Quadratic and Polynomial Functions
Quadratic Functions and the Parabola
The parabola as the graph of a second-degree function, how the leading coefficient decides its direction, and reading the range from the turning point.
What this page covers
- Recognise a quadratic function and name its coefficients
- Predict the direction of opening from the leading coefficient
- State the domain and range of a quadratic
- Read the vertex from a completed-square form
The general form
The condition a ≠ 0 is what makes it quadratic; with a = 0 the function collapses to a straight line. The graph is always a parabola.
Evaluating — the source's example
For f(x) = 2x2 - 5x + 2:
Both evaluate to zero, so 12 and 2 are the roots. The function factors as (2x - 1)(x - 2).
The simplest parabola
The source starts from y = x2, tabulates it, and reads off every feature. Everything else is a transformation of this curve.
| x | −4 | −2 | 0 | 2 | 4 |
|---|---|---|---|---|---|
| x2 | 16 | 4 | 0 | 4 | 16 |
- Vertex
- (0, 0), a minimum
- Domain
- All real x
- Range
- y ≥ 0, that is [0, ∞)
- Axis of symmetry
- The y-axis, x = 0
- Opens
- Upward, because a = 1 > 0
The source then negates it. For y = -x2 the table values all change sign, so the curve flips: maximum at x = 0, domain all real x, range y ≤ 0.
a > 0: opens upward, vertex is a minimum, range is [k, ∞). a < 0: opens downward, vertex is a maximum, range is (-∞, k].
Shifting the parabola
The source works four transformed cases in succession, each illustrating one movement.
| Function | Vertex | Opens | Range |
|---|---|---|---|
| y = x2 + 4 | (0, 4) | Up | [4, ∞) |
| y = x2 - 7 | (0, -7) | Up | [-7, ∞) |
| y = (x - 1)2 + 4 | (1, 4) | Up | [4, ∞) |
| y = 7 - (x + 2)2 | (-2, 7) | Down | (-∞, 7] |
For y = x2 - 7 the source finds the roots by setting y = 0: x = ±√7. Note the two roots sit symmetrically about the axis x = 0, as they must.
For y = 7 - (x + 2)2 the source reasons directly: has maximum, because -x2; max is 7 when x = -2. The subtracted square can only reduce the value below 7, and it contributes nothing when x = -2.
(x - 1)2 shifts the graph right by 1, and (x + 2)2 shifts it left by 2. The vertex sits where the bracket is zero, so the sign inside is opposite to the direction of movement.
Vertex form
The argument is short and worth internalising. (x - h)2 ≥ 0 always, and equals zero only at x = h. So when a > 0 the term a(x - h)2 is non-negative, and y is smallest exactly when it vanishes — giving y = k at x = h.
Expanding back — the source's check
For y = 7 - 2(x - 5)2:
Maximum 7 at x = 5, since a = -2 < 0. The expanded form hides that entirely, which is the argument for keeping the vertex form when the turning point matters.
The vertex is at x = -b2a = -20-4 = 5 ✓, and y = -2(25) + 100 - 43 = 7 ✓.
Domain, range and symmetry
| Feature | Value | Reason |
|---|---|---|
| Domain | All real x | A polynomial is defined everywhere |
| Range (a > 0) | [k, ∞) | The minimum is k |
| Range (a < 0) | (-∞, k] | The maximum is k |
| Axis of symmetry | x = h = -b2a | The parabola reflects onto itself |
| y-intercept | (0, c) | Set x = 0 in ax2 + bx + c |
| Number of x-intercepts | 0, 1 or 2 | Decided by the discriminant |
The domain is never the interesting part; it is always everything. The range is where the information is, and it comes entirely from the vertex and the sign of a.
If a parabola has two roots, their average is h. This is a useful check: for y = x2 - 7 the roots ±√7 average to 0, which is the axis.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| (x - 1)2 shifts left | It shifts right | The vertex is where the bracket vanishes |
| Restricting the domain | It is always all real x | Polynomials are defined everywhere |
| Assuming the vertex is a minimum | Check the sign of a | a < 0 gives a maximum |
| Reading the vertex y from the expanded form | Complete the square first | c is the y-intercept, not the vertex |
| Giving the range as all real y | It is bounded on one side | The vertex bounds it |
| -2(x - 5)2 expanded as -2x2 - 50 | -2x2 + 20x - 50 | The middle term is not optional |
Frequently asked questions
What is the domain of a quadratic?
Every real number. A polynomial is defined everywhere, so the restriction is always on the range, never the domain.
How do I know whether the turning point is a maximum or a minimum?
By the sign of a. Positive means the parabola opens upward and the turning point is a minimum; negative means it opens downward and the turning point is a maximum.
Why is the vertex form so useful?
Because (x - h)2 is never negative and is zero only at x = h. So y = a(x - h)2 + k has its extreme value k at x = h, with no further work needed.
Where is the axis of symmetry?
The vertical line through the vertex, x = h. Equivalently x = -b2a. Every parabola is symmetric about it, which is why its two roots are equidistant from it.
Source. Handwritten teaching notes, Week 6, pages 1-2.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.