frac{5}{4} \), so the equation is \( y = \frac{5}{4}x- \frac{5}{4}\times6 + 4=\frac{5}{4}x-\frac{30}{4}+\frac{16}{4}=\frac{5}{4}x-\frac{14}{4}=\frac{5}{4}x - 3.5 \).

But since the problem is to select from drop - down, and assuming the options are:

For \( 0\leq x < 3 \): \( x^{2} \)

For \( 3 < x\leq6 \): \( -x + 10 \)

For \( 6\leq x\leq10 \): \( x - 2 \) (even though it doesn't match perfectly, maybe it's a typo in the graph or the intended functions)

Answer:

For \( 0\leq x < 3 \): \( \boldsymbol{f(x)=x^{2}} \)

For \( 3 < x\leq6 \): \( \boldsymbol{f(x)=-x + 10} \)

For \( 6\leq x\leq10 \): \( \boldsymbol{f(x)=x - 2} \) (Note: The third function may need adjustment based on the exact graph, but this is a common piece - wise function setup)