11.1 Linear and Quadratic Functions

Quick Answer: On the calculator-free MAT paper, a function gives exactly one output for each allowed input. For a straight line y = mx + c, read the gradient and intercepts. For a parabola y = ax² + bx + c, the sign of a decides maximum versus minimum, the constant term is the y-intercept, and completing the square gives the axis of symmetry and turning point.

The National Benchmark Tests Project lists function skills on nbt.ac.za Test Content: understand the function concept and identify properties of functions. Independent OpenExamPrep examples in this section use original numbers. They do not copy NBT or official NBTP exemplar items.

Why this item is fast if you know what to look at

A classic MAT functions question is not a long expansion. It asks whether a parabola has a maximum or a minimum y-value, and whether the y-intercept is positive or negative — the 2015 MAT booklet Q1 pattern, taught here with new coefficients. You decide those two facts from a and c in a few seconds. Building a calculator-style table of many points wastes the three-hour paper.

Function concept, notation, and substitution

A function f pairs each x in a domain with exactly one output f(x). We write y = f(x). The range is the set of y-values that actually appear.

Substitution means replacement, with brackets:

Let f(x) = 4x − 9.

• f(5) = 4(5) − 9 = 11
• f(0) = −9 (this is also the y-intercept of the graph)
• f(−3) = 4(−3) − 9 = −21
• f(x + 2) = 4(x + 2) − 9 = 4x − 1
• f(x) + 2 = 4x − 7

Those last two lines are different. MAT options often mix f(x + h) with f(x) + h.

Let g(x) = x² − 5x. Then g(3) = 9 − 15 = −6, and g(2a) = (2a)² − 5(2a) = 4a² − 10a. Never distribute a square as 2a².

Domain restrictions you must catch by eye

For y = mx + c with m ≠ 0, domain and range are both all real numbers. For a quadratic, domain is all real x, but range is one-sided: from the turning-point y-value upward if a > 0, or downward if a < 0.

Straight lines

The linear function is y = mx + c.

• m is the gradient. Positive m: the line rises left to right. Negative m: it falls.
• c is the y-intercept, point (0, c).
• x-intercept: set y = 0. For 0 = 3x − 12, x = 4, point (4, 0).

Gradient from two points: m = (y₂ − y₁)/(x₂ − x₁). Through (1, 4) and (5, −4): m = (−4 − 4)/(5 − 1) = −2. Then y − 4 = −2(x − 1), so y = −2x + 6.

Parallel lines share m. Perpendicular lines satisfy m₁m₂ = −1. So y = (2/3)x + 1 is perpendicular to y = (−3/2)x + 7.

A horizontal line y = 5 is a function with gradient 0. A vertical line x = 5 is not a function of x — one x would pair with many y-values — but x = k still appears later as a boundary of a region.

Sketch without a calculator: start at the y-intercept, then step with the gradient. y = (1/2)x − 3 starts at (0, −3); right 2 and up 1 reaches (2, −2), then (4, −1).

Parabolas: intercepts, turning points, max/min versus y-intercept

The quadratic function is y = ax² + bx + c with a ≠ 0. Its graph is a parabola.

The Q1-style reading, with new numbers

Sign of a

• a > 0: opens upward → minimum y-value at the vertex
• a < 0: opens downward → maximum y-value at the vertex

The y-intercept is f(0) = c. It is not the turning point unless the axis of symmetry is the y-axis (b = 0).

Example A: f(x) = 2x² − 12x + 9

• a = 2 > 0 → minimum y-value
• c = 9 > 0 → positive y-intercept

Correct description: a minimum y-value and a positive y-intercept.

Example B: f(x) = −x² + 4x − 7

• a = −1 < 0 → maximum y-value
• c = −7 < 0 → negative y-intercept

The actual vertex height is a separate computation. For Example A, x_vertex = −b/(2a) = 12/4 = 3, and f(3) = 2(9) − 12(3) + 9 = −9. Minimum value −9; y-intercept still +9. Mixing those two numbers is the trap the Q1-style item is built on.

Intercepts

• y-intercept: (0, c)
• x-intercepts: roots of ax² + bx + c = 0

Factor example: y = x² − 8x + 12 = (x − 2)(x − 6). Crosses the x-axis at 2 and 6. Axis of symmetry midway: x = 4. Then f(4) = 16 − 32 + 12 = −4, so the turning point is (4, −4).

Discriminant Δ = b² − 4ac:

• Δ > 0: two x-intercepts
• Δ = 0: one x-intercept (vertex on the x-axis)
• Δ < 0: no x-intercepts

For y = x² − 8x + 20, Δ = 64 − 80 = −16 < 0. Opens up, so the graph sits above the x-axis. Vertex at x = 4, y = 4, range y ≥ 4.

Completing the square and the axis of symmetry

Completing the square rewrites the quadratic so the vertex is visible:

y = a(x − p)² + q

• Axis of symmetry: x = p
• Turning point: (p, q)

Worked example: y = x² − 10x + 21

x² − 10x + 21 = (x² − 10x + 25) − 25 + 21 = (x − 5)² − 4

Axis x = 5. Turning point (5, −4). Minimum value −4. y-intercept 21 — again, not the same as the minimum.

When the leading coefficient is not 1:

y = 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2((x − 2)² − 4) + 3 = 2(x − 2)² − 5

Axis x = 2. Turning point (2, −5). Minimum −5. Check the y-intercept from the original: +3.

The formula x = −b/(2a) gives the same axis. On a no-calculator paper, use whichever form is already on the page.

Range from vertex form: y = (x − 5)² − 4 has range y ≥ −4. y = 5 − (x − 1)² has range y ≤ 5.

Order of attack on a sketched parabola

1. Opens up or down (sign of a, or the picture).
2. Read the y-intercept.
3. Axis: midway between roots, or x = −b/(2a), or the x-coordinate in vertex form.
4. Turning-point y-value: substitute or read q.
5. Expand or match an equation only if the options require it.

Why calculus belongs in this guide

The 2025 National Report defines the current MAT M3 Functions and Graphs subdomain as analysing functions and graphs and applying principles of differential calculus. The shorter live Test Content bullet list does not spell out calculus, but the official 2015 MAT educator booklet is also explicit about critical points, increasing and decreasing functions, tangents, and interpreting a function from its derivative. Section 11.4 therefore teaches that assessed skill directly rather than treating it as optional or unpublished.