1. **State the problem:** Find the vertex and intercepts of the quadratic function $y = x^2 - 8x + 7$. 2. **Formula and rules:** The vertex of a parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. The y-intercept is at $x=0$, and x-intercepts are found by solving $y=0$. 3. **Find the vertex:** Here, $a=1$, $b=-8$, $c=7$. Calculate $x$-coordinate of vertex: $$x = -\frac{-8}{2 \times 1} = \frac{8}{2} = 4$$ Calculate $y$-coordinate: $$y = 4^2 - 8 \times 4 + 7 = 16 - 32 + 7 = -9$$ Vertex is at $(4, -9)$. 4. **Find y-intercept:** Substitute $x=0$: $$y = 0^2 - 8 \times 0 + 7 = 7$$ Y-intercept is $(0,7)$. 5. **Find x-intercepts:** Solve $x^2 - 8x + 7 = 0$. Calculate discriminant: $$\Delta = (-8)^2 - 4 \times 1 \times 7 = 64 - 28 = 36$$ Since $\Delta > 0$, two real roots exist: $$x = \frac{8 \pm \sqrt{36}}{2} = \frac{8 \pm 6}{2}$$ Roots: $$x_1 = \frac{8 + 6}{2} = 7$$ $$x_2 = \frac{8 - 6}{2} = 1$$ X-intercepts are $(7,0)$ and $(1,0)$. **Final answer:** - Vertex: $(4, -9)$ - Y-intercept: $(0,7)$ - X-intercepts: $(1,0)$ and $(7,0)$