Question

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

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

$

Explanation:

Step1: Analyze the first piece \( f(x)=x^2 \) for \( x\leq2 \)

The function \( y = x^2 \) is a parabola opening upwards with vertex at the origin \((0,0)\). For \( x\leq2 \), we include the part of the parabola where \( x \) is less than or equal to 2. At \( x = 2 \), \( f(2)=2^2 = 4 \), so there should be a closed dot at \( (2,4) \) for this piece.

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

The function \( y=x + 2 \) is a linear function with a slope of 1 and y - intercept at 2. For \( x>2 \), when \( x = 2 \), \( f(2)=2 + 2=4 \), but since \( x>2 \), we use an open dot at \( (2,4) \) for this piece (because \( x = 2 \) is not included in this sub - domain). Now we check the three graphs:

• The first graph: The parabola \( y = x^2 \) for \( x\leq2 \) (with a closed dot at \( x = 2 \)) and the line \( y=x + 2 \) for \( x>2 \) (with an open dot at \( x = 2 \)) matches our analysis.
• The second graph: The parabola is opening downwards, which is not \( y=x^2 \), so it is incorrect.
• The third graph: The line part has a closed dot at \( x = 2 \), but for \( y=x + 2 \) with \( x>2 \), we need an open dot at \( x = 2 \), so it is incorrect.

Answer:

The first graph (the one with the upward - opening parabola for \( x\leq2 \) and the line with an open dot at \( x = 2 \) for \( x>2 \))