Quadratic Models Revision Notes for HSC SSCE Mathematics Advanced

Quadratic Models

Quadratic models are powerful mathematical tools that help us understand and solve real-world problems involving parabolic relationships. In this topic, you will learn how to construct quadratic functions to represent practical situations, interpret key features of these models, and use them to solve optimisation problems.

Understanding quadratic models

A quadratic function has the general form $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants and $a \neq 0$ . These functions create parabolic graphs that can model many real-world scenarios, such as:

The path of a projectile (like a ball being thrown)
Calculating area or perimeter
Determining profit margins in business
Finding optimal solutions to practical problems

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The key feature of quadratic models is their parabolic shape, which allows them to represent situations where values increase to a maximum (or decrease to a minimum) and then change direction. This characteristic makes them ideal for modeling turning point behavior in real-world applications.

Forms of quadratic functions

Quadratic functions can be written in three different forms, and each form makes certain features easier to identify. Understanding when to use each form is essential for efficient problem-solving.

Standard form

$f(x) = ax^2 + bx + c$

When to use: This form is particularly useful for:

Finding the $y$ -intercept, which occurs at the point $(0, c)$
Solving for the roots using the quadratic formula
Identifying the coefficients needed for calculations

Vertex form

$f(x) = a(x - h)^2 + k$

When to use: This form directly reveals:

The vertex of the parabola at the point $(h, k)$
Whether the vertex represents a maximum or minimum value
The axis of symmetry at $x = h$

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The vertex gives you the maximum value when $a < 0$ (parabola opens downward) or the minimum value when $a > 0$ (parabola opens upward). This is crucial for optimization problems.

Factorised form

$f(x) = a(x - x_1)(x - x_2)$

When to use: This form immediately shows:

The $x$ -intercepts (roots) at $x_1$ and $x_2$
Where the function crosses the horizontal axis
Break-even points in business applications

Finding the vertex

When you have a quadratic function in standard form, you can find the $x$ -coordinate of the vertex using the following formula:

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Vertex Formula:

$x = -\frac{b}{2a}$

Once you have the $x$ -coordinate, substitute it back into the original function to find the corresponding $y$ -coordinate. This vertex point $(h, k)$ represents either the maximum or minimum value of the function.

Domain and range in quadratic models

When using quadratic functions to model real-world situations, it's essential to consider realistic constraints that limit the mathematical domain and range.

Domain

The domain represents all possible input values. While mathematically a quadratic function has domain $\mathbb{R}$ (all real numbers), in practical applications the domain is often restricted by context.

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Common Domain Restrictions:

Time values are typically non-negative ( $x \geq 0$ )
Physical dimensions must be positive
Values between break-even points represent profitable regions
Manufacturing or production constraints may limit quantities

Always consider the real-world context when determining the appropriate domain.

Range

The range represents all possible output values and depends on:

The concavity of the parabola (whether it opens upward or downward)
The $y$ -coordinate of the vertex

For a parabola:

Opening upward ( $a > 0$ ): Range is $[k, \infty)$ where $k$ is the $y$ -coordinate of the vertex
Opening downward ( $a < 0$ ): Range is $(-\infty, k]$ where $k$ is the $y$ -coordinate of the vertex

Worked example: projectile motion

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Worked Example: Modeling a Projectile

Problem: A ball is thrown upward from a height of 2 metres with velocity 10 m/s. Its height is modelled by $f(x) = -4.9x^2 + 10x + 2$ , where $x$ is time in seconds and $f(x)$ is height in metres.

Part a: Finding when the ball hits the ground

When the ball hits the ground, its height is zero, so we need to solve $f(x) = 0$ .

Method: Use the quadratic formula to find when height equals zero.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Identify the coefficients: $a = -4.9$ , $b = 10$ , $c = 2$

$x = \frac{-10 \pm \sqrt{10^2 - 4(-4.9)(2)}}{2(-4.9)}$

$x = \frac{-10 \pm \sqrt{100 + 39.2}}{-9.8}$

$x = \frac{-10 \pm \sqrt{139.2}}{-9.8}$

This gives two solutions:

$x = \frac{-10 + \sqrt{139.2}}{-9.8} \text{ or } x = \frac{-10 - \sqrt{139.2}}{-9.8}$

$x = -0.184 \text{ or } x = 2.224$

Since time cannot be negative in this context, we discard the negative solution.

Answer: The ball hits the ground at $x = 2.224$ seconds.

Part b: Finding the maximum height

The maximum height occurs at the vertex of the parabola. Since $a = -4.9 < 0$ , the parabola opens downward, confirming the vertex gives a maximum.

