Question

unit test
unit test complete
a function, h(x), is defined as shown.
\\( h(x) = \

$$\begin{cases} \\frac{1}{4}x - 4, & x \\leq 0 \\\\ \\frac{1}{3}x - 3, & 0 < x \\leq 3 \\\\ \\frac{1}{2}x - 2, & x \\geq 4 \\end{cases}$$

\\)
which graph represents h(x)?
three graphs are shown, omitted here as per instruction

Explanation:

Step1: Analyze \( h(x) \) for \( x \leq 0 \)

The function is \( \frac{1}{4}x - 4 \) when \( x \leq 0 \). Let's find the y - intercept (when \( x = 0 \)): \( h(0)=\frac{1}{4}(0)-4=-4 \)? Wait, no, wait. Wait, when \( x = 0 \), for the first piece \( x\leq0 \), \( h(0)=\frac{1}{4}(0)-4=-4 \)? Wait, no, maybe I made a mistake. Wait, let's check the domain again. The first piece is \( x\leq0 \), second \( 0 < x\leq3 \), third \( x\geq4 \). Wait, when \( x = 0 \), we use the first piece: \( h(0)=\frac{1}{4}(0)-4=-4 \)? But let's check the slope. The slope of the first piece is \( \frac{1}{4} \), which is a positive slope, but a small positive slope.

Step2: Analyze \( 0 < x\leq3 \)

The function is \( \frac{1}{3}x - 3 \). Let's find \( h(x) \) at \( x = 0^+ \) (approaching 0 from the right). \( h(0^+)=\frac{1}{3}(0)-3=-3 \). At \( x = 3 \), \( h(3)=\frac{1}{3}(3)-3=1 - 3=-2 \).

Step3: Analyze \( x\geq4 \)

The function is \( \frac{1}{2}x - 2 \). At \( x = 4 \), \( h(4)=\frac{1}{2}(4)-2=2 - 2=0 \). The slope here is \( \frac{1}{2} \), which is steeper than the first two slopes.

Now let's check the graphs:

- For the first piece (\( x\leq0 \)): slope \( \frac{1}{4} \), y - intercept at \( x = 0 \) is \( - 4 \)? Wait, no, maybe I messed up the first piece. Wait, the first piece is \( \frac{1}{4}x-4 \). Let's take \( x=-4 \): \( h(-4)=\frac{1}{4}(-4)-4=-1 - 4=-5 \). \( x = 0 \): \( h(0)=-4 \). So the line for \( x\leq0 \) has a slope of \( \frac{1}{4} \), passing through \( (0, - 4) \) (closed circle at \( x = 0 \) for the first piece? Wait, no, the domain of the first piece is \( x\leq0 \), so at \( x = 0 \), we use the first piece. The second piece starts at \( x>0 \), so at \( x = 0 \), the first piece is used, and the second piece starts at \( x>0 \) with an open circle at \( x = 0 \)? Wait, no, the function is defined as:

\( h(x)=\begin{cases}\frac{1}{4}x - 4, & x\leq0\\\frac{1}{3}x - 3, & 0 < x\leq3\\\frac{1}{2}x - 2, & x\geq4\end{cases} \)

So at \( x = 0 \), \( h(0)=\frac{1}{4}(0)-4=-4 \) (closed circle for the first piece). At \( x = 0^+ \), \( h(0^+)=\frac{1}{3}(0)-3=-3 \) (open circle for the second piece? Wait, no, the second piece is \( 0 < x\leq3 \), so at \( x = 0 \), the first piece is used, and at \( x>0 \), the second piece is used. So the graph for \( x\leq0 \) is a line with slope \( \frac{1}{4} \), passing through \( (0, - 4) \). For \( 0 < x\leq3 \), it's a line with slope \( \frac{1}{3} \), starting at \( x = 0^+ \) with \( h(0^+)=-3 \) (open circle at \( x = 0 \) for the second piece) and ending at \( x = 3 \) with \( h(3)=-2 \) (closed circle). For \( x\geq4 \), it's a line with slope \( \frac{1}{2} \), starting at \( x = 4 \) with \( h(4)=0 \) (closed circle) and increasing.

Now let's check the middle graph:

- The left - most line (for \( x\leq0 \)): has a slope, and when \( x = 0 \), it should be at \( y=-4 \)? Wait, maybe the middle graph: the left - hand line (for \( x\leq0 \)) has a slope, and the middle line (for \( 0 < x\leq3 \)) has a slope of \( \frac{1}{3} \), and the right - hand line (for \( x\geq4 \)) has a slope of \( \frac{1}{2} \).

Wait, maybe the correct graph is the middle one? Wait, let's re - evaluate:

First piece (\( x\leq0 \)): \( y=\frac{1}{4}x - 4 \). Let's take \( x=-4 \): \( y=\frac{1}{4}(-4)-4=-1 - 4=-5 \). \( x = 0 \): \( y=-4 \). So it's a line with slope \( \frac{1}{4} \), going from \( (x=-4,y = - 5) \) to \( (x = 0,y=-4) \).

Second piece (\( 0 < x\leq3 \)): \( y=\frac{1}{3}x - 3 \). At \( x = 1 \): \( y=\frac{1}{3}(1)-3=\frac{1}{3}-3=-\frac{8}{3}\approx - 2.67 \). At \( x = 3 \): \( y=-2 \).

Third piece (\( x\geq4 \)): \( y=\frac{1}{2}x - 2 \). At \( x = 4 \): \( y = 0 \), \( x = 5 \): \( y=\frac{1}{2}(5)-2=\frac{5}{2}-2=\frac{1}{2}=0.5 \), \( x = 6 \): \( y=\frac{1}{2}(6)-2=3 - 2 = 1 \).

Now looking at the three graphs:

- The first graph: The left - hand line (for \( x\leq0 \)) has a y - intercept at \( y = 1 \)? No, that's wrong.

- The middle graph: The left - hand line (for \( x\leq0 \)) goes down, slope positive but small, the middle line (for \( 0 < x\leq3 \)) has a slope, and the right - hand line (for \( x\geq4 \)) starts at \( x = 4 \) with \( y = 0 \) and goes up.

- The third graph: The lines are all above the x - axis, which can't be right because for \( x = 0 \), \( y=-4 \) in the first piece.

So the middle graph is the correct one.

Answer:

The middle graph (the one with three line segments: one for \( x\leq0 \), one for \( 0 < x\leq3 \), and one for \( x\geq4 \) with the appropriate slopes and y - values at the endpoints).