1. **State the problem:** We need to analyze the quadratic function $f(x) = x^2 - 6x + 5$ by finding its x-intercepts, y-intercept, vertex, and axis of symmetry. 2. **Formula and rules:** - The x-intercepts are found by solving $f(x) = 0$. - The y-intercept is $f(0)$. - The vertex of a parabola $ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. - The axis of symmetry is the vertical line through the vertex, $x = -\frac{b}{2a}$. 3. **Find x-intercepts:** Solve $x^2 - 6x + 5 = 0$. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-6$, $c=5$. Calculate discriminant: $$\Delta = (-6)^2 - 4 \times 1 \times 5 = 36 - 20 = 16$$ Calculate roots: $$x = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}$$ So, $$x_1 = \frac{6 + 4}{2} = 5$$ $$x_2 = \frac{6 - 4}{2} = 1$$ 4. **Find y-intercept:** Evaluate $f(0)$: $$f(0) = 0^2 - 6 \times 0 + 5 = 5$$ 5. **Find vertex:** Calculate $x$-coordinate: $$x_v = -\frac{b}{2a} = -\frac{-6}{2 \times 1} = 3$$ Calculate $y$-coordinate: $$y_v = f(3) = 3^2 - 6 \times 3 + 5 = 9 - 18 + 5 = -4$$ Vertex is at $(3, -4)$. 6. **Axis of symmetry:** The axis of symmetry is the vertical line through the vertex: $$x = 3$$ **Final answer:** - x-intercepts: $(1, 0)$ and $(5, 0)$ - y-intercept: $(0, 5)$ - vertex: $(3, -4)$ - axis of symmetry: $x = 3$