Summary Lecture (Quadratic Functions)

Based on the “Summary lecture”, this video summarises the topics covered regarding quadratic equations and functions.

The lecture begins by positioning quadratic functions as a generalisation of the concept of a straight line or a linear function, which is typically in the form of mx + b. A quadratic function is defined in the form f(x) = ax² + bx + c, with the crucial condition that a is not equal to 0. If ‘a’ were equal to 0, the term with x² would disappear, and it would simply reduce to a linear function. The name “quadratic” is related to the term “square”.

The first major topic summarised is how to graph this quadratic function. The lecture provides a “cookbook recipe” or a “three-point strategy” for graphing:

1. Find the axis of symmetry, which is given by the equation x = -b/(2a). The lecture notes that the derivation for this equation was covered in previous videos, and remembers that ‘c’ does not play any role in this formula, only ‘a’ and ‘b’ do. This axis of symmetry is a vertical line.
2. Figure out the y-intercept of the function. This is found by setting x = 0 in the function, which gives the point (0, c).
3. Choose any arbitrary point (x₁, y₁) on the graph by substituting an x₁ value into the function to find y₁. Then, using the symmetry property of the parabola around the axis of symmetry, figure out another point (x₂, y₂). Because of the symmetry, y₂ will be equal to y₁. The corresponding x₂ value will be symmetrically positioned on the opposite side of the axis of symmetry from x₁.

Once these points are figured out, you plot them on a coordinate plane and draw a smooth curve joining these points in a symmetric fashion to get the graph of the function, which is a parabola. An example, f(x) = x² + 4x + 4, is used to illustrate this process, identifying the axis of symmetry at x = -2, the y-intercept at (0, 4), and finding points like (-4, 4) using symmetry from (0, 4) around the x = -2 axis. The lecture also notes that the point on the axis of symmetry where the parabola meets it is the vertex, and for f(x) = x² + 4x + 4, the vertex is at (-2, 0).

The second important concept summarised is the slope of a quadratic function. Unlike the slope of a straight line, which is constant, the slope of a quadratic function is variable. The lecture states that the slope of a quadratic function f(x) = ax² + bx + c at any point x is given by the formula 2ax + b. This formula depends on the coefficients ‘a’ and ‘b’, but not on the constant term ‘c’.

A significant application of the slope concept is finding where the slope is equal to zero. Setting 2ax + b = 0 and solving for x gives x = -b/(2a). The lecture highlights that this is the same x-value as the axis of symmetry and therefore corresponds to the x-coordinate of the vertex. Where the slope is zero, the function is changing its behaviour from increasing to decreasing or decreasing to increasing. This point, the vertex, is where the quadratic function attains its minimum or maximum value. The direction the parabola opens (upward for a > 0 or downward for a < 0) determines whether this vertex is a minimum or a maximum point.