Question

which of the following is the graph for this piecewise defined function?
$f(x) = \

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

$

Explanation:

Step1: Analyze the first piece \( f(x) = x \) for \( x \leq 0 \)

This is a linear function with a slope of 1. When \( x = 0 \), \( f(0)=0 \). For \( x < 0 \), it's a straight line passing through the origin with a slope of 1, so it should include the point \( (0,0) \) (since \( x \leq 0 \) includes \( x = 0 \)).

Step2: Analyze the second piece \( f(x) = x^2 \) for \( x > 0 \)

This is a quadratic function opening upwards. For \( x > 0 \), we consider the right - hand side of the parabola \( y = x^2 \). At \( x = 0 \), the limit of \( x^2 \) as \( x \) approaches 0 from the right is \( 0^2=0 \), but since the domain for \( x^2 \) is \( x > 0 \), the point \( (0,0) \) is not included in the graph of \( y=x^2 \) for \( x > 0 \), but the graph approaches \( (0,0) \) from the right.

Step3: Compare with the given graphs

- The first graph: The left - hand side is a parabola (which would correspond to \( y = x^2 \) for \( x\leq0 \)), but our first piece is \( y = x \) for \( x\leq0 \), so this graph is incorrect.
- The second graph: The left - hand side is a line with slope 1 passing through the origin (correct for \( y = x, x\leq0 \)) and the right - hand side is the parabola \( y = x^2 \) for \( x > 0 \) with an open circle at \( (0,0) \)? No, wait, the left - hand side includes \( x = 0 \) (since \( x\leq0 \)), so the point \( (0,0) \) should be included in the left - hand line and the right - hand parabola approaches \( (0,0) \). Wait, no, the left - hand function \( f(x)=x \) at \( x = 0 \) is \( f(0) = 0 \), and the right - hand function \( f(x)=x^2 \) at \( x = 0 \) is not included (since \( x>0 \)), but the graph of the left - hand function includes \( (0,0) \). The second graph has a closed circle at \( (0,0) \) for the left - hand line and the right - hand parabola starts at \( (0,0) \)? No, the right - hand function \( x^2 \) for \( x > 0 \) has a graph that starts just to the right of 0. Wait, let's re - examine:
The third graph: The left - hand side is a line with slope 1 (correct for \( y = x, x\leq0 \)) with a closed circle at \( (0,0) \)? No, the third graph has an open circle at \( (0,0) \) for the left - hand line? No, wait the left - hand function \( f(x)=x \) for \( x\leq0 \) includes \( x = 0 \), so the point \( (0,0) \) should be on the left - hand line. The second graph: left - hand line (slope 1, passes through (0,0)) and right - hand parabola ( \( y=x^2,x > 0 \) ) with a closed circle? No, the right - hand function's domain is \( x>0 \), so the point \( (0,0) \) is not in the domain of \( x^2 \) (for \( x > 0 \)), but the left - hand function's domain includes \( x = 0 \). Wait, maybe I made a mistake. Let's check the three graphs again.
Wait, the function is:
- For \( x\leq0 \), \( y=x \): this is a straight line with slope 1, passing through the origin, and including the point \( (0,0) \) (since \( x = 0 \) is in the domain).
- For \( x>0 \), \( y=x^2 \): this is a parabola opening upwards, with the point \( (0,0) \) not included (since \( x>0 \)), but the graph approaches \( (0,0) \) from the right.

