Quadratic Functions and Their Properties

Quadratic functions are polynomial functions of degree two, typically in the form:

$$ f(x) = ax^2 + bx + c $$

where \( a \), \( b \), and \( c \) are constants. The graph of a quadratic function is a parabola, which can open either upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)). Understanding the properties of quadratic functions, such as the vertex, axis of symmetry, and direction of opening, helps us analyze their behavior.

Key Properties of Quadratic Functions

  • Vertex: The vertex is the highest or lowest point of the parabola. For the function \( f(x) = ax^2 + bx + c \), the vertex occurs at:

    $$ x = -\frac{b}{2a} $$

    After finding this \( x \)-value, substitute it back into \( f(x) \) to find the \( y \)-coordinate of the vertex.
  • Axis of Symmetry: The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two symmetric halves. Its equation is:

    $$ x = -\frac{b}{2a} $$

  • Direction of Opening: If \( a > 0 \), the parabola opens upwards (like a "U"), and if \( a < 0 \), it opens downwards (like an "n").
  • Y-intercept: The y-intercept is the point where the graph intersects the y-axis, given by \( f(0) = c \).

Example

Consider the quadratic function:

$$ f(x) = 2x^2 - 4x + 1 $$

To find the vertex, use \( x = -\frac{b}{2a} = -\frac{-4}{2 \cdot 2} = 1 \).

Substitute \( x = 1 \) into \( f(x) \):

$$ f(1) = 2(1)^2 - 4(1) + 1 = -1 $$

The vertex is at \( (1, -1) \), the axis of symmetry is \( x = 1 \), and since \( a = 2 > 0 \), the parabola opens upwards.

Practice Questions

  1. Find the vertex, axis of symmetry, and direction of opening for the function \( f(x) = -x^2 + 6x - 5 \).
    Solution

    For \( f(x) = -x^2 + 6x - 5 \):

    • Vertex: \( x = -\frac{6}{2(-1)} = 3 \)
    • Substitute \( x = 3 \) into \( f(x) \): \( f(3) = -(3)^2 + 6(3) - 5 = 4 \)
    • The vertex is \( (3, 4) \).
    • Axis of symmetry: \( x = 3 \)
    • Direction of opening: Downwards (since \( a = -1 < 0 \))
  2. Determine the y-intercept and vertex for the function \( f(x) = 3x^2 + 12x + 7 \).
    Solution

    For \( f(x) = 3x^2 + 12x + 7 \):

    • Y-intercept: \( f(0) = 7 \)
    • Vertex: \( x = -\frac{12}{2 \cdot 3} = -2 \)
    • Substitute \( x = -2 \) into \( f(x) \): \( f(-2) = 3(-2)^2 + 12(-2) + 7 = -5 \)
    • The vertex is \( (-2, -5) \).
    • Direction of opening: Upwards (since \( a = 3 > 0 \))
  3. For the function \( f(x) = 4x^2 - 8x + 3 \), find the vertex and determine if it is a maximum or minimum point.
    Solution

    For \( f(x) = 4x^2 - 8x + 3 \):

    • Vertex: \( x = -\frac{-8}{2 \cdot 4} = 1 \)
    • Substitute \( x = 1 \) into \( f(x) \): \( f(1) = 4(1)^2 - 8(1) + 3 = -1 \)
    • The vertex is \( (1, -1) \), which is a minimum point since the parabola opens upwards.
→Factoring Quadratics