Question

which of the following is the expression of the piecewise - defined function below? (chart of the piecewise - defined function is shown, with x - axis and y - axis, and the function graph composed of different parts) last saved 10:44:55 pm questions filter (11) *

Explanation:

Step1: Analyze the left part (x ≤ 0)

The left part of the graph (for \( x \leq 0 \)) looks like a parabola opening upwards? Wait, no, when \( x \leq 0 \), the curve seems to be a parabola \( y = x^2 \)? Wait, no, let's check the vertex. Wait, the left part: when \( x = 0 \), the point is (0,0)? Wait, no, the graph at \( x = 0 \) has a point, and for \( x \leq 0 \), the curve is a parabola? Wait, maybe \( y = x^2 \) for \( x \leq 0 \)? Wait, no, let's check the right part.

Step2: Analyze the right part (x ≥ 0)

The right part (for \( x \geq 0 \)) is a line. Let's find the slope. The line passes through (0, 0) and (1, 2)? Wait, no, the point at \( x = 0 \) is (0, 2)? Wait, the graph shows a point at (0, 2) and then a line going up. Wait, maybe the right part is \( y = x + 2 \) for \( x \geq 0 \)? Wait, no, let's re-examine.

Wait, the graph has two parts: left of x=0 (maybe a parabola \( y = x^2 \) but shifted? No, maybe \( y = x^2 \) for \( x \leq 0 \) and \( y = x + 2 \) for \( x \geq 0 \)? Wait, no, let's check the coordinates. Wait, the left curve: when x=-2, y=4? If it's \( y = x^2 \), then at x=-2, y=4, which matches. The right line: when x=0, y=2; x=1, y=3? Wait, no, the right line starts at (0, 2) and goes up with slope 1? So equation \( y = x + 2 \) for \( x \geq 0 \). And left part: \( y = x^2 \) for \( x \leq 0 \). So the piecewise function is:

\( f(x) =

$$\begin{cases} x^2 & \text{if } x \leq 0 \\ x + 2 & \text{if } x > 0 \end{cases}$$

\)

Wait, but need to confirm the points. At x=0, both parts? Wait, the graph at x=0 has a closed dot, so for x ≤ 0, the function is \( x^2 \), and for x ≥ 0, it's a line. Wait, maybe the left part is \( y = x^2 \) (since at x=-2, y=4; x=-1, y=1; x=0, y=0? Wait, no, the graph at x=0 has a point at (0, 2)? Wait, I think I misread the graph. Let's look again. The graph has a point at (0, 2) and then a line going up, and the left curve comes down to (0, 0)? No, the image is a bit unclear, but typically, a piecewise function with left as \( y = x^2 \) (for x ≤ 0) and right as \( y = x + 2 \) (for x ≥ 0) or similar. Wait, maybe the correct piecewise function is:

\( f(x) =

$$\begin{cases} x^2 & \text{when } x \leq 0 \\ x + 2 & \text{when } x > 0 \end{cases}$$

\)

But since the options are not provided, but the process is to identify the piecewise function by analyzing each segment.

Answer:

The piecewise function is likely \( f(x) =

$$\begin{cases} x^2 & x \leq 0 \\ x + 2 & x > 0 \end{cases}$$

\) (assuming the graph's left is \( y = x^2 \) for \( x \leq 0 \) and right is a line \( y = x + 2 \) for \( x \geq 0 \)).