Quadratic functions do more than draw graphs: they are a tool for finding when a quantity is largest or smallest.
A parabola is highest or lowest at its vertex, so if a > 0 the y-coordinate of the vertex is the minimum, and if a < 0 it is the maximum.
When x is restricted to a range, though, the vertex may lie outside it, so the values at both ends of the range must be compared as well.
Finding the shape of rectangle with the largest area that 20 m of fencing can enclose comes down to a single vertex once the area is written as a quadratic function of the width.
Maximum and minimum over a restricted range
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Where the maximum and minimum of y = (x − 1)² − 2 occur as the range of x changes
-1
4
📊 How to find the maximum and minimum
①If the vertex lies inside the range of x, the function has its minimum (when a > 0) or maximum (when a < 0) there.
②If the vertex lies outside the range, compare the values at both ends.
③The candidates are the vertex and the two end points.
④For the starting range −1 ≤ x ≤ 4 in the picture, the minimum −2 is at the vertex (1, −2) and the maximum 7 is at the right end x = 4.
The fence with the largest area
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The width x and area S of a rectangle with perimeter 20 m, and the graph of S = x(10 − x)
5
🏠 Solving the fence problem
①A perimeter of 20 m means width plus height is 10 m, so if the width is x the height is 10 − x.
②The area is S = x(10 − x) = −x² + 10x, a quadratic function of x.
③S = −(x − 5)² + 25, so the maximum area, 25 m², occurs at x = 5.
④The area is largest for the 5 m by 5 m square; the vertex of the graph on the right marks exactly that moment.
Quadratic functions and quadratic equations
Linking function and equation
setting y = 0 in y = ax² + bx + c gives ax² + bx + c = 0
The x-coordinates where the graph meets the x-axis are the roots of the quadratic equation.
🔗 Three links
①The x-coordinates where the graph of a quadratic function meets the x-axis are the roots of the quadratic equation.
②The sign of the discriminant D tells how many points the graph shares with the x-axis.
③The points where two graphs meet are the solutions of the system formed by their equations.
Finding the equation of a quadratic function
Vertex and one other point
vertex (p, q) and a point (x₁, y₁) → substitute into y = a(x − p)² + q
Once the vertex is known, only a remains to be found.
Two x-intercepts and one other point
meets the x-axis at x = α, β → substitute a point into y = a(x − α)(x − β)
When the x-intercepts are known, use the factorised form.
📝 Strategy for finding the equation
①If the vertex is given, use vertex form.
②If the two points on the x-axis are given, use the factorised form.
③If three points on the graph are given, substitute them into y = ax² + bx + c and solve the system.
Worked examples
Example 1
Find the minimum value of y=x²−4x+7.
1
Complete the square: x² − 4x + 7 = (x − 2)² + 3.
y = (x−2)²+3
2
a > 0, so the minimum is at the vertex, and its value is 3.
min = 3 (at x=2)
▸ 3
When a > 0, the minimum is at the vertex.
Example 2
Find the maximum value of y=−(x−1)²+5.
1
The vertex is (1, 5) and a = −1 < 0.
y = −(x−1)²+5
2
a < 0, so the maximum is at the vertex, and its value is 5.
max = 5 (at x=1)
▸ 5
When a < 0, the maximum is at the vertex.
Exam checklist
Key ideas in applications
maximum and minimum → vertex and end points · meeting the x-axis → discriminant · finding the equation → conditions in the right form
Most application problems are one of these three types.
G9 school-exam type
What is the minimum value of y=2x²−8x+11?
①1
②3
③5
④8
⑤11
▸ ② 3
1
Complete the square: 2x²−8x+11 = 2(x−2)²+3.
y = 2(x−2)²+3
2
Since a=2>0, the minimum at x=2 is 3.
min = 3
🎯 Exam points
①The maximum or minimum occurs at the vertex: a minimum when a > 0, a maximum when a < 0.
②If x is restricted, always compare the values at the end points too.
③For the largest area or distance, express it as a quadratic function and find the vertex.
④The x-coordinates where the graph meets the x-axis are the roots of the quadratic equation.
⑤When finding an equation, choose the form that fits the given conditions.