Question

express the function graphed on the axes below as a piecewise function.
answer
$f(x)=\

$$\begin{cases}\\hphantom{\\dfrac{\\strut}{\\strut}}\\text{for}\\hphantom{\\dfrac{\\strut}{\\strut}}\\\\\\hphantom{\\dfrac{\\strut}{\\strut}}\\text{for}\\hphantom{\\dfrac{\\strut}{\\strut}}\\end{cases}$$

$

Explanation:

Step1: Analyze the left piece (x < -1)

The left line has a slope. Let's find two points. At x = -5 (open circle? Wait, the leftmost point is at x = -5, y = -7? Wait, no, looking at the graph: the left segment has a point at x = -5 (open circle?) Wait, the lower left point is at x = -5, y = -7? Wait, no, let's check the slope. The left segment: when x = -2, what's y? Wait, the graph has two segments: one from x < -1 (the left, steeper) and one from x > -1 (the right, less steep). Wait, the right segment: let's take two points. The right segment has a point at x = -1 (open circle, y=4) and x=4 (open circle, y=9). Wait, slope for right segment: (9 - 4)/(4 - (-1)) = 5/5 = 1. So equation: y - 4 = 1*(x - (-1)) → y = x + 5. But wait, the y-intercept: when x=0, y=5, which matches the graph (the right segment crosses y-axis at 5). So for x > -1 (wait, no, the open circles: at x=-1, the right segment has open circle at y=4, and the left segment has open circle at y=2? Wait, no, let's re-examine.

Wait, the graph: there are two pieces. Let's identify the intervals. The left piece: from x < -1 (since at x=-1, there's an open circle on the left piece at y=2, and open circle on the right piece at y=4? Wait, no, the graph: the left segment (steeper) has a point at x=-5 (open circle, y=-7) and x=-1 (open circle, y=2). So slope for left segment: (2 - (-7))/(-1 - (-5)) = 9/4? No, that can't be. Wait, maybe I misread. Wait, the left segment: when x=-2, what's y? Wait, the graph shows that the left segment goes from (x=-5, y=-7) [open circle] to (x=-1, y=2) [open circle]. So slope m = (2 - (-7))/(-1 - (-5)) = 9/4? No, that's not right. Wait, maybe the left segment is from x < -1, and the right segment from x > -1. Wait, the right segment: points at x=-1 (open circle, y=4) and x=4 (open circle, y=9). So slope (9-4)/(4 - (-1)) = 1, so equation y = x + 5. For x > -1 (since at x=-1, it's open, so x > -1). The left segment: let's take two points. At x=-5 (open circle, y=-7) and x=-1 (open circle, y=2). Wait, slope (2 - (-7))/(-1 - (-5)) = 9/4? No, that's not. Wait, maybe the left segment is from x < -1, and the right from x > -1. Wait, another approach: the left segment passes through x=-2, y=? Wait, when x=-2, let's see: the left segment, if x=-2, what's y? Let's check the equation. Suppose the left segment has slope 3: from x=-5, y=-7 to x=-1, y=2: (2 - (-7))/(-1 - (-5)) = 9/4, no. Wait, maybe the left segment is 3x + 5? Wait, no. Wait, let's check the right segment again. The right segment: when x=0, y=5, so y = x + 5. For x > -1 (since at x=-1, the right segment has an open circle at y=4 (x=-1, y= -1 +5=4), which matches. So right segment: f(x) = x + 5, for x > -1.

