1. **State the problem:** We are given the quadratic function $f(x) = -x^2 + 2x - 2$ and want to analyze it. 2. **Formula and rules:** A quadratic function is generally $f(x) = ax^2 + bx + c$. - The vertex form is $f(x) = a(x-h)^2 + k$ where $(h,k)$ is the vertex. - The vertex $h$ can be found by $h = -\frac{b}{2a}$. - The function opens downward if $a < 0$ and upward if $a > 0$. 3. **Find the vertex:** Here, $a = -1$, $b = 2$, $c = -2$. Calculate $h = -\frac{2}{2(-1)} = -\frac{2}{-2} = 1$. 4. **Find $k$ by evaluating $f(h)$:** $$k = f(1) = -(1)^2 + 2(1) - 2 = -1 + 2 - 2 = -1$$ 5. **Vertex form:** $$f(x) = - (x - 1)^2 - 1$$ 6. **Interpretation:** - The vertex is at $(1, -1)$. - Since $a = -1 < 0$, the parabola opens downward, so the vertex is a maximum point. 7. **Find intercepts:** - **y-intercept:** Set $x=0$, $f(0) = -0 + 0 - 2 = -2$, so $(0, -2)$. - **x-intercepts:** Solve $-x^2 + 2x - 2 = 0$ or $x^2 - 2x + 2 = 0$. Calculate discriminant $\Delta = b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4 < 0$. No real roots, so no x-intercepts. **Final answer:** The vertex is at $(1, -1)$, the parabola opens downward, y-intercept is $(0, -2)$, and there are no real x-intercepts.