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A quadratic passes through (5,0), has symmetry axis x=2 and minimum −9. Find a,b,c.

Read the idea, work independently, then explain what changed.

TOPIC 01

2022 JM01

The minimum specifies the vertex ordinate, not the constant term c.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

A quadratic passes through (5,0), has symmetry axis x=2 and minimum −9. Find a,b,c.

f(x)=ax2+bx+cf(x)=ax^2+bx+c

Official paper · jm01-2022 · II.1(a) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Start with vertex form.
Hint 2
Use the point (5,0) to find the scale.
Worked solution
  1. Write f(x)=a(x−2)²−9 with a>0.

    0=9a−9  ⟹  a=10=9a-9\implies a=1
  2. Expand the result.

    f(x)=(x−2)2−9=x2−4x−5f(x)=(x-2)^2-9=x^2-4x-5

a=1, b=−4, c=−5.

Checks and common pitfalls: The minimum specifies the vertex ordinate, not the constant term c.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Write f(x)=a(x−2)²−9 with a>0.
    0=9a−9  ⟹  a=10=9a-9\implies a=1
  • Expand the result.
    f(x)=(x−2)2−9=x2−4x−5f(x)=(x-2)^2-9=x^2-4x-5

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Curriculum and source notes ↗