1. The problem is to identify the graph that represents the piecewise function: $$f(x) = \begin{cases} \frac{1}{2}x + 2 & \text{for } x \leq 1 \\ -\frac{5}{4}x + 1 & \text{for } x > 1 \end{cases}$$ 2. Let's analyze each piece: - For $x \leq 1$, the function is $f(x) = \frac{1}{2}x + 2$. This is a line with slope $\frac{1}{2}$ and y-intercept 2. - For $x > 1$, the function is $f(x) = -\frac{5}{4}x + 1$. This is a line with slope $-\frac{5}{4}$ and y-intercept 1. 3. Evaluate the function at the boundary $x=1$: - From the first piece: $f(1) = \frac{1}{2}(1) + 2 = \frac{1}{2} + 2 = 2.5$ - From the second piece: $f(1) = -\frac{5}{4}(1) + 1 = -1.25 + 1 = -0.25$ 4. Since the function is defined as $\frac{1}{2}x + 2$ for $x \leq 1$, the value at $x=1$ is $2.5$. The second piece applies only for $x > 1$, so the function jumps down to $-0.25$ immediately after $x=1$. 5. This means the graph should have a point at $(1, 2.5)$ and then a jump down to a line starting just after $x=1$ at $y = -0.25$ with a negative slope. 6. Graph A shows a line increasing gently (slope about 0.5) up to near $(1, 2.5)$ and then a steeper line continuing upward, which does not match the negative slope after $x=1$. 7. Graph B shows an inverted V shape: increasing to a peak near $(1, 2.5)$ and then descending steeply, matching the negative slope after $x=1$. 8. Therefore, the correct graph representing the piecewise function is Graph B. **Final answer:** Graph B matches the piecewise function.