Piecewise Graph Ab7B38

1. **State the problem:** We need to sketch the graph of the piecewise function: $$f(x) = \begin{cases} -x & \text{if } x \leq 0 \\ 9 - x^2 & \text{if } 0 < x \leq 3 \\ x - 3 & \text{if } x > 3 \end{cases}$$ 2. **Analyze each piece:** - For $$x \leq 0$$, the function is $$f(x) = -x$$, which is a line with slope $$-1$$ reflected over the y-axis. - For $$0 < x \leq 3$$, the function is $$f(x) = 9 - x^2$$, a downward-opening parabola with vertex at $$(0,9)$$ but restricted to $$x$$ between 0 and 3. - For $$x > 3$$, the function is $$f(x) = x - 3$$, a line with slope 1 starting just after $$x=3$$. 3. **Find key points and continuity:** - At $$x=0$$, from the left, $$f(0) = -0 = 0$$. - At $$x=0$$ from the right, $$f(0) = 9 - 0^2 = 9$$ but since the parabola is defined for $$0 < x$$, the value at 0 is not included, so the function jumps from 0 to values near 9 just right of 0. - At $$x=3$$, $$f(3) = 9 - 3^2 = 9 - 9 = 0$$. - For $$x > 3$$, $$f(3)$$ is not defined but the line starts just after 3 with $$f(3^+) = 3 - 3 = 0$$, so the function is continuous at 3. 4. **Sketch each part:** - Draw the line $$y = -x$$ for $$x \leq 0$$ ending at the origin (0,0). - Draw the parabola $$y = 9 - x^2$$ from just right of 0 to $$x=3$$, starting near 9 and going down to 0. - Draw the line $$y = x - 3$$ for $$x > 3$$ starting just right of (3,0) and increasing. 5. **Summary:** - The graph has a jump discontinuity at $$x=0$$. - The graph is continuous at $$x=3$$. Final answer: The piecewise graph consists of a line $$y = -x$$ for $$x \leq 0$$, a downward parabola $$y = 9 - x^2$$ for $$0 < x \leq 3$$, and a line $$y = x - 3$$ for $$x > 3$$, with a jump at $$x=0$$ and continuity at $$x=3$$.