Step 1: Find the $x$ -coordinate of the vertex:

$x = -\frac{b}{2a}$

$x = -\frac{10}{2(-4.9)}$

$x = 1.02$

Step 2: Substitute $x = 1.02$ into the function to find the maximum height:

$f(1.02) = -4.9(1.02)^2 + 10(1.02) + 2$

$f(1.02) = -4.9(1.0404) + 10.2 + 2$

$f(1.02) = 7.10$

Answer: At $x = 1.02$ seconds, the maximum height is 7.10 metres.

Part c: Stating the domain and range

Domain: Time starts at zero and continues until the ball hits the ground at $x = 2.224$ seconds. Therefore, the domain is $[0, 2.224]$ .

Range: The ball starts at height 2 metres ( $f(0) = 2$ ), reaches a maximum of 7.10 metres at the vertex, then returns to height zero. Since $a = -4.9 < 0$ , the parabola opens downward. Therefore, the range is $[0, 7.10]$ .

Visual verification:

The parabola confirms our calculations, showing the motion from $x = 0$ to $x = 2.224$ , with the maximum at $(1.02, 7.10)$ .

Optimisation with quadratic models

Quadratic functions are particularly useful for solving optimisation problems where you need to find maximum or minimum values. Common applications include maximising area, minimising cost, or maximising profit.

Break-even point

In business applications, a break-even point is the point at which income from production equals the cost of production. At this point, profit is zero.

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To find break-even points, solve the equation $f(x) = 0$ where $f(x)$ represents profit. The roots of the quadratic function indicate where profit changes from negative (loss) to positive (profit) or vice versa.

Worked example: profit optimisation

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Worked Example: Maximizing Business Profit

Problem: A company's profit is modelled by $f(x) = -2x^2 + 40x - 150$ , where $x$ is units sold (in hundreds) and $f(x)$ is profit in thousands of dollars.

Part a: Finding break-even points

Break-even occurs when profit equals zero, so we solve $f(x) = 0$ .

Method: Use factorisation to find the roots.

$-2x^2 + 40x - 150 = 0$

Factor out $-2$ :

$-2(x^2 - 20x + 75) = 0$

Divide both sides by $-2$ :

$x^2 - 20x + 75 = 0$

Factorise:

$(x - 15)(x - 5) = 0$

Using the null factor law:

$x = 5 \text{ or } x = 15$

Answer: The break-even points are at $x = 5$ and $x = 15$ , representing 500 and 1500 units respectively.

Part b: Finding maximum profit

The maximum profit occurs at the vertex of the parabola. Since $a = -2 < 0$ , the parabola opens downward, so the vertex represents a maximum.

Step 1: Find the $x$ -coordinate of the vertex:

$x = -\frac{b}{2a}$

$x = -\frac{40}{2(-2)}$

$x = 10$

Step 2: Substitute $x = 10$ into the function:

$f(10) = -2(10)^2 + 40(10) - 150$

$f(10) = -2(100) + 400 - 150$

$f(10) = -200 + 400 - 150$

$f(10) = 50$

Answer: At $x = 10$ (1000 units), the maximum profit is 50 thousand dollars ($50,000).

Part c: Stating the domain and range

Domain: Profit is non-negative between the break-even points. Therefore, the domain is $[5, 15]$ .

Range: The profit ranges from zero (at the break-even points) to the maximum at the vertex. Since $a = -2 < 0$ , the parabola opens downward with minimum at $f(5) = f(15) = 0$ and maximum at 50. Therefore, the range is $[0, 50]$ .

Visual verification:

The graph confirms that profit is positive when $5 < x < 15$ , with the maximum at $(10, 50)$ .

Exam tips

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Common Mistakes to Avoid:

Always check whether the context requires restrictions on the domain (e.g., time cannot be negative, physical dimensions must be positive)
When finding maximum or minimum values, first check the sign of $a$ to determine whether the parabola opens upward or downward
For break-even problems, remember that these occur where profit equals zero, not where profit is maximised
Label your axes clearly on graphs and mark key points such as the vertex and intercepts
Always express your final answer in the context of the problem (e.g., time in seconds, profit in dollars)

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Key Points to Remember:

Quadratic functions model parabolic relationships such as projectile motion, area calculations, and profit analysis
Three forms serve different purposes: standard form for finding $y$ -intercepts and using the quadratic formula, vertex form for identifying maximum/minimum values, and factorised form for finding $x$ -intercepts
The vertex gives the optimal value: located at $x = -\frac{b}{2a}$ , it represents either the maximum (when $a < 0$ ) or minimum (when $a > 0$ ) value of the function
Domain depends on context: while mathematically unrestricted, real-world applications often require non-negative values or values within specific boundaries
Break-even points occur where profit equals zero: found by solving $f(x) = 0$ , these indicate the boundaries between profit and loss

Quadratic Models (HSC SSCE Mathematics Advanced): Revision Notes