Quadratic Models
Master quadratic models for NSW Year 12 Mathematics Standard 2. In this topic you use a parabola \(y=ax^{2}+bx+c\) to model practical situations — the height of a thrown ball or a water jet, the area of an enclosure, or the price that maximises profit — and read its shape, intercepts and vertex.
You will learn to interpret the \(y\)-intercept as the initial value and the \(x\)-intercepts as the zeros, find the maximum or minimum at the vertex using \(x=-\dfrac{b}{2a}\) or the midpoint of the \(x\)-intercepts, and state the valid domain and limitations of a model — a core Standard 2 skill for solving real-world quadratic problems.
Theory
Quadratic models use a parabola \(y=ax^{2}+bx+c\) to describe practical relationships. This Year 12 Standard 2 (NSW) guide shows how to read the parabola's shape, interpret the \(y\)-intercept and \(x\)-intercepts, find the vertex (the maximum or minimum) with \(x=-\dfrac{b}{2a}\) or the midpoint of the intercepts, and state the valid domain and limitations of the model.
A quadratic model describes a practical relationship with a parabola \(y=ax^{2}+bx+c\) (with \(a\neq 0\)). The graph is a smooth U-shape (or upside-down U), symmetric about a vertical axis of symmetry.
The coefficient \(a\) sets the shape: if \(a>0\) the parabola opens upwards and the vertex is a minimum; if \(a<0\) it opens downwards and the vertex is a maximum. The \(y\)-intercept is \(c\), the value when \(x=0\) — the initial value of the model. The \(x\)-intercepts (zeros) are where \(y=0\), such as where a height model returns to the ground.
The vertex lies on the axis of symmetry \(x=-\dfrac{b}{2a}\), which is also the midpoint of the two \(x\)-intercepts. Substituting that \(x\) back gives the maximum or minimum value. Because a model only makes sense over sensible values, you also state a valid domain and note the model's limitations. This is a core NSW Year 12 Mathematics Standard 2 skill.
A quadratic model is a parabola:
The vertex sits on the axis of symmetry, half-way between the \(x\)-intercepts:
The \(y\)-intercept is the initial value, read at \(x=0\):
How to solve a quadratic-model problem
- Read the sign of \(a\) — decide whether the vertex is a maximum (\(a<0\)) or a minimum (\(a>0\)).
- Intercepts — the \(y\)-intercept \(c\) is the initial value; the \(x\)-intercepts (set \(y=0\)) are the zeros of the model.
- Find the vertex — use \(x=-\dfrac{b}{2a}\), or the midpoint of the two \(x\)-intercepts.
- Get the value — substitute that \(x\) back to find the maximum or minimum \(y\).
- Interpret and check the domain — state what the answer means and the range of \(x\) over which the model is valid.
The release height is the \(h\)-intercept (\(t=0\)); then substitute \(t=1\).
| \(h(0)\) | \(=\) | \(15 + 10(0) - 5(0)^{2}\) |
| \(\) | \(=\) | \(15\text{ m}\) |
| \(h(1)\) | \(=\) | \(15 + 10(1) - 5(1)^{2}\) |
| \(\) | \(=\) | \(15 + 10 - 5 = 20\text{ m}\) |
The vertex is the midpoint of the two \(x\)-intercepts; substitute it back for the area.
| \(x(8-x)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0 \ \text{ or } \ x = 8\) |
| \(x_{\text{vertex}}\) | \(=\) | \(\dfrac{0+8}{2} = 4\) |
| \(A_{\max}\) | \(=\) | \(4(8-4) = 16\text{ m}^{2}\) |
Use \(x=-\dfrac{b}{2a}\) with \(a=-3,\ b=24\), then read the zeros for the domain.
| \(x\) | \(=\) | \(-\dfrac{24}{2(-3)} = 4\) |
| \(P_{\max}\) | \(=\) | \(24(4) - 3(4)^{2}\) |
| \(\) | \(=\) | \(96 - 48 = 48\) |
| \(\text{zeros}\) | \(:\) | \(x = 0,\ 8 \Rightarrow 0 \le x \le 8\) |
Maximum profit \(\$48\) at a price of \(\$4\), valid for \(0 \le x \le 8\).
The peak is at \(x=-\dfrac{b}{2a}\); the landing point is the positive \(x\)-intercept.
| \(x\) | \(=\) | \(-\dfrac{8}{2(-2)} = 2\) |
| \(h_{\max}\) | \(=\) | \(8(2) - 2(2)^{2} = 8\text{ m}\) |
| \(2x(4-x)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0 \ \text{ or } \ x = 4\) |
Greatest height \(8\) m; the water lands \(4\) m away. Valid domain \(0 \le x \le 4\).
Common pitfalls
Frequently asked questions
What is a quadratic model?
It is a practical relationship described by a parabola, written y equals a x squared plus b x plus c. It is used when a quantity rises to a maximum and then falls, or falls to a minimum and then rises, such as the height of a thrown ball, the area of an enclosure, or the profit at a given price.
How do you find the maximum or minimum of a quadratic model?
The maximum or minimum is at the vertex. Find its x-value with x equals minus b over 2a, or take the midpoint of the two x-intercepts, then substitute that x back into the model to get the actual maximum or minimum value.
What do the x-intercepts and y-intercept mean?
The y-intercept is the value of the model when x is zero, which is the constant c and the initial value. The x-intercepts are the zeros, where the quantity is zero, for example the times when a ball is at ground level.
How can you tell if a parabola opens up or down?
Look at the sign of a, the coefficient of x squared. If a is positive the parabola opens upwards and the vertex is a minimum. If a is negative it opens downwards and the vertex is a maximum.
What is the valid domain of a quadratic model?
It is the set of x-values over which the model makes sense in context. For a ball thrown into the air, time runs from release until it lands, so the domain might be zero to five seconds. Values outside that range, or negative outputs, have no real meaning.
Why does a quadratic model have limitations?
A parabola continues without bound in both directions, but the real situation does not. Beyond a sensible range the model gives impossible answers, such as negative heights or profits, so it should only be used over its valid domain.