AOS1 Topic 3: Quadratic Functions

A quadratic function is a type of polynomial function that can be written in the standard form:

\[ f(x) = ax^2 + bx + c \]

where:

  • a, b, and c are constants, with \(a \neq 0\).
  • The highest exponent of the variable \(x\) is 2, making it a second-degree polynomial.

Key Features of Quadratic Functions

1. Parabola Shape

The graph of a quadratic function is a U-shaped curve called a parabola. Depending on the sign of a, the parabola opens:

  • Upwards if \(a > 0\)
  • Downwards if \(a < 0\)

2. Vertex

The vertex is the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards. It represents the maximum or minimum value of the quadratic function.

The vertex \((h, k)\) can be found using the formula:

\[ h = -\frac{b}{2a} \]

and substituting \(h\) into the function to find \(k\).

3. Axis of Symmetry

The parabola is symmetric about a vertical line called the axis of symmetry, which passes through the vertex. The equation of the axis of symmetry is:

\[ x = h = -\frac{b}{2a} \]

4. Y-Intercept

The y-intercept is the point where the graph crosses the y-axis. It occurs when \(x = 0\), so the y-intercept is \(f(0) = c\).

5. X-Intercepts (Roots)

The x-intercepts are the points where the parabola crosses the x-axis. These can be found by solving the quadratic equation \(ax^2 + bx + c = 0\). The solutions (roots) can be found using the quadratic formula:

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

The discriminant, \(b^2 - 4ac\), determines the nature of the roots:

  • If \(b^2 - 4ac > 0\), there are two real and distinct roots.
  • If \(b^2 - 4ac = 0\), there is one real, repeated root.
  • If \(b^2 - 4ac < 0\), there are no real roots (the parabola does not cross the x-axis).

Forms of Quadratic Functions

1. Standard Form

The general form of a quadratic function is:

\[ f(x) = ax^2 + bx + c \]

This form is useful for identifying the y-intercept and for using the quadratic formula to find the roots.

2. Vertex Form

The vertex form of a quadratic function is:

\[ f(x) = a(x - h)^2 + k \]

where \((h, k)\) is the vertex of the parabola. This form makes it easy to identify the vertex and the axis of symmetry.

3. Factored Form

The factored form of a quadratic function is:

\[ f(x) = a(x - r_1)(x - r_2) \]

where \(r_1\) and \(r_2\) are the roots (x-intercepts) of the quadratic equation. This form is helpful for finding the x-intercepts directly.

Domain and Range of Quadratic Function

Domain of Quadratic Function

A quadratic function is a polynomial function that is defined for all real values of \( x \). So, the domain of a quadratic function is the set of real numbers, that is, \( \mathbb{R} \). In interval notation, the domain of any quadratic function is \( (-\infty, \infty) \).

Range of Quadratic Function

The range of the quadratic function depends on the graph's opening side and vertex. So, look for the lowermost and uppermost \( f(x) \) values on the graph of the function to determine the range of the quadratic function. The range of any quadratic function with vertex \( (h, k) \) and the equation \( f(x) = a(x - h)^2 + k \) is:

  • \( y \geq k \) (or) \([ k, \infty )\) when \( a > 0 \) (as the parabola opens up when \( a > 0 \)).
  • \( y \leq k \) (or) \(( -\infty, k ]\) when \( a < 0 \) (as the parabola opens down when \( a < 0 \)).

    • Applications of Quadratic Functions

    Quadratic functions have numerous real-world applications, including:

    • Projectile motion: The path of an object thrown or propelled is parabolic, and its height as a function of time can be modeled by a quadratic function.
    • Optimization problems: In many scenarios, quadratic functions are used to maximize or minimize a quantity, such as maximizing profit or minimizing cost.
    • Architecture and engineering: Parabolic shapes, such as arches, are modeled using quadratic functions.
    Example 1

    Example 1: Find the vertex, axis of symmetry, and intercepts for the quadratic function:

    \[ f(x) = 2x^2 - 4x + 1 \]

    Solution:

    Vertex:

    Using the formula \(h = -\frac{b}{2a} = -\frac{-4}{2(2)} = 1\).

