Piecewise Parabolas Fb3Ea5

1. **State the problem:** We are given a piecewise function defined as: $$f(x) = \begin{cases} x^2 & \text{if } x \geq 1 \\ -x^2 + 1 & \text{if } x < 0 \end{cases}$$ 2. **Understand the function:** This function has two parts: - For $x \geq 1$, the function is $f(x) = x^2$, which is an upward-opening parabola. - For $x < 0$, the function is $f(x) = -x^2 + 1$, which is a downward-opening parabola shifted up by 1. 3. **Note about domain gaps:** The function is not defined for $0 \leq x < 1$ based on the given definition. 4. **Graph shape and position hints:** - The upward parabola $x^2$ starts at $x=1$ and extends to the right. - The downward parabola $-x^2 + 1$ covers all $x < 0$. - There is a gap between $0$ and $1$ where the function is not defined. 5. **Plot points for $x \geq 1$:** - At $x=1$, $f(1) = 1^2 = 1$. - At $x=2$, $f(2) = 4$. 6. **Plot points for $x < 0$:** - At $x=0$, $f(0) = -0 + 1 = 1$ (though $x=0$ is not included since $x<0$). - At $x=-1$, $f(-1) = -1 + 1 = 0$. - At $x=-2$, $f(-2) = -4 + 1 = -3$. 7. **Summary:** The graph consists of two separate parabola segments: - A downward parabola on the left side ($x<0$) starting near $(0,1)$ going downwards. - An upward parabola on the right side ($x \geq 1$) starting at $(1,1)$ going upwards. - No values between $0$ and $1$. **Final answer:** The piecewise function graph has two disconnected parabola segments as described above.