1. The problem asks us to form a quadratic function $f(x)$ and solve $f(x)=0$, find its minimum or maximum point, and sketch the quadratic. 2. A quadratic function has the form $$f(x) = ax^2 + bx + c$$ where $a \neq 0$. 3. To have two different roots or two equal roots, the discriminant $$\Delta = b^2 - 4ac$$ determines the nature of roots: - If $\Delta > 0$, two different real roots. - If $\Delta = 0$, two equal real roots. 4. Let's choose an example quadratic with two different roots: $$f(x) = x^2 - 4x + 3$$. 5. Solve $f(x) = 0$: $$x^2 - 4x + 3 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$$ So the roots are: $$x = \frac{4 + 2}{2} = 3$$ $$x = \frac{4 - 2}{2} = 1$$ 6. Find the vertex (minimum or maximum point): The vertex $x$-coordinate is given by: $$x = -\frac{b}{2a} = -\frac{-4}{2 \times 1} = 2$$ Calculate $f(2)$: $$f(2) = 2^2 - 4 \times 2 + 3 = 4 - 8 + 3 = -1$$ Since $a=1 > 0$, the parabola opens upwards and the vertex is a minimum point at $(2, -1)$. 7. Sketching the quadratic: - Roots at $x=1$ and $x=3$ (where $f(x)=0$) - Vertex at $(2, -1)$ minimum point - Parabola opens upwards This completes the solution for the quadratic function with two different roots.