1. The problem is to understand the quadratic function $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 2. The general form of a quadratic function is $f(x) = ax^2 + bx + c$ where $a \neq 0$. 3. Important rules: - The graph of $f(x)$ is a parabola. - If $a > 0$, the parabola opens upwards; if $a < 0$, it opens downwards. - The vertex of the parabola is at $x = -\frac{b}{2a}$. - The axis of symmetry is the vertical line $x = -\frac{b}{2a}$. 4. To find the vertex, substitute $x = -\frac{b}{2a}$ into $f(x)$: $$f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c$$ 5. Simplify the expression: $$= a \cdot \frac{b^2}{4a^2} - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{b^2}{2a} + c$$ 6. Combine terms: $$= \frac{b^2}{4a} - \frac{2b^2}{4a} + c = -\frac{b^2}{4a} + c$$ 7. So the vertex is at: $$\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$$ 8. This vertex represents the maximum or minimum point of the parabola depending on the sign of $a$. 9. The y-intercept is at $f(0) = c$. 10. The x-intercepts (roots) can be found using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives the points where the parabola crosses the x-axis. Final answer: The quadratic function $f(x) = ax^2 + bx + c$ has vertex at $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$, y-intercept at $c$, and x-intercepts given by the quadratic formula.