grade 9 Applications of Quadratic Functions

Applications of Quadratic Functions

A quadratic is not just a graph; it is also how you find a maximum or a minimum. 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. Slide the domain ends and the fence width and the maximum and minimum appear.

Max & Min on a Domain

Domain start-1

Domain end4

📊 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. This graph is fixed as y=(x−1)²−2. On the default domain −1≤x≤4 the max is 7 and the min is −2. Intersections and finding the rule stay in the written boxes

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

← Previous

Graphs of Quadratic Functions

Trigonometric Ratios

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