$$\DeclareMathOperator{cosec}{cosec}$$

# Algebra and Functions

## Furthermore

The quadratic function $$f(x) = ax^2 + bx + c$$ is a fundamental concept in algebra. Its graph forms a parabola, a symmetric curve that can open upwards or downwards depending on the coefficient $$a$$. The y-intercept of this function is the point where the graph crosses the y-axis, which occurs at $$(0, c)$$. An important feature of the parabola is its axis of symmetry, a vertical line that divides the graph into two mirror images. This axis passes through the vertex of the parabola, a point representing the maximum or minimum value of the function.

$$\text{ The axis of symmetry is } x = - \dfrac{b}{2a}$$

In the factored form $$f(x) = a(x - p)(x - q)$$, the function is expressed as a product of two linear terms, where $$p$$ and $$q$$ are the x-intercepts of the graph, the points where the graph crosses the x-axis. These intercepts are found at $$(p, 0)$$ and $$(q, 0)$$. In the vertex (completing the square) form $$f(x) = a(x - h)^2 + k$$, the function is represented in a way that highlights its vertex, located at $$(h, k)$$. This form is particularly useful for easily identifying the vertex and for graphing the parabola.

Key Formulae:

$$\text{Standard Form: } f(x) = ax^2 + bx + c \\ \text{Factored Form: } f(x) = a(x - p)(x - q) \\ \text{Vertex Form: } f(x) = a(x - h)^2 + k$$

Example:

Consider the quadratic function $$f(x) = 2x^2 - 8x + 6$$. Its standard form is already given. To find its factored form, we need to factorise the quadratic equation:

$$f(x) = 2(x - 1)(x - 3)$$

The x-intercepts are $$(1, 0)$$ and $$(3, 0)$$. To express this function in vertex form, we complete the square:

$$f(x) = 2(x - 2)^2 - 2$$

The vertex of this parabola is at $$(2, -2)$$.

The solution of quadratic equations is a fundamental concept in algebra. A quadratic equation is typically in the form $$ax^2 + bx + c = 0$$, where $$a$$, $$b$$, and $$c$$ are coefficients. The solutions to these equations can be found using the quadratic formula:

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

This formula calculates the roots of the quadratic equation. The nature of these roots is determined by the discriminant $$\Delta = b^2 - 4ac$$. The discriminant reveals:

• Two distinct real roots if $$\Delta > 0$$
• Two equal real roots (or one real root) if $$\Delta = 0$$
• No real roots if $$\Delta < 0$$

Example: Consider the quadratic equation $$2x^2 - 4x - 6 = 0$$. To find the roots:

1. Identify the coefficients: $$a = 2$$, $$b = -4$$, and $$c = -6$$.
2. Compute the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$.
3. Since $$\Delta > 0$$, there are two distinct real roots.
$$x = \frac{{-(-4) \pm \sqrt{{64}}}}{{2 \times 2}} = \frac{{4 \pm 8}}{4}$$

Hence, the solutions are $$x = 3$$ and $$x = -1$$.

This video on Factorising Quadratic Functions and Equations is from Revision Village and is aimed at students taking the IB Maths AA Standard level course.

This video on Completing The Square is from Revision Village and is aimed at students taking the IB Maths AA Standard level course

This video on Discriminant Test (Quadratics) is from Revision Village and is aimed at students taking the IB Maths AA Standard level course

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