Quadratic Functions: Complete Guide with Interactive Graphing

Master the general form \( f(x) = ax^2 + bx + c \) through interactive exploration

Quadratic functions are fundamental in algebra with wide applications in physics, engineering, and economics. This interactive tutorial explores their graphs, properties, and transformations using a live graphing calculator.

Explore the relationship between the \(x\)-intercepts of a quadratic function's graph and the solutions to \(f(x) = 0\) by adjusting coefficients \(a\), \(b\), and \(c\) in real-time.

Further learning: Quadratic Functions Tutorial • Graphing Guide • Free Graph Paper

Interactive Quadratic Function Grapher

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Determines parabola direction: \(a > 0\) opens upward, \(a < 0\) opens downward

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Affects horizontal position and vertex location

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Y-intercept of the parabola: point \((0, c)\)

Purple curve: quadratic function • Red dot: vertex • Values update in real-time

A. Definition and Basic Properties

A quadratic function has the general form:

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

where \(a\), \(b\), and \(c\) are real numbers with \(a \neq 0\). Its graph is a parabola—a symmetric U-shaped curve.

Examples:

  1. \( f(x) = -2x^2 + x - 1 \) (opens downward, \(a < 0\))
  2. \( f(x) = x^2 + 3x + 2 \) (opens upward, \(a > 0\))

Interactive Exploration 1:

Enter the example coefficients above into the grapher. Observe how the sign of \(a\) determines direction. Try different values to see how each coefficient affects the shape.

B. Standard Form and Vertex

Every quadratic function can be rewritten in standard (vertex) form:

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

where \((h, k)\) is the vertex. Complete the square to convert:

  1. Start with \( f(x) = ax^2 + bx + c \)
  2. Factor \(a\) from quadratic and linear terms: \[ f(x) = a\left[x^2 + \frac{b}{a}x\right] + c \]
  3. Add and subtract \(\left(\frac{b}{2a}\right)^2\): \[ f(x) = a\left[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c \]
  4. Complete the square: \[ f(x) = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c \]
  5. Simplify to vertex form: \[ f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \]

Thus:

\[ h = -\frac{b}{2a}, \quad k = c - \frac{b^2}{4a} \]

Example: Convert \( f(x) = -2x^2 + 4x + 1 \) to vertex form

  1. Factor \(-2\): \( f(x) = -2(x^2 - 2x) + 1 \)
  2. Complete square: \( f(x) = -2(x^2 - 2x + 1 - 1) + 1 \)
  3. Simplify: \( f(x) = -2[(x - 1)^2 - 1] + 1 \)
  4. Final form: \( f(x) = -2(x - 1)^2 + 3 \)
  5. Vertex: \( (1, 3) \)

Interactive Exploration 2:

Set \(a = -2\), \(b = 4\), \(c = 1\) in the grapher. Verify the vertex is at \((1,3)\), a maximum point. Try \(a = 1\), \(b = -2\), \(c = 0\) to see a minimum point at \((1,-1)\).

C. \(x\)-Intercepts (Roots/Zeros)

The \(x\)-intercepts solve \( ax^2 + bx + c = 0 \). The discriminant \(D = b^2 - 4ac\) determines the nature of roots:

When \(D \geq 0\), intercepts are:

\[ x_1 = \frac{-b + \sqrt{D}}{2a}, \quad x_2 = \frac{-b - \sqrt{D}}{2a} \]

Examples:

  1. \(f(x) = x^2 + 2x - 3\)
    \(D = 2^2 - 4(1)(-3) = 16 > 0\)
    \(x_1 = \frac{-2 + 4}{2} = 1\), \(x_2 = \frac{-2 - 4}{2} = -3\)
    Intercepts: \((1,0)\) and \((-3,0)\)
  2. \(g(x) = -x^2 + 2x - 1\)
    \(D = 4 - 4(-1)(-1) = 0\)
    \(x = -\frac{b}{2a} = \frac{-2}{-2} = 1\)
    Single intercept: \((1,0)\)
  3. \(h(x) = -2x^2 + 2x - 2\)
    \(D = 4 - 4(-2)(-2) = -12 < 0\)
    No real x-intercepts

Interactive Exploration 3:

Test these cases in the grapher. Observe how the discriminant affects intercepts. Try finding intercepts for:

  1. \(f(x) = x^2 + x - 2\)
  2. \(g(x) = 4x^2 + x + 1\)
  3. \(h(x) = x^2 - 4x + 4\)

D. \(y\)-Intercept

The \(y\)-intercept occurs at \(x = 0\):

\[ f(0) = a(0)^2 + b(0) + c = c \]

Thus, the \(y\)-intercept is always at point \((0, c)\).

Examples:

  1. \(f(x) = x^2 + 2x - 3\) → Intercept: \((0, -3)\)
  2. \(g(x) = 4x^2 - x + 1\) → Intercept: \((0, 1)\)
  3. \(h(x) = -x^2 + 4x + 4\) → Intercept: \((0, 4)\)

Interactive Exploration 4:

Vary \(c\) in the grapher while keeping \(a\) and \(b\) fixed. Observe how the entire parabola shifts vertically without changing shape.

E. Practice: Find Equation from Graph

Given a parabola's key points, determine its equation \(f(x) = ax^2 + bx + c\).

Example Problem:

Parabola with vertex (-2,-2) and y-intercept (0,6)

Find the quadratic function for this parabola.

Method 1: Using Intercepts

X-intercepts at \((-3,0)\) and \((-1,0)\), Y-intercept at \((0,6)\):

  1. Use intercept form: \( f(x) = a(x + 3)(x + 1) \)
  2. Plug \((0,6)\): \( 6 = a(0+3)(0+1) \) → \( a = 2 \)
  3. Result: \( f(x) = 2(x+3)(x+1) = 2x^2 + 8x + 6 \)

Method 2: Using Vertex Form

Vertex at \((-2,-2)\), Y-intercept at \((0,6)\):

  1. Vertex form: \( f(x) = a(x + 2)^2 - 2 \)
  2. Plug \((0,6)\): \( 6 = a(0+2)^2 - 2 \) → \( a = 2 \)
  3. Result: \( f(x) = 2(x+2)^2 - 2 = 2x^2 + 8x + 6 \)

Method 3: System of Equations

Using points \((-3,0)\), \((-1,0)\), \((0,6)\):

  1. General form: \( f(x) = ax^2 + bx + c \)
  2. From \((0,6)\): \( c = 6 \)
  3. From \((-3,0)\): \( 9a - 3b + 6 = 0 \)
  4. From \((-1,0)\): \( a - b + 6 = 0 \)
  5. Solve system: \( a = 2, b = 8, c = 6 \)
  6. Result: \( f(x) = 2x^2 + 8x + 6 \)
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