A look into Quadratic Functions

A quadratic function is a type of polynomial function that is defined by a quadratic equation. The general form of a quadratic function is:

where (a), (b), and (c) are constants with (a \neq 0). The variable (x) is the independent variable, and (f(x)) represents the value of the function at (x).

Key Characteristics

1. Parabola Shape

The graph of a quadratic function is a parabola. Depending on the sign of (a):

• If (a > 0), the parabola opens upwards (U-shaped).
• If (a < 0), the parabola opens downwards (∩-shaped).

2. Vertex

The vertex of the parabola is the highest or lowest point, depending on the direction it opens. The vertex can be found using the formula:

To find the (y)-coordinate of the vertex, substitute (x) back into the quadratic function:

So, the vertex is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.

3. Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. It has the equation:

4. Y-Intercept

The y-intercept is the point where the parabola crosses the y-axis. It occurs when (x = 0). For the quadratic function $f(x) = ax^2 + bx + c$, the y-intercept is:

5. X-Intercepts (Roots)

The x-intercepts are the points where the parabola crosses the x-axis. They can be found by solving the quadratic equation $ax^2 + bx + c = 0$ using the quadratic formula:

where $\Delta = b^2 - 4ac$ is called the discriminant. The nature of the roots is determined by the discriminant:

• If $\Delta > 0$, there are two distinct real roots.
• If $\Delta = 0$, there is exactly one real root (the vertex is on the x-axis).
• If $\Delta < 0$, there are no real roots (the parabola does not intersect the x-axis).

Example

Consider the quadratic function:

• Vertex: Use $x = -\frac{b}{2a} = -\frac{-4}{2 \cdot 2} = 1$

  • Substitute (x = 1) into $f(x)$: $f(1) = 2(1)^2 - 4(1) + 1 = -1$
  • Vertex: $(1, -1)$

• Axis of Symmetry: $x = 1$

• Y-Intercept: $f(0) = 1$

• X-Intercepts: Solve $2x^2 - 4x + 1 = 0$ using the quadratic formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 - 8}}{4}$

  $= \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$

  So, the x-intercepts are $x = 1 + \frac{\sqrt{2}}{2}$ and $x = 1 - \frac{\sqrt{2}}{2}$