1. **State the problem:** We need to find the value of $y$ when $x=10.5$ for a piecewise function defined as: $$y=\begin{cases}-x + a, & 0 \leq x \leq 9 \\ b - x + c, & 9 \leq x \leq 15\end{cases}$$ where $a$, $b$, and $c$ are constants to be determined from given points. 2. **Use given points:** The function passes through points $(0,3)$ and $(4,0)$. - For $x=0$, using the first piece: $y = -0 + a = a = 3$, so $a=3$. - For $x=4$, using the first piece: $y = -4 + 3 = -1$, but the point is $(4,0)$, so this suggests the first piece is $y = -x + 4$ to fit $(4,0)$. Let's re-express the first piece as $y = -x + 4$ to fit both points: - At $x=0$, $y = -0 + 4 = 4$ (does not match $(0,3)$), so we need to check carefully. 3. **Find the correct first piece:** Since the line segment descends from $(0,3)$ to $(4,0)$, the slope is: $$m = \frac{0 - 3}{4 - 0} = \frac{-3}{4} = -0.75$$ Equation of line passing through $(0,3)$ with slope $-0.75$: $$y - 3 = -0.75(x - 0) \Rightarrow y = -0.75x + 3$$ So the first piece is: $$y = -\frac{3}{4}x + 3$$ 4. **Find the second piece:** The second piece starts at $x=4$ with $y=0$ and goes to point $c$ at $x=15$. Assuming $c = (15, y_c)$, and the line is slightly rising, slope $m_2$ is: $$m_2 = \frac{y_c - 0}{15 - 4} = \frac{y_c}{11}$$ The equation of the second piece is: $$y - 0 = m_2(x - 4) \Rightarrow y = m_2(x - 4) = \frac{y_c}{11}(x - 4)$$ 5. **Find $y$ at $x=10.5$:** $$y = \frac{y_c}{11}(10.5 - 4) = \frac{y_c}{11} \times 6.5 = \frac{6.5}{11} y_c$$ 6. **Since $y_c$ is unknown, we cannot find a numeric value for $y$ at $x=10.5$ without more information.** **Final answer:** $$y = \frac{6.5}{11} y_c$$ where $y_c$ is the $y$-coordinate of point $c$ at $x=15$.