Quadratic Functions: General Form, Vertex Form, and Intercepts

Quadratic functions are polynomial functions of degree 2. This tutorial explains their graphs, core properties (vertex, axis of symmetry, and intercepts), and the relationship between a function's graph and the solutions to its corresponding equation.

For a hands-on learning experience, try our Interactive Quadratic Function Applet before continuing to graphing quadratic functions.

A - Definition of a Quadratic Function

A quadratic function \(f\) is defined as:

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

where \(a\), \(b\), and \(c\) are real numbers and \(a \neq 0\). Its graph forms a parabola—a symmetrical U-shaped curve that opens upward if \(a > 0\) and downward if \(a < 0\). The magnitude \(|a|\) controls the width: larger \(|a|\) produces a narrower parabola, while smaller \(|a|\) produces a wider one.

Examples:

B - Vertex Form and the Vertex of a Quadratic Function

Any quadratic function can be rewritten in vertex form (also called standard form in some textbooks):

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

where the point \((h, k)\) is the vertex of the parabola. The vertex is the minimum point when the parabola opens upward (\(a > 0\)), and the maximum point when it opens downward (\(a < 0\)).

The vertical line \( x = h \) is the axis of symmetry: the parabola is a mirror image on either side of this line.

Derivation of the Vertex Coordinates

Starting from the general form \( f(x) = ax^{2} + bx + c \), complete the square to derive \(h\) and \(k\):

  1. Factor \(a\) from the quadratic and linear terms:
    \[ f(x) = a\left[x^{2} + \frac{b}{a}x\right] + c \]
  2. Add and subtract \(\left(\frac{b}{2a}\right)^{2}\) inside the brackets to complete the square:
    \[ f(x) = a\left[x^{2} + \frac{b}{a}x + \left(\frac{b}{2a}\right)^{2} - \left(\frac{b}{2a}\right)^{2}\right] + c \]
  3. Rewrite the perfect square trinomial:
    \[ f(x) = a\left[\left(x + \frac{b}{2a}\right)^{2} - \left(\frac{b}{2a}\right)^{2}\right] + c \]
  4. Distribute \(a\) and simplify to reach vertex form:
    \[ f(x) = a\left(x + \frac{b}{2a}\right)^{2} - \frac{b^{2}}{4a} + c \]

Comparing with \( f(x) = a(x - h)^{2} + k \) gives the vertex and axis of symmetry:

\[ h = -\frac{b}{2a}, \qquad k = c - \frac{b^{2}}{4a} = \frac{4ac - b^{2}}{4a} = f(h) \]

Axis of symmetry: \( x = -\dfrac{b}{2a} \)

Example: Convert to Vertex Form and Find the Vertex

Rewrite \( f(x) = -2x^{2} + 4x + 1 \) in vertex form and identify its vertex.

Show Solution

Method 1: Completing the Square

  1. Factor \(-2\) from the \(x\) terms:
    \( f(x) = -2(x^{2} - 2x) + 1 \)
  2. Complete the square inside the parentheses (half of \(-2\) is \(-1\), squared is \(1\)):
    \( f(x) = -2(x^{2} - 2x + 1 - 1) + 1 \)
  3. Regroup and distribute:
    \( f(x) = -2[(x - 1)^{2} - 1] + 1 \)
    \( f(x) = -2(x - 1)^{2} + 2 + 1 \)
    \( f(x) = -2(x - 1)^{2} + 3 \)

The vertex form is \( f(x) = -2(x - 1)^{2} + 3 \), giving a vertex of \((1, 3)\) and axis of symmetry \( x = 1 \).

Method 2: Using the Vertex Formula

  • \( h = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1 \)
  • \( k = f(1) = -2(1)^2 + 4(1) + 1 = 3 \)

Vertex: \((1, 3)\).

C - X-Intercepts of a Quadratic Function

The x-intercepts (also called roots or zeros) occur where the graph crosses the x-axis, that is, where \(f(x) = 0\):

\[ ax^{2} + bx + c = 0 \]

These are found using the quadratic formula:

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

The expression under the square root, the discriminant \( D = b^{2} - 4ac \), determines the number of real x-intercepts:

Examples: Finding X-Intercepts

Example 1: Find the x-intercepts of \( f(x) = x^{2} + 2x - 3 \).

Show Solution

Set \( x^{2} + 2x - 3 = 0 \).

Calculate the discriminant: \( D = (2)^{2} - 4(1)(-3) = 4 + 12 = 16 \).

Since \(D > 0\), there are two distinct intercepts:

\[ x_{1} = \frac{-2 + \sqrt{16}}{2(1)} = \frac{-2 + 4}{2} = 1 \] \[ x_{2} = \frac{-2 - \sqrt{16}}{2(1)} = \frac{-2 - 4}{2} = -3 \]

X-intercepts: \((1, 0)\) and \((-3, 0)\).

Example 2: Find the x-intercepts of \( g(x) = -x^{2} + 2x - 1 \).

Show Solution

Set \( -x^{2} + 2x - 1 = 0 \).

Calculate the discriminant: \( D = (2)^{2} - 4(-1)(-1) = 4 - 4 = 0 \).

Since \(D = 0\), there is exactly one repeated intercept:

\[ x = \frac{-b}{2a} = \frac{-2}{2(-1)} = 1 \]

X-intercept: \((1, 0)\) (a double root; the vertex is on the x-axis).

Example 3: Find the x-intercepts of \( h(x) = -2x^{2} + 2x - 2 \).

Show Solution

Set \( -2x^{2} + 2x - 2 = 0 \).

Calculate the discriminant: \( D = (2)^{2} - 4(-2)(-2) = 4 - 16 = -12 \).

Since \(D < 0\), there are no real x-intercepts.

Practice: Find the X-Intercepts

  1. \( f(x) = x^{2} + x - 2 \)
    Show Solution

    \( D = 1^{2} - 4(1)(-2) = 9 \).

    \( x = \frac{-1 \pm 3}{2} \implies x = 1 \text{ or } x = -2 \).

    X-intercepts: \((1, 0)\) and \((-2, 0)\).

  2. \( g(x) = 4x^{2} + x + 1 \)
    Show Solution

    \( D = 1^{2} - 4(4)(1) = 1 - 16 = -15 \).

    Because the discriminant is negative, there are no real x-intercepts.

  3. \( h(x) = x^{2} - 4x + 4 \)
    Show Solution

    \( D = (-4)^{2} - 4(1)(4) = 16 - 16 = 0 \).

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

    X-intercept: \((2, 0)\) (a double root).

D - Y-Intercept of a Quadratic Function

The y-intercept is the point where the graph crosses the y-axis, which occurs when \(x = 0\). Evaluating the general form gives:

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

Thus, the parabola always intersects the y-axis at the point \((0, c)\).

Examples:

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