Finding the equation from a graph

← Previous3 of 4Next →

What this lesson covers

  • Roots suggest y = k(x − a)(x − b).
  • A vertex suggests y = k(x − h)² + c.
  • One extra point finds k.
Key idea

Choose the form that fits

  • If you can see the roots, use the factorised form.
  • If you can see the turning point, use the vertex form.
  • Either way, substitute one more point to find k.
Worked example

Roots 1 and 5, through (0, 10)

  • The roots go in the brackets.
  1. y = k(x − 1)(x − 5)
  2. 10 = k(−1)(−5) = 5kuse (0, 10)
  3. k = 2
  4. y = 2(x − 1)(x − 5)
Worked example

Vertex (2, 3), through (0, 11)

  • The vertex goes in the bracket and on the end.
  1. y = k(x − 2)² + 3
  2. 11 = 4k + 3
  3. k = 2
  4. y = 2(x − 2)² + 3
Worked example

Cubic with roots −1, 2 and 3, through (0, 12)

  • The same idea with a third bracket.
  1. y = k(x + 1)(x − 2)(x − 3)
  2. 12 = k(1)(−2)(−3) = 6k
  3. k = 2
  4. y = 2(x + 1)(x − 2)(x − 3)

Common mistakes

  • Leaving k out and assuming it is 1.
    Always test with another point on the graph.
  • Writing (x − 2) for a root at −2.
    A root at −2 gives the bracket (x + 2).

Your turn

  • 1. Find the parabola with roots −2 and 4 through (0, −24).
  • 2. Find the parabola with vertex (1, −3) through (3, −11).
Show answer

1. y = 3(x + 2)(x − 4) 2. y = −2(x − 1)² − 3

Summary

What to take away

  • Pick the form that matches what you can see.
  • One extra point always gives k.

More in AS 91257

Families of graphsNotesTransforming graphsNotesFinding the equation from a graphNotesModelling with the right graphNotes

Is your child on track?

Take the free placement test