1. **Problem Statement:** Find the axis of symmetry, vertex, y-intercept, and x-intercepts of the quadratic function $y = ax^2 + bx + c$. Also, show how to find the x-intercepts and provide the graph. 2. **Formula and Important Rules:** - Axis of symmetry: $x = -\frac{b}{2a}$ - Vertex: $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$ - Y-intercept: The value of $y$ when $x=0$, which is $c$. - X-intercepts: Solve $ax^2 + bx + c = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Example:** Consider $y = 2x^2 - 4x + 1$. 4. **Axis of Symmetry:** $$x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1$$ 5. **Vertex:** Calculate $y$ at $x=1$: $$y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1$$ Vertex is at $(1, -1)$. 6. **Y-intercept:** At $x=0$: $$y = 2(0)^2 - 4(0) + 1 = 1$$ So, y-intercept is $(0,1)$. 7. **X-intercepts:** Solve $2x^2 - 4x + 1 = 0$: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 2 \times 1}}{2 \times 2} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4}$$ Simplify $\sqrt{8} = 2\sqrt{2}$: $$x = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$$ So, x-intercepts are $\left(1 + \frac{\sqrt{2}}{2}, 0\right)$ and $\left(1 - \frac{\sqrt{2}}{2}, 0\right)$. 8. **Summary:** - Axis of symmetry: $x=1$ - Vertex: $(1, -1)$ - Y-intercept: $(0,1)$ - X-intercepts: $\left(1 + \frac{\sqrt{2}}{2}, 0\right)$ and $\left(1 - \frac{\sqrt{2}}{2}, 0\right)$