Quadratic Functions from the Product of Two Linear Functions

Q: Why is the product of two linear functions a quadratic function?

Multiplying (ax + b)(Ax + B) gives aA x^2 + (aB + bA)x + bB. Because a and A are both non-zero, the coefficient aA of x^2 is non-zero, so the result is a quadratic function.

Q: Where is the vertex of the parabola (ax + b)(Ax + B)?

The parabola is zero at the roots of the two linear factors: x = -b/a and x = -B/A. The vertex lies on the axis of symmetry, the vertical line through the midpoint of these two roots.

Q: What happens if a = 0 or A = 0?

If a = 0 or A = 0, one of the factors is constant. The product then has degree 1 or 0, so it is no longer a quadratic function. The applet detects this case and labels the product as linear or constant.

Using fundamental algebra, we can prove that multiplying two linear functions produces a quadratic function. This relationship is explored interactively below.

Let \( h(x) \) and \( g(x) \) be two linear functions:

\[ h(x) = ax + b, \qquad g(x) = Ax + B \]

where \( a \) and \( A \) are non-zero constants. Their product is

\[ f(x) = h(x)\,g(x) = (ax + b)(Ax + B) \]

Expanding,

\[ f(x) = aA\,x^{2} + (aB + bA)\,x + bB \]

Because \( a \neq 0 \) and \( A \neq 0 \), the coefficient \( aA \) of \( x^2 \) is non-zero, so \( f \) is a genuine quadratic function. Adjust the coefficients below to see how the graphs change in real time, and note how the x-intercepts of \( h(x) \) and \( g(x) \) become the x-intercepts of \( f(x) \).

Interactive Grapher

a (slope of h): 1.00

b (y-intercept of h): 2.00

A (slope of g): 1.00

B (y-intercept of g): 0.00

h(x) = 1.00x + 2.00

g(x) = 1.00x + 0.00

f(x) = h(x) × g(x) = 1.00x² + 2.00x + 0.00

Note: Because one of the slopes is zero, the product is no longer a quadratic function. The graph shown is linear or constant.

X-Intercepts:
\( h(x) = 0 \) when \( x = \) -2.00
\( g(x) = 0 \) when \( x = \) 0.00
\( f(x) = 0 \) when \( x = \) -2.00, 0.00

Hover over graph

Observations and Analysis

X-intercept inheritance: The quadratic \( f(x) \) is zero exactly where either \( h(x) = 0 \) or \( g(x) = 0 \). This is because a product of two numbers is zero if and only if at least one factor is zero.
Y-intercept: \( f(0) = h(0)\,g(0) = b \cdot B \). The y-intercept of the parabola is the product of the y-intercepts of the two lines.
Vertex position: The vertex lies on the axis of symmetry, which passes through the midpoint of the two distinct x-intercepts: \( x = \dfrac{1}{2}\left(-\dfrac{b}{a} - \dfrac{B}{A}\right) \).
Double root: When \( h(x) \) and \( g(x) \) share the same x-intercept (that is, when \( -\frac{b}{a} = -\frac{B}{A} \), or equivalently \( aB = bA \)), the parabola touches the x-axis at a single point — a double root. In this case the discriminant is \((aB - bA)^2 = 0\).
Opening direction: The parabola opens upward if \( aA > 0 \) and downward if \( aA < 0 \).
Degenerate case: If \( a = 0 \) or \( A = 0 \), one factor is constant, so the product has degree 1 or 0. It is then linear or constant, not quadratic.

Frequently Asked Questions

Why is the product of two linear functions a quadratic function?

Multiplying \( (ax+b)(Ax+B) \) gives \( aA x^2 + (aB+bA)x + bB \). Because \( a \) and \( A \) are both non-zero, the coefficient \( aA \) of \( x^2 \) is non-zero, so the result is a quadratic function.

Where is the vertex of the parabola?

The parabola is zero at the roots of the two linear factors, \( x = -b/a \) and \( x = -B/A \). The vertex lies on the axis of symmetry, the vertical line through the midpoint of these two roots.

What happens if \( a = 0 \) or \( A = 0 \)?

If either slope is zero, one of the factors is constant and the product has degree 1 or 0. It is then linear or constant, not quadratic. The applet detects this case and labels it accordingly.