Quadratic Functions: Complete Guide with Interactive Graphing

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

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

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 that opens upward if \(a > 0\) and downward if \(a < 0\). The magnitude \(|a|\) controls the width: larger \(|a|\) gives a narrower parabola.

Examples:

Interactive Quadratic Function Grapher

Explore the effects of the coefficients \( a , b \) and \(c \) on the vertex, the x and y intercepts

Determines opening direction (up if \(a>0\), down if \(a<0\)).

Shifts the axis of symmetry \(x = -b/(2a)\).

The y-intercept of the parabola is \((0, c)\).

Zoom Controls

B. Vertex Form and the Vertex

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

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

where \((h, k)\) is the vertex. Converting from general form to vertex form uses the technique of completing the square.

The vertex \((h, k)\) and the axis of symmetry are given by:

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

Axis of symmetry: the vertical line \( x = -\dfrac{b}{2a} \), about which the parabola is symmetric.

The vertex is a minimum if \(a > 0\) and a maximum if \(a < 0\).

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

Show Solution
  1. Factor out the coefficient \(a = -2\) 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. Group the perfect square trinomial:
    \( f(x) = -2[(x - 1)^2 - 1] + 1 \)
  4. Distribute the \(-2\) back to the constants:
    \( f(x) = -2(x - 1)^2 + 2 + 1 \)
  5. Final form: \( f(x) = -2(x - 1)^2 + 3 \)
  6. Vertex: \( (1, 3) \); axis of symmetry: \( x = 1 \)

Check with the interactive grapher: set \(a = -2\), \(b = 4\), \(c = 1\) to verify the maximum vertex is at \((1, 3)\).

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

The \(x\)-intercepts solve the equation \( ax^2 + bx + c = 0 \).

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

When \(D \geq 0\), the intercepts are given by the quadratic formula:

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

Find the intercepts for \(f(x) = x^2 + 2x - 3\)

Show Solution

Calculate the discriminant \(D\):

\[ D = 2^2 - 4(1)(-3) = 4 + 12 = 16 > 0 \]

Because \(D > 0\), there are two distinct x-intercepts:

\[ x_1 = \frac{-2 + \sqrt{16}}{2(1)} = \frac{-2 + 4}{2} = 1 \] \[ x_2 = \frac{-2 - \sqrt{16}}{2(1)} = \frac{-2 - 4}{2} = -3 \]

Intercepts: \((1, 0)\) and \((-3, 0)\)

Find the intercepts for \(g(x) = -x^2 + 2x - 1\)

Show Solution

Calculate the discriminant \(D\):

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

Because \(D = 0\), there is exactly one repeated intercept (at the vertex):

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

Single intercept: \((1, 0)\) (a double root).

Find the intercepts for \(h(x) = -2x^2 + 2x - 2\)

Show Solution

Calculate the discriminant \(D\):

\[ D = 2^2 - 4(-2)(-2) = 4 - 16 = -12 < 0 \]

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

D. \(y\)-Intercept

The \(y\)-intercept occurs where the graph crosses the y-axis (at \(x = 0\)):

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

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

E. Practice: Find the Equation from a Graph

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

Parabola with vertex (-2,-2), y-intercept (0,6), and x-intercepts (-3,0) and (-1,0)

Find the quadratic function for this parabola passing through the vertex \((-2,-2)\), y-intercept \((0,6)\), and x-intercepts \((-3,0)\) and \((-1,0)\).

Show Solution (Method 1: Using Intercepts)

Use the x-intercepts \((-3,0)\) and \((-1,0)\), and the y-intercept \((0,6)\):

  1. Start with the intercept form: \( f(x) = a(x - r_1)(x - r_2) \)
  2. Substitute the roots: \( f(x) = a(x + 3)(x + 1) \)
  3. Plug in the y-intercept \((0,6)\) to find \(a\):
    \( 6 = a(0 + 3)(0 + 1) \)
    \( 6 = 3a \implies a = 2 \)
  4. Result: \( f(x) = 2(x+3)(x+1) = 2x^2 + 8x + 6 \)
Show Solution (Method 2: Using Vertex Form)

Use the vertex \((-2,-2)\) and the y-intercept \((0,6)\):

  1. Start with the vertex form: \( f(x) = a(x - h)^2 + k \)
  2. Substitute the vertex: \( f(x) = a(x + 2)^2 - 2 \)
  3. Plug in the y-intercept \((0,6)\) to find \(a\):
    \( 6 = a(0 + 2)^2 - 2 \)
    \( 6 = 4a - 2 \implies 8 = 4a \implies a = 2 \)
  4. Result: \( f(x) = 2(x+2)^2 - 2 = 2(x^2 + 4x + 4) - 2 = 2x^2 + 8x + 6 \)
Show Solution (Method 3: System of Equations)

Using the points \((-3,0)\), \((-1,0)\), and \((0,6)\) in the general form \( f(x) = ax^2 + bx + c \):

  1. From \((0,6)\): \( a(0)^2 + b(0) + c = 6 \implies c = 6 \)
  2. From \((-3,0)\): \( a(-3)^2 + b(-3) + 6 = 0 \implies 9a - 3b + 6 = 0 \)
  3. From \((-1,0)\): \( a(-1)^2 + b(-1) + 6 = 0 \implies a - b + 6 = 0 \)
  4. Solve the system:
    From the third equation, \( a = b - 6 \). Substitute into the second equation:
    \( 9(b - 6) - 3b + 6 = 0 \)
    \( 9b - 54 - 3b + 6 = 0 \)
    \( 6b = 48 \implies b = 8 \)
    Thus, \( a = 8 - 6 = 2 \).
  5. Result: \( f(x) = 2x^2 + 8x + 6 \)

F. Frequently Asked Questions

What is the vertex formula for a quadratic function?

For \( f(x) = ax^2 + bx + c \), the vertex is \( (h, k) \) with \( h = -\dfrac{b}{2a} \) and \( k = f(h) = c - \dfrac{b^2}{4a} = \dfrac{4ac - b^2}{4a} \). The vertical line \( x = h \) is the axis of symmetry.

How do you find the x-intercepts of a quadratic function?

Set \( f(x) = 0 \) and solve \( ax^2 + bx + c = 0 \) with the quadratic formula. The discriminant \( D = b^2 - 4ac \) tells you whether there are two, one, or no real x-intercepts.

What does the discriminant tell you?

The discriminant \( D = b^2 - 4ac \) determines the number of real roots: \( D > 0 \) gives two distinct real x-intercepts, \( D = 0 \) gives one repeated (double) root, and \( D < 0 \) gives no real x-intercepts.

Where is the y-intercept of a quadratic function?

At \( x = 0 \), so \( f(0) = c \). The parabola always crosses the y-axis at \( (0, c) \).

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