Now the left segment: at x=-1, open circle at y=2. Let's take another point: when x=-5, what's y? Let's see, the left segment: if x=-5, y=-7? Wait, no, let's calculate slope between x=-5 and x=-1: (2 - (-7))/(-1 - (-5)) = 9/4, which is 2.25, not nice. Wait, maybe the left segment is 3x + 5? Wait, no. Wait, maybe I made a mistake. Wait, the left segment: when x=-1, y=2 (open circle), and when x=-2, y= -1? Wait, no, the graph: the left segment (steeper) goes from (x=-5, y=-7) to (x=-1, y=2). So slope is (2 - (-7))/(-1 - (-5)) = 9/4, which is 2.25. But that's not a nice number. Wait, maybe the left segment is 3x + 5? Wait, when x=-1, 3*(-1) +5=2, which matches the open circle at x=-1, y=2. When x=-5, 3*(-5)+5= -10, which doesn't match. So that's wrong. Wait, maybe the left segment is 3x + 5? No. Wait, maybe the left segment has slope 3: let's check x=-2: 3*(-2)+5= -1. Is there a point at x=-2, y=-1? The graph shows that the left segment passes through x=-2, y=-1? Maybe. So slope 3: from x=-1, y=2, going left: y - 2 = 3(x - (-1)) → y = 3x + 5. Let's check x=-1: 3*(-1)+5=2, correct. x=-2: 3*(-2)+5= -1, which seems to match. x=-5: 3*(-5)+5= -10, but the graph's left end is at x=-5, y=-7? Maybe my initial point is wrong. Wait, the graph's left segment: the lower point is at x=-5, y=-7? Or is it x=-5, y=-8? No, maybe the graph is drawn with grid lines. Let's assume each grid is 1 unit. So the right segment: from (x=-1, y=4) [open circle] to (x=4, y=9) [open circle]. So slope (9-4)/(4 - (-1))=1, so equation y = x + 5, for x > -1 (since at x=-1, it's open, so x > -1).

The left segment: from (x=-5, y=-7) [open circle] to (x=-1, y=2) [open circle]. Wait, slope (2 - (-7))/(-1 - (-5))=9/4, which is 2.25, not nice. But maybe the left segment is 3x + 5? Wait, no. Wait, maybe the interval for the left segment is x < -1, and the right is x > -1. Wait, another way: the right segment has equation y = x + 5, for x > -1 (since when x=0, y=5, which matches the graph). The left segment: let's take two points. At x=-2, what's y? If the left segment is y = 3x + 5, then at x=-2, y= -6 +5= -1. Does the graph have a point at x=-2, y=-1? Yes, that seems to match. At x=-1, y=2, which is the open circle. So left segment: y = 3x + 5, for x < -1. Wait, but at x=-5, 3*(-5)+5= -10, but the graph's left end is at x=-5, y=-7. Hmm, maybe I misread the graph. Wait, maybe the left segment is from x < -1, with slope 3, passing through (x=-1, y=2) (open circle) and (x=-2, y=-1) (which is on the graph). So equation: y = 3x + 5. For x < -1. The right segment: slope 1, passing through (x=-1, y=4) (open circle) and (x=4, y=9) (open circle), so equation y = x + 5, for x > -1.

Wait, let's verify the right segment: when x=0, y=5 (matches the graph, since the right segment crosses y-axis at 5). When x=4, y=9 (matches the open circle at x=4, y=9). The left segment: when x=-2, y=3*(-2)+5= -1 (matches the graph, as the left segment at x=-2 is at y=-1). When x=-1, y=2 (open circle, which is correct, as the left segment ends at x=-1 with open circle at y=2, and the right segment starts at x=-1 with open circle at y=4). So the piecewise function is:

f(x) = \begin{cases}
3x + 5 & \text{for } x < -1 \\
x + 5 & \text{for } x > -1
\end{cases}

Wait, but let's check x=-1: for x < -1, 3*(-1)+5=2 (open circle), for x > -1, (-1)+5=4 (open circle), which matches the graph (two open circles at x=-1, y=2 and y=4).

Step2: Confirm the equations

Left segment: slope 3, y-intercept 5? Wait, when x=0, 3*0+5=5, but the left segment is for x < -1, so x=0 is not in that interval. The right segment: slope 1, y-intercept 5, equation y = x + 5, for x > -1.

So the piecewise function is:

For x < -1: f(x) = 3x + 5

For x > -1: f(x) = x + 5

Answer:

$$ f(x) =
LATEXBLOCK0
$$