    Substitute \(x = 1\) into the function to find \(k\):

    \[ f(1) = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1 \]

    So, the vertex is \((1, -1)\).

    Axis of Symmetry:

    The axis of symmetry is \(x = 1\).

    Y-Intercept:

    Substitute \(x = 0\) into the function:

    \[ f(0) = 2(0)^2 - 4(0) + 1 = 1 \]

    So, the y-intercept is \((0, 1)\).

    X-Intercepts:

    Use the quadratic formula to find the roots:

    \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} \]

    \[ x = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2} \]

    So, the x-intercepts are approximately \(x = 1.71\) and \(x = 0.29\).

    Example 2

    Solve the quadratic equation using the quadratic formula

    Given the quadratic equation:

    \[ 3x^2 - 5x + 2 = 0 \]

    Solution:

    Here, \(a = 3\), \(b = -5\), and \(c = 2\). We can use the quadratic formula:

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

    Substitute the values of \(a\), \(b\), and \(c\):

    \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(2)}}{2(3)} \]

    \[ x = \frac{5 \pm \sqrt{25 - 24}}{6} \]

    \[ x = \frac{5 \pm \sqrt{1}}{6} \]

    \[ x = \frac{5 \pm 1}{6} \]

    So, the two solutions are:

    • \[ x = \frac{5 + 1}{6} = 1 \]
    • \[ x = \frac{5 - 1}{6} = \frac{2}{3} \]

    Therefore, the solutions are \(x = 1\) and \(x = \frac{2}{3}\).

    Example 3

    Find the maximum value of the quadratic function

    Given the quadratic function:

    \[ f(x) = -2x^2 + 8x - 3 \]

    Solution:

    The function opens downwards because \(a = -2\). Therefore, it has a maximum value at its vertex.

    First, find the vertex:

    \[ h = -\frac{b}{2a} = -\frac{8}{2(-2)} = 2 \]

    Substitute \(x = 2\) into the function to find \(k\):

    \[ f(2) = -2(2)^2 + 8(2) - 3 = -8 + 16 - 3 = 5 \]

    So, the maximum value of the function is 5, and it occurs at \(x = 2\).

    Example 4

    Calculating Domain and Range of Quadratic Function:

    Consider the quadratic function \( f(x) = x^2 - 4x + 3 \). What is the domain and range of this function?

    Solution:

    Domain

    The given quadratic function is \( f(x) = x^2 - 4x + 3 \), which is a polynomial function. Polynomial functions are defined for all real numbers, so the domain of this quadratic function is all real numbers.

    Therefore, the domain of \( f(x) \) is:

    \( \text{Domain of } f(x): (-\infty, \infty) \)

    Range

    First, we need to find the vertex of the quadratic function. The standard form of the quadratic function is:

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

    For \( f(x) = x^2 - 4x + 3 \), we have \( a = 1 \), \( b = -4 \), and \( c = 3 \).

    The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) can be found using the formula:

    \( x = -\frac{b}{2a} \)

    Substitute the values of \( a \) and \( b \):

    \( x = -\frac{-4}{2(1)} = 2 \)

    Now, substitute \( x = 2 \) into the function to find the corresponding \( y \)-value:

    \( f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \)

    The vertex of the parabola is \( (2, -1) \). Since the coefficient of \( x^2 \) is positive (\( a = 1 \)), the parabola opens upwards. Therefore, the range is all values greater than or equal to the \( y \)-coordinate of the vertex, which is -1.

    Thus, the range of \( f(x) \) is:

    \( \text{Range of } f(x): [-1, \infty) \)

    Exercise &&1&& (&&1&& Question)

    What is the vertex of the quadratic function:

    \( f(x) = 2x^2 - 8x + 3 \)?

    1
    Submit

    Exercise &&2&& (&&1&& Question)

    Find the axis of symmetry for the quadratic equation:

    \( y = -x^2 + 6x - 4 \).

    2
    Submit

    Exercise &&3&& (&&1&& Question)

    What is the maximum value of the quadratic function:

    \( f(x) = -x^2 + 4x + 1 \)?

    3
    Submit

    Exercise &&4&& (&&1&& Question)

    Find the solutions of the quadratic equation:

    \( 2x^2 - 3x = 5 \).

    4
    Submit