Linear and quadratic functions

Review: output, difference and rate

A function associates each admissible input with an output. For distinct inputs x₁ and x₂, average rate is [f(x₂) − f(x₁)]/(x₂ − x₁). The quotient distinguishes a total change from a change per input unit. If cost rises by 600 Kz when quantity increases three units, average rate is 200 Kz per unit.

Substituting a value and solving a condition are different tasks. For C(n) = 200n + 500, C(4) = 1300 calculates the cost of four units. Solving C(n) = 1300 seeks the corresponding quantity and gives n = 4. A graph collects these pairs, but interpretation depends on axes and units.

A comparison requires variables to have the same meaning. We cannot directly compare a daily-cost function with a monthly-cost function without aligning input and output. Before calculating, write what x measures, what y measures and which x values make sense. This discipline makes parameters understandable.

The linear function and the meaning of slope

An affine function is written f(x) = mx + p, with real m and p. Parameter p = f(0) is the y-intercept. For any distinct inputs a and b, [f(b) − f(a)]/(b − a) = [m(b − a)]/(b − a) = m. Thus m is the constant rate of change and the graph is a straight line. If p = 0, the function is also directly proportional, often called linear under this convention.

If m > 0, the line increases; if m < 0, it decreases; if m = 0, the function is constant. For m ≠ 0, the zero solves mx + p = 0, so x = −p/m. A vertical line, x = a, is not the graph of a function y of x because one x-coordinate permits several y-coordinates.

Coefficients have contextual meaning. In C(n) = 150n + 800, with C in Kz and n in units, 800 Kz is a fixed cost and 150 Kz/unit the cost per additional unit. A y-intercept may be a physically meaningless extrapolation if zero does not belong to the situation's actual domain; always read it in the light of the model.

Quadratic variation and second differences

For a quadratic f(x) = ax² + bx + c, a ≠ 0, average rate between x and x + h is a(2x + h) + b, for h ≠ 0. Simply expand f(x + h) − f(x) and divide by h. The presence of x shows why no constant slope exists for the entire parabola. Coefficient a does not play the same role as m in a line.

For equal steps of size h, first differences vary and differences between those differences are constant, equal to 2ah². For f(x) = x² at x = 0, 1, 2, 3, outputs are 0, 1, 4, 9; first differences are 1, 3, 5 and second differences 2, 2. This pattern helps recognize data compatible with a quadratic model.

Compatible does not mean proved for every input. A few points can belong to several curves; measured data include errors. For output in m² and input in m, the x² coefficient may be dimensionless, while other term coefficients need units making every term an area. Quantities with incompatible dimensions cannot be added.

Transformations and vertex form

Writing f(x) = a(x − h)² + k shows a reference parabola shifted horizontally to h and vertically to k, with vertical scale |a| and reflection if a < 0. The vertex is (h, k). The sign inside parentheses needs care: (x + 3)² = (x − (−3))² shifts the vertex to x = −3.

To move from expanded to vertex form, complete the square. For example, 2x² − 12x + 11 = 2(x² − 6x) + 11 = 2[(x − 3)² − 9] + 11 = 2(x − 3)² − 7. Hence the vertex is (3, −7), axis x = 3 and range on domain ℝ is [−7, +∞[.

Changing c while keeping a and b shifts the curve vertically. Changing b while keeping a and c moves the axis and also changes vertex height. It is therefore unhelpful to memorize that each coefficient “moves the curve” in just one direction. The formulas h = −b/(2a) and k = f(h) explain the combined effect.

Zeros, signs and quadratic inequalities

The discriminant Δ = b² − 4ac indicates how many real zeros exist. When Δ > 0, call the ordered roots r₁ and r₂, with r₁ < r₂. In a(x − r₁)(x − r₂), we can determine each factor's sign on each interval. If a > 0, the function is positive outside the roots and negative between them. If a < 0, signs reverse.

If Δ = 0, the expression is a(x − r)²: away from the root its sign matches a and does not change when crossing r. If Δ < 0, the function never touches the horizontal axis and keeps a's sign throughout ℝ. These conclusions follow from factoring or vertex form, not merely memorizing a drawing.

Solving f(x) ≥ 0 asks for intervals where output is nonnegative; roots are included. Solving f(x) > 0 excludes them. Then intersect the algebraic solution with the problem domain. If x represents objects, only integers in the interval may be retained. An inequality's answer is a collection of inputs.

Intersections and choosing models

Two graphs intersect where they have the same input and output. To compare f and g, solve f(x) = g(x), or study the sign of f(x) − g(x). Comparing a line and parabola generally leads to a quadratic equation with zero, one or two real intersections. The condition f(x) > g(x) corresponds to f(x) − g(x) > 0.

In an area, optimization or revenue problem, the vertex can give an extremum. But if the domain is [a, b], we must know whether the vertex's x-coordinate lies inside and compare relevant boundaries too. For a discrete domain, integers nearest the vertex deserve checking. A continuous maximum at x = 4,5 does not authorize selling half an object.

A linear function assumes constant absolute increase; a quadratic assumes a specific form of change in that increase. Neither should be extended indefinitely just because it fits some data. Model choice needs a reason, such as a geometric relationship, defined tariff or pattern supported by enough information.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Every affine function is direct proportion.

Direct proportion requires a zero y-intercept.

A quadratic's coefficient a is its constant slope.

Average rate depends on the interval; a controls curvature and concavity.

Only roots are presented for a quadratic inequality.

Roots delimit intervals; signs must be studied and endpoints included or excluded.

The continuous maximum always solves a counting problem.

A count requires integer inputs; compare admissible ones near the vertex.