a function, $h(x)$, is defined as shown.\n$h(x)=\\begin{cases}\\frac{1}{4}x - 4, & x\\leq0\\\\\\frac{1}{3}x…

a function, $h(x)$, is defined as shown.\n$h(x)=\\begin{cases}\\frac{1}{4}x - 4, & x\\leq0\\\\\\frac{1}{3}x - 3, & 0 < x\\leq3\\\\\\frac{1}{2}x - 2, & x\\geq4\\end{cases}$\nwhich graph represents $h(x)$?

a function, $h(x)$, is defined as shown.\n$h(x)=\\begin{cases}\\frac{1}{4}x - 4, & x\\leq0\\\\\\frac{1}{3}x - 3, & 0 < x\\leq3\\\\\\frac{1}{2}x - 2, & x\\geq4\\end{cases}$\nwhich graph represents $h(x)$?

Answer

Answer:

We need to analyze the function (h(x)) for different intervals to determine the correct graph.

  1. For (x\leq0), (h(x)=\frac{1}{4}x - 4).
    • When (x = 0), (h(0)=\frac{1}{4}(0)-4=-4). The slope of this line is (\frac{1}{4}), and it is a decreasing - sloped line for (x\leq0).
  2. For (0\lt x\leq3), (h(x)=\frac{1}{3}x - 3).
    • When (x = 0) (the right - hand limit as we approach from the positive side), (h(0^{+})=\frac{1}{3}(0)-3=-3). When (x = 3), (h(3)=\frac{1}{3}(3)-3=1 - 3=-2). The slope of this line is (\frac{1}{3}).
  3. For (x\geq4), (h(x)=\frac{1}{2}x - 2).
    • When (x = 4), (h(4)=\frac{1}{2}(4)-2=2 - 2 = 0). The slope of this line is (\frac{1}{2}).

We look for a graph that has a line with slope (\frac{1}{4}) and (y) - intercept (-4) for (x\leq0), a line with slope (\frac{1}{3}) starting at an open - circle at (x = 0) (since (x>0) for this part) and ending at (x = 3), and a line with slope (\frac{1}{2}) starting at (x = 4).

Without seeing the full set of options clearly, we can analyze the key - points and slopes. The first line (y=\frac{1}{4}x - 4) has a (y) - intercept of (-4) and a gentle positive slope for (x\leq0). The second line (y=\frac{1}{3}x - 3) has a steeper slope than the first one and starts just above (y=-3) (open - circle at (x = 0)) and goes to (y=-2) at (x = 3). The third line (y=\frac{1}{2}x - 2) has the steepest slope among the three and starts at (y = 0) when (x = 4).

We need to check the graphs for these characteristics: the correct graph will have a line starting at ((0,-4)) with a slope of (\frac{1}{4}) for (x\leq0), then a line starting just above ((0,-3)) (open - circle) with a slope of (\frac{1}{3}) for (0\lt x\leq3), and then a line starting at ((4,0)) with a slope of (\frac{1}{2}) for (x\geq4).

Explanation:

Step1: Analyze (h(x)) for (x\leq0)

For (h(x)=\frac{1}{4}x - 4), when (x = 0), (h(0)=-4), slope (m=\frac{1}{4}).

Step2: Analyze (h(x)) for (0\lt x\leq3)

For (h(x)=\frac{1}{3}x - 3), (h(0^{+})=-3) (right - hand limit), (h(3)=-2), slope (m = \frac{1}{3}).

Step3: Analyze (h(x)) for (x\geq4)

For (h(x)=\frac{1}{2}x - 2), (h(4)=0), slope (m=\frac{1}{2}).

Step4: Match with graphs

Check graphs for correct slopes and endpoints.