Quadratic Models Revision Notes for VCE SSCE Mathematical Methods

Quadratic Models

Introduction to quadratic models

Quadratic models allow us to use quadratic functions to solve real-world problems. These models are particularly useful when we need to find the maximum or minimum value of something in a practical situation.

infoNote

Common applications of quadratic models include:

Maximising area with limited materials (like fencing)
Modelling the path of projectiles (like balls or rockets)
Optimising profit or cost in business scenarios

The key skill is translating a word problem into a quadratic function, then using algebraic techniques to find the optimal value.

Finding maximum area with limited fencing

Let's explore how to use quadratic functions to solve optimization problems. The technique involves defining appropriate variables, expressing the quantity to be optimized as a function, and then using completing the square to find the maximum or minimum value.

lightbulbExample

Worked Example: Vegetable Garden Fence

Problem: Jenny has 20 metres of fencing wire to enclose a rectangular vegetable garden. She will use the wire for three sides of the garden, with an existing timber fence forming the fourth side. What is the maximum area she can enclose?

Solution:

First, we need to define our variables carefully:

Let $A$ represent the area of the rectangular garden (in square metres)
Let $x$ represent the length of the garden (the side parallel to the timber fence)

Since Jenny uses 20 metres of fencing for three sides (two widths and one length), we can work out the width:

Width = $\frac{20 - x}{2} = 10 - \frac{x}{2}$

Now we can express the area as a function of $x$ :

$A = \text{length} \times \text{width}$

A = x \left(10 - \frac{x}{2}\right)

Expanding this expression:

A = 10x - \frac{x^2}{2}

To find the maximum area, we need to convert this into vertex form by completing the square. First, we'll factor out the coefficient of $x^2$ :

A = -\frac{1}{2}(x^2 - 20x)

Now we complete the square inside the brackets. We add and subtract $100$ (which is half of 20, squared):

A = -\frac{1}{2}(x^2 - 20x + 100 - 100)

A = -\frac{1}{2}(x^2 - 20x + 100) + 50

The expression $(x^2 - 20x + 100)$ is a perfect square:

A = -\frac{1}{2}(x - 10)^2 + 50

This is now in vertex form, $a(x - h)^2 + k$ , where the vertex is at $(h, k)$ .

Since the coefficient of the squared term is negative ( $-\frac{1}{2}$ ), the parabola opens downward, meaning the vertex represents a maximum point.

Answer: The maximum area is 50 m², which occurs when $x = 10$ metres.

Interpretation: The optimal dimensions are a length of 10 metres (parallel to the existing fence) and a width of 5 metres, giving a total area of 50 square metres.

chatImportant

Remember: When completing the square, the sign of the leading coefficient tells us whether we have a maximum or minimum:

Negative coefficient → parabola opens downward → maximum value
Positive coefficient → parabola opens upward → minimum value

Modelling projectile motion

Quadratic functions are ideal for modelling the path of objects moving through the air under gravity. The trajectory of a projectile follows a parabolic path, which can be described by a quadratic equation of the form $y = ax^2 + bx + c$ .

lightbulbExample

Worked Example: Cricket Ball Trajectory

Problem: A fielder throws a cricket ball. It leaves his hand at a height of 2 metres above the ground. The wicketkeeper catches it 60 metres away, also at a height of 2 metres. After travelling 25 metres horizontally, the ball reaches a height of 15 metres. The path follows a parabola with equation $y = ax^2 + bx + c$ .

Part a: Find the values of $a$ , $b$ , and $c$

We have three points on the parabola: $(0, 2)$ , $(25, 15)$ , and $(60, 2)$ .

