1. **State the problem:** We are given the quadratic function $y = -x^2 - 2x + 3$ and want to analyze it. 2. **Formula and rules:** The general form of a quadratic function is $y = ax^2 + bx + c$. - Here, $a = -1$, $b = -2$, and $c = 3$. - Since $a < 0$, the parabola opens downward. 3. **Find the vertex:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a}$. $$x = -\frac{-2}{2 \times -1} = -\frac{-2}{-2} = -1$$ Substitute $x = -1$ into the function to find $y$: $$y = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4$$ So the vertex is at $(-1, 4)$. 4. **Find the $y$-intercept:** Set $x=0$: $$y = -0^2 - 2(0) + 3 = 3$$ So the $y$-intercept is $(0, 3)$. 5. **Find the $x$-intercepts:** Set $y=0$ and solve for $x$: $$0 = -x^2 - 2x + 3$$ Multiply both sides by $-1$ to simplify: $$0 = x^2 + 2x - 3$$ Factor the quadratic: $$0 = (x + 3)(x - 1)$$ So the solutions are $x = -3$ and $x = 1$. 6. **Summary:** - Vertex: $(-1, 4)$ - $y$-intercept: $(0, 3)$ - $x$-intercepts: $(-3, 0)$ and $(1, 0)$ - Parabola opens downward because $a = -1 < 0$. Final answer: The vertex is at $(-1, 4)$, the parabola opens downward, and the intercepts are $(0, 3)$, $(-3, 0)$, and $(1, 0)$.