Applications of Quadratic Functions

Quadratic functions do more than draw graphs; they help you find maximum and minimum values. A parabola peaks or bottoms out at its vertex, so problems like maximizing an area become simple. With a fence of fixed length, one quadratic shows exactly when the rectangle has the greatest area. Here you can slide the domain ends and change the fence width to watch where the maximum and minimum appear.

Max & Min on a Domain

Start (lo)-1

End (hi)4

📊 Strategy for Max/Min

①If the vertex is inside the domain → vertex value is max or min

②If outside → compare the two endpoints

③Candidates: vertex, left endpoint, right endpoint

Maximum Area — Fence Problem

Width x (m)5

🏠 Solving the Fence Problem

①Perimeter = 20m → width x, height (10−x)

②Area S = x(10−x) = −x²+10x

③Maximum at the vertex of the parabola

④x = 5 → S = 25m² (a square!)

Quadratic Functions ↔ Quadratic Equations

Linking Function and Equation

y = ax²+bx+c, set y=0 → ax²+bx+c = 0

x-intercepts of the graph = roots of the equation

🔗 The Connection

①Where the graph meets the x-axis → roots of the equation

②Discriminant D tells number of intersections

③Two graphs meeting → system of equations

Finding the Quadratic Equation

Vertex + One Point

Vertex (p,q) + point (x₁,y₁) → plug into y=a(x−p)²+q

Once vertex is known, only a remains

Two x-intercepts + One Point

Intercepts α, β → y=a(x−α)(x−β); plug a point

When roots are known, use factored form

📝 Strategy for Finding the Equation

①Vertex given → use vertex form

②x-intercepts given → use factored form

③Three points given → use general form y=ax²+bx+c with a system

Work It Out

Example 1

Find the minimum value of y=x²−4x+7.

Complete the square: x²−4x+7 = (x−2)²+3.

y = (x−2)²+3

Since a>0, the minimum is at the vertex: 3.

min = 3 (at x=2)

▸ 3

When a>0, the vertex gives the minimum value.

Example 2

Find the maximum value of y=−(x−1)²+5.

The vertex is (1, 5) and a=−1<0.

y = −(x−1)²+5

Since a<0, the maximum is at the vertex: 5.

max = 5 (at x=1)

▸ 5

When a<0, the vertex gives the maximum value.

Exam Wrap-up

Application Core

Max/Min → vertex, intersections → discriminant, find equation → match conditions

Three core application types

G9 school-exam type

What is the minimum value of y=2x²−8x+11?

①1

②3

③5

④8

⑤11

▸ ② 3

Complete the square: 2x²−8x+11 = 2(x−2)²+3.

y = 2(x−2)²+3

Since a=2>0, the minimum at x=2 is 3.

min = 3

🎯 Exam Points

①Max/Min lives at the vertex (min if a>0, max if a<0)

②With a restricted domain, also check both endpoints

③Maximize area/distance → model as a quadratic → vertex

④x-intercepts = roots of the related equation

⑤Pick the form that matches the given conditions

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Quadratic Function Graph

Trigonometric Ratio

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