Each point gives us an equation when we substitute into $y = ax^2 + bx + c$ :

From point $(0, 2)$ :

2 = a(0)^2 + b(0) + c

2 = c \quad \text{... equation (1)}

From point $(25, 15)$ :

15 = a(25)^2 + b(25) + c

15 = 625a + 25b + c \quad \text{... equation (2)}

From point $(60, 2)$ :

2 = a(60)^2 + b(60) + c

2 = 3600a + 60b + c \quad \text{... equation (3)}

Now we have a system of three equations with three unknowns. We can substitute equation (1) into equations (2) and (3):

Substituting $c = 2$ into equation (2):

15 = 625a + 25b + 2

13 = 625a + 25b \quad \text{... equation (2')}

Substituting $c = 2$ into equation (3):

2 = 3600a + 60b + 2

0 = 3600a + 60b \quad \text{... equation (3')}

We can simplify equation (3') by dividing both sides by 60:

0 = 60a + b \quad \text{... equation (3')}

Rearranging: $b = -60a$

Now multiply equation (3') by 25:

0 = 1500a + 25b

Subtract this from equation (2'):

13 = 625a + 25b

0 = 1500a + 25b

Subtracting:

13 = -875a

Therefore:

a = -\frac{13}{875}

Substituting back:

b = -60 \times \left(-\frac{13}{875}\right) = \frac{780}{875} = \frac{156}{175}

Answer: The path of the ball has equation $y = -\frac{13}{875}x^2 + \frac{156}{175}x + 2$

Part b: Find the maximum height of the ball

For a quadratic in the form $y = ax^2 + bx + c$ , the maximum occurs at $x = -\frac{b}{2a}$ .

x = -\frac{\frac{156}{175}}{2 \times (-\frac{13}{875})} = -\frac{\frac{156}{175}}{-\frac{26}{875}} = 30

Substituting $x = 30$ into the equation:

y = -\frac{13}{875}(30)^2 + \frac{156}{175}(30) + 2

y = -\frac{13}{875}(900) + \frac{4680}{175} + 2

y = -\frac{11700}{875} + \frac{4680}{175} + 2

y = \frac{538}{35} \text{ metres}

Answer: The maximum height is $\frac{538}{35} \text{ metres}$ (approximately 15.4 metres).

Part c: Find the height when the ball is 5 metres before the wicketkeeper

The wicketkeeper is at $x = 60$ , so 5 metres before is at $x = 55$ .

Substituting $x = 55$ into the equation:

y = -\frac{13}{875}(55)^2 + \frac{156}{175}(55) + 2

y = \frac{213}{35} \text{ metres}

Answer: The height of the ball is $\frac{213}{35}$ metres (approximately 6.1 metres).

infoNote

When working with projectile motion, remember that three data points are needed to determine a unique quadratic equation. This is because we have three unknown coefficients ( $a$ , $b$ , and $c$ ) in the general form $y = ax^2 + bx + c$ .

Key techniques for quadratic models

Understanding these core techniques will help you tackle any quadratic modelling problem effectively.

Completing the square

This technique converts a quadratic from standard form to vertex form:

Standard form: $y = ax^2 + bx + c$
Vertex form: $y = a(x - h)^2 + k$

The vertex $(h, k)$ gives us the maximum (if $a < 0$ ) or minimum (if $a > 0$ ) value directly. This is why completing the square is so powerful for optimization problems—it reveals the optimal value immediately.

Using simultaneous equations

When you have multiple data points, follow this systematic approach:

Substitute each point into the general quadratic equation
Create a system of equations
Solve systematically by substitution or elimination
Check your answer makes sense in the context

chatImportant

Always work carefully through each step of solving simultaneous equations. Common mistakes include:

Forgetting to substitute known values into all equations
Sign errors when rearranging equations
Arithmetic errors when simplifying fractions

Interpreting results

Always remember to:

State what your variables represent
Include appropriate units
Explain what the mathematical answer means in the real-world context
Check if your answer is reasonable

The mathematical solution is only valuable when you can interpret its meaning in the original problem context.

bookmarkSummary

Key Points to Remember:

Quadratic models are powerful tools for solving optimization and trajectory problems in real-world contexts
Completing the square converts a quadratic to vertex form, making it easy to identify maximum or minimum values
A negative leading coefficient indicates a maximum (parabola opens downward), while a positive leading coefficient indicates a minimum (parabola opens upward)
Three points are needed to determine a unique quadratic equation—this creates three simultaneous equations to solve
Always interpret your answer in the context of the problem, including appropriate units and explanations

Quadratic Models (VCE SSCE Mathematical Methods): Revision Notes