1. **State the problem:** We are given the quadratic function $f(x) = -x^2 + 8x - 1$ and need to find the intercepts, vertex, and axis of symmetry. 2. **Find the intercepts:** - The y-intercept is found by evaluating $f(0)$: $$f(0) = -(0)^2 + 8(0) - 1 = -1$$ So the y-intercept is at $(0, -1)$. - The x-intercepts are found by solving $f(x) = 0$: $$-x^2 + 8x - 1 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}x^2 + \cancel{8}x - \cancel{1} = \cancel{0} \Rightarrow x^2 - 8x + 1 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-8$, $c=1$: $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(1)}}{2(1)} = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm \sqrt{60}}{2}$$ Simplify $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$: $$x = \frac{8 \pm 2\sqrt{15}}{2} = 4 \pm \sqrt{15}$$ So the x-intercepts are at: $$\left(4 - \sqrt{15}, 0\right) \text{ and } \left(4 + \sqrt{15}, 0\right)$$ 3. **Find the vertex:** The vertex formula for a quadratic $ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Given $a = -1$, $b = 8$: $$x = -\frac{8}{2(-1)} = -\frac{8}{-2} = 4$$ Evaluate $f(4)$: $$f(4) = -(4)^2 + 8(4) - 1 = -16 + 32 - 1 = 15$$ So the vertex is at $(4, 15)$. 4. **Find the axis of symmetry:** The axis of symmetry is the vertical line through the vertex: $$x = 4$$ **Final answers:** - Y-intercept: $(0, -1)$ - X-intercepts: $\left(4 - \sqrt{15}, 0\right)$ and $\left(4 + \sqrt{15}, 0\right)$ - Vertex: $(4, 15)$ - Axis of symmetry: $x = 4$