The second graph: The left - hand line is \( y = x \) with \( x\leq0 \) (includes \( (0,0) \), closed circle) and the right - hand parabola \( y=x^2 \) with \( x > 0 \) (open circle at \( (0,0) \))? No, the second graph has a closed circle at \( (0,0) \) for the right - hand parabola? No, looking at the graphs:
- First graph: symmetric about y - axis, left and right both parabola (wrong, left should be line).
- Second graph: left is line (slope 1, through (0,0), closed circle), right is parabola (starts at (0,0), closed circle? No, the right - hand function is for \( x>0 \), so the point (0,0) is not in its domain. Wait, no, the value of the function at \( x = 0 \) is given by the first piece (\( f(0)=0 \)). The graph of the function at \( x = 0 \) is the point (0,0) from the first piece. The right - hand piece is for \( x>0 \), so its graph is the parabola \( y = x^2 \) for \( x>0 \), which approaches (0,0) as \( x \) approaches 0 from the right.
- Third graph: left - hand line has an open circle at (0,0), but the first piece \( f(x)=x \) for \( x\leq0 \) includes \( x = 0 \), so the left - hand line should have a closed circle at (0,0). The second graph has a closed circle at (0,0) for the left - hand line (correct, since \( x = 0 \) is in the domain of \( f(x)=x \)) and the right - hand parabola starts at (0,0) with a closed circle? No, the right - hand function's domain is \( x>0 \), so the point (0,0) is not in its domain, but the function value at \( x = 0 \) is 0 (from the first piece). So the graph should have a closed circle at (0,0) from the left - hand line and the right - hand parabola approaches (0,0) (open circle at (0,0) for the right - hand parabola? No, the graph of a function at a point is determined by the piece whose domain includes that point. Since \( x = 0 \) is in the domain of \( f(x)=x \) ( \( x\leq0 \) ), the point (0,0) is on the line \( y = x \). The right - hand parabola \( y=x^2 \) for \( x>0 \) will have points like (1,1), (2,4), etc., and as \( x \) approaches 0 from the right, \( x^2 \) approaches 0. So the correct graph is the second one? Wait, no, let's check the three graphs again.

Wait, the third graph: left - hand line has an open circle at (0,0), which is wrong because \( f(0)=0 \) (from \( f(x)=x, x\leq0 \)). The first graph: left - hand side is a parabola, wrong. The second graph: left - hand side is a line with slope 1, passing through (0,0) (correct for \( f(x)=x, x\leq0 \)) and right - hand side is the parabola \( y = x^2 \) for \( x>0 \) (correct, since for \( x>0 \), \( f(x)=x^2 \)). So the second graph is correct. Wait, but let's check the circles: the left - hand function includes \( x = 0 \), so the point (0,0) should be a closed circle for the left - hand line. The right - hand function does not include \( x = 0 \), but the graph of the right - hand function approaches (0,0). In the second graph, the left - hand line has a closed circle at (0,0) (correct) and the right - hand parabola starts at (0,0) with a closed circle? No, the right - hand function's domain is \( x>0 \), so the point (0,0) is not in its domain, but the graph of the right - hand function can have a closed circle? No, actually, the function's value at \( x = 0 \) is given by the first piece, so the graph at \( x = 0 \) is the point from the first piece. The second graph has a closed circle at (0,0) for the left - hand line (correct, since \( f(0)=0 \) from \( f(x)=x \)) and the right - hand parabola has a closed circle at (0,0)? No, that would imply that \( f(0)=0^2 = 0 \), but the right - hand piece is for \( x>0 \). However, in the context of graphing piecewise functions, when two pieces meet at a point, if one piece includes the point and the other approaches it, we draw the included point with a closed circle. Since the first piece includes \( x = 0 \), we draw a closed circle at (0,0) for the first piece, and the second piece approaches (0,0), so the graph of the second piece can have an open circle at (0,0), but in the second graph, it seems like the right - hand parabola has a closed circle? Wait, maybe I am overcomplicating. Let's list the key features:

- Left part: \( y = x \), \( x\leq0 \): line with slope 1, passes through (0,0), includes (0,0).
- Right part: \( y=x^2 \), \( x > 0 \): parabola opening up, for \( x>0 \), so points like (1,1), (2,4), etc., and as \( x ightarrow0^+ \), \( y ightarrow0 \).

The second graph has the left - hand line (slope 1, through (0,0)) and the right - hand parabola (correct shape for \( y = x^2, x>0 \)). The third graph has an open circle at (0,0) for the left - hand line, which is wrong because \( f(0)=0 \) (from the first piece). The first graph has the left - hand side as a parabola, which is wrong. So the correct graph is the second one.

Answer:

The second graph (the middle one)