Question

graph the piecewise - defined function
$f(x)=\begin{cases}-\frac{1}{4}x^{2}+2&\text{for }xleq2\\frac{1}{4}x&\text{for }x > 2end{cases}$
choose the correct graph
a.
graph a
b.
graph b
c.
graph c
d.
graph d

Explanation:

Step1: Analyze the first piece \( f(x) = -\frac{1}{4}x^2 + 2 \) for \( x \leq 2 \)

This is a quadratic function (a parabola) with a negative leading coefficient, so it opens downward. The vertex of \( y = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Here, \( a = -\frac{1}{4} \), \( b = 0 \), so the vertex is at \( x = 0 \). Plugging \( x = 0 \) in, we get \( y = 2 \), so the vertex is \( (0, 2) \). When \( x = 2 \), \( f(2) = -\frac{1}{4}(2)^2 + 2 = -\frac{1}{4}(4) + 2 = -1 + 2 = 1 \), so the point \( (2, 1) \) is on this part (and it's a closed dot since \( x \leq 2 \)).

Step2: Analyze the second piece \( f(x) = \frac{1}{4}x \) for \( x > 2 \)

This is a linear function with a positive slope (\( \frac{1}{4} \)). When \( x = 2 \), \( f(2) = \frac{1}{4}(2) = \frac{1}{2} \), but since \( x > 2 \), the point at \( x = 2 \) for this part is an open dot (value \( \frac{1}{2} \), but the first part has a closed dot at \( (2, 1) \)). For \( x > 2 \), as \( x \) increases, \( y \) increases. For example, when \( x = 4 \), \( f(4) = \frac{1}{4}(4) = 1 \), so the point \( (4, 1) \) is on this line.

Step3: Compare with the options

• Option A: Likely doesn't match the vertex or the linear part.
• Option B: The downward - opening parabola for \( x \leq 2 \) (with vertex at \( (0, 2) \) and closed dot at \( (2, 1) \)) and the linear part \( y=\frac{1}{4}x \) for \( x > 2 \) (with open dot at \( x = 2 \), \( y=\frac{1}{2} \) and increasing slope) matches our analysis.
• Option C: The linear part seems to have a negative slope or incorrect behavior, so eliminate.
• Option D: The parabola might not open downward or have incorrect vertex, eliminate.

Answer:

B