Difficulty: Intermediate | Prerequisites: familiarity with quadratic functions, standard form, and the concept of a parabola's vertex.
Quadratic functions crop up whenever a real-world quantity depends on a squared variable, and this section is about learning to recognise those situations and set up the algebra. You have already graphed parabolas and found vertices; now the task is to translate a word problem into a quadratic model and then use the vertex to answer an optimisation question ("what value of x makes this the largest or smallest?"). This is the bridge between abstract algebra and applied problem-solving, and it will reappear in calculus when you study optimisation with derivatives. If you are comfortable with the vertex formula x = -b/(2a) and can identify what a, b, and c represent in a given context, you are ready for this material.
Many real-world problems (maximising area, minimising cost, finding peak height) boil down to quadratic functions. Write the quantity you want to optimise as a function of one variable, identify a, b, and c, then plug into x = -b/(2a) to find the input that gives the maximum or minimum. If the parabola opens downward (a < 0), the vertex is a maximum; if it opens upward (a > 0), the vertex is a minimum.
Quadratic function
A function of the form f(x) = ax² + bx + c, where a ≠ 0. The graph is a parabola. In simple terms, any expression where the highest power of x is x-squared.
Standard form
The arrangement f(x) = ax² + bx + c, which makes it straightforward to read off a, b, and c for the vertex formula. Think of it as the "ready to plug in" form.
Vertex
The highest or lowest point on a parabola, located at x = -b/(2a). In simple terms, the turning point where the curve changes direction.
Axis of symmetry
The vertical line x = -b/(2a) that passes through the vertex. The parabola is a mirror image on either side of this line.
Optimisation (optimization)
The process of finding the maximum or minimum value of a function. In this context, it means finding the vertex of a quadratic to answer "what is the best possible value?"
Constraint equation
An equation that ties two or more variables together based on a fixed condition (e.g. a fixed perimeter or a fixed total cost). Think of it as the rule that limits your choices before you optimise.
Leading coefficient (a)
The coefficient of x² in standard form. Its sign tells you the direction the parabola opens: positive means it opens upward (minimum), negative means it opens downward (maximum).
Decide what quantity you want to maximise or minimise (area, cost, profit, height).
Identify the variables. Typically the problem gives you two related quantities and one fixed constraint.
Translate the fixed condition into algebra. For example, a 200-metre perimeter becomes 2x + πy = 200.
Solve the constraint for one variable in terms of the other so you can substitute.
Substitute the constraint into the objective. This should reduce everything to a single-variable quadratic in standard form.
Rearrange into f(x) = ax² + bx + c so you can identify a, b, and c clearly.
Compute x = -b/(2a). This is the x-value that produces the optimum.
Substitute back into the original function to find the optimum value (the y-coordinate of the vertex).
Check the sign of a to confirm whether you found a maximum (a < 0) or a minimum (a > 0).
State the answer in the units the problem asked for. Include both the optimal input and the optimal output if the question asks for dimensions or amounts.
Verify the answer makes physical sense (e.g. lengths should be positive).
Standard form of a quadratic
f(x) = ax² + bx + c
Vertex x-coordinate (the optimisation formula)
x = -b / (2a)
Vertex y-coordinate (the optimum value)
Substitute x = -b/(2a) back into f(x).
Perimeter of a rectangle
P = 2l + 2w
Circumference of a circle
C = 2πr
Area of a rectangle
A = l × w
These geometric formulas appear repeatedly in modelling problems because they supply the constraint equations.
Engineers use quadratic optimisation to design structures that enclose the most space with limited material, such as fencing a yard or laying out a running track. Businesses use the same maths to find the production level that minimises cost or the advertising spend that maximises profit.
Students often forget to check the sign of a before declaring a maximum or minimum. A positive a means the vertex is a minimum, not a maximum. Always check.
Plugging into x = -b/(2a) gives the x-value, not the optimum value itself. You still need to substitute back into the original function to find the actual maximum or minimum output.
When a problem involves a circle or semicircle, students sometimes use the diameter where the formula calls for the radius (or vice versa). Read the geometry carefully.
Students sometimes forget that the constraint equation must be solved for one variable before substituting. Jumping straight to the area or cost formula with two unknowns leaves you stuck.
⚠️ The vertex formula x = -b/(2a) is the single most-tested tool in this unit. Know it cold.
⚠️ Word problems on the unit test will require you to set up both a constraint and an objective, then combine them into a quadratic. Practice the translation from English to algebra.
⚠️ You may be asked to state whether the vertex is a max or min and justify your answer by referencing the sign of a.
⚠️ Problems often ask for the dimensions (both x and y), not just the optimum value. Remember to solve for the second variable after finding x.
True or false: If a > 0, the vertex of the parabola is a maximum. (Answer: False, it is a minimum.)
Fill in the blank: The x-coordinate of the vertex is found using x = ____. (Answer: -b/(2a))
True or false: To find the optimum value of the function, you substitute x = -b/(2a) back into f(x). (Answer: True.)
Fill in the blank: Before you can write a single-variable quadratic, you need to use the ____ equation to eliminate one variable. (Answer: constraint)
True or false: A quadratic model for profit with a negative leading coefficient has a maximum profit at its vertex. (Answer: True.)
Q: A farmer has 120 metres of fencing and wants to enclose a rectangular area against a barn wall (so only three sides need fencing). Write the area as a function of the width x, then find the dimensions that maximise the area.
A: The constraint is 2x + y = 120, so y = 120 - 2x. Area = x(120 - 2x) = -2x² + 120x. Using x = -b/(2a) = -120/(2 × -2) = 30, we get x = 30 m, y = 60 m. Maximum area = 1,800 m².
Q: Given f(x) = -0.5x² + 20x + 230, find the value of x that maximises f(x) and state the maximum value.
A: x = -20/(2 × -0.5) = -20/(-1) = 20. f(20) = -0.5(400) + 20(20) + 230 = -200 + 400 + 230 = 430. Maximum value is 430, occurring at x = 20.
Q: A ball is thrown upward with height given by h(t) = -16t² + 64t + 5. What is the maximum height?
A: t = -64/(2 × -16) = -64/(-32) = 2 seconds. h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69 feet.
Q: The cost function for a manufacturer is C(x) = 0.045x² - 110x + 100,000. How many units minimise cost?
A: x = -(-110)/(2 × 0.045) = 110/0.09 ≈ 1,222 units. The parabola opens upward (a > 0), so the vertex is a minimum.
This material connects directly to Section 2.2 on transformations of quadratic functions, since vertex form f(x) = a(x - h)² + k is another way to read off the optimum. In calculus, the vertex formula is replaced by setting the derivative equal to zero, but the underlying logic is identical. Constraint-based optimisation also reappears in linear programming and multivariable calculus (Lagrange multipliers).
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