1. **State the problem:** We are given the function $$f(x) = 8 - 2x - x^2$$ and need to find the coordinates of points a, b, and c on its graph. 2. **Understand the function:** This is a quadratic function in standard form $$f(x) = -x^2 - 2x + 8$$ which is a downward-opening parabola because the coefficient of $$x^2$$ is negative. 3. **Find the vertex (point c):** The vertex of a parabola $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. Here, $$a = -1$$ and $$b = -2$$, so $$x_c = -\frac{-2}{2 \times -1} = -\frac{-2}{-2} = -1$$. 4. **Calculate $$y$$ at vertex:** $$f(-1) = 8 - 2(-1) - (-1)^2 = 8 + 2 - 1 = 9$$. So, point c is $$(-1, 9)$$. 5. **Find point a:** Point a is on the positive y-axis above the vertex, so $$x=0$$. Calculate $$f(0)$$: $$f(0) = 8 - 2(0) - 0^2 = 8$$. So, point a is $$(0, 8)$$. 6. **Find point b:** Point b is on the negative x-axis below the x-axis, so $$y=0$$ and $$x < 0$$. Set $$f(x) = 0$$: $$8 - 2x - x^2 = 0$$ Rewrite as: $$-x^2 - 2x + 8 = 0$$ Multiply both sides by $$-1$$ to simplify: $$\cancel{-}x^2 - 2x + 8 = 0 \Rightarrow x^2 + 2x - 8 = 0$$ 7. **Solve quadratic equation:** $$x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-8)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2}$$ $$x = \frac{-2 \pm 6}{2}$$ Two solutions: - $$x = \frac{-2 + 6}{2} = 2$$ (positive, discard since point b is on negative x-axis) - $$x = \frac{-2 - 6}{2} = -4$$ (negative, valid) 8. **Find $$y$$ at $$x = -4$$:** $$f(-4) = 8 - 2(-4) - (-4)^2 = 8 + 8 - 16 = 0$$ So, point b is $$(-4, 0)$$. **Final answers:** - Point a: $$(0, 8)$$ - Point b: $$(-4, 0)$$ - Point c: $$(-1, 9)$$