Lesson library
A.5(E)·Algebra II·Conic SectionsSupporting

Completing the Square

Use completing the square to rewrite equations of parabolas and other conics.

Students complete the square with a nonunit leading coefficient to rewrite a parabola equation and verify that both forms describe the same graph.

Lesson plan and worked examples by TEKSForge · not a TEA-issued lesson. Videos by independent educators.

Archived course: TEKS code and readiness label from TEA's 2013 Algebra II key. Algebra II is not one of Texas's current required STAAR EOC assessments, and this lesson does not change the Biology and Algebra I Study Plans.

Learn the concept

Prepare the equation

Isolate y, then factor a from the quadratic and linear terms: y=a(x2+bax)+cy=a(x^2+\frac ba x)+c. This makes the coefficient of the squared term inside the parentheses equal to 1.

Create a perfect square

For x2+pxx^2+px, add (p/2)2(p/2)^2 inside. Since the outside multiplier is a, compensate by subtracting a(p/2)2a(p/2)^2 outside. The expression’s value must remain unchanged.

Read and verify vertex form

Rewrite the trinomial as a squared binomial to obtain y=a(x−h)2+ky=a(x-h)^2+k. The vertex is (h,k)(h,k). Expand the result to verify every coefficient, not just the vertex.

Worked example

Write 3x2−18x−y+10=03x^2-18x-y+10=0 in vertex form.

  1. Isolate y: y=3x2−18x+10y=3x^2-18x+10.
  2. Factor: y=3(x2−6x)+10y=3(x^2-6x)+10.
  3. Add 9 inside and subtract 27 outside: y=3(x2−6x+9)+10−27y=3(x^2-6x+9)+10-27.
  4. Factor the square: y=3(x−3)2−17y=3(x-3)^2-17. Expanding returns 3x2−18x+103x^2-18x+10.

y=3(x−3)2−17y=3(x-3)^2-17; vertex (3,−17)(3,-17).

Watch the lesson

Texas public-school, Texas educator-created, and TEKS-aligned lessons, hosted by their creators. Each card names its source.

Texas public school video

Algebra 2 Connecting Standard Form to Vertex Form

UT IPSI · Kid2Kid / Texas Education Agency

Connect completing the square to equivalent standard and vertex forms.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

  1. 1. Rewrite y=4x2+16x+3y=4x^2+16x+3 in vertex form.

    Show answer and reasoning

    y=4(x+2)2−13y=4(x+2)^2-13

    4(x2+4x+4)+3−16=4(x+2)2−134(x^2+4x+4)+3-16=4(x+2)^2-13.

  2. 2. Which equals y=2x2−16x+1y=2x^2-16x+1: 2(x−4)2−312(x-4)^2-31 or (2x−4)2−15(2x-4)^2-15?

    Show answer and reasoning

    y=2(x−4)2−31y=2(x-4)^2-31

    The first expands to 2x2−16x+12x^2-16x+1. The second expands to 4x2−16x+14x^2-16x+1, with the wrong leading coefficient.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Algebra II EOC · 2013 · Item 32 · 1 point

Which equation has the same graph as 2x2−4x−y+11=02x^2 - 4x - y + 11 = 0?

  1. y=(2x−1)2+9y = (2x - 1)^2 + 9
  2. y=(2x−1)2+13y = (2x - 1)^2 + 13
  3. y=2(x−1)2+9y = 2(x - 1)^2 + 9
  4. y=2(x−1)2+13y = 2(x - 1)^2 + 13
Show STAAR answer and explanation

y=2(x−1)2+9y = 2(x - 1)^2 + 9

Solve for y: y=2x2−4x+11=2(x2−2x)+11y=2x^2-4x+11=2(x^2-2x)+11. Complete the square: 2(x2−2x+1)+11−2=2(x−1)2+92(x^2-2x+1)+11-2=2(x-1)^2+9.

y=(2x−1)2+9y = (2x - 1)^2 + 9: (2x−1)2+9(2x-1)^2+9 expands to 4x2−4x+104x^2-4x+10; its leading coefficient is 4, not 2.

y=(2x−1)2+13y = (2x - 1)^2 + 13: (2x−1)2+13(2x-1)^2+13 expands to 4x2−4x+144x^2-4x+14.

y=2(x−1)2+13y = 2(x - 1)^2 + 13: Adding 2(1)=22(1)=2 inside the parentheses must be balanced by subtracting 2. This expands to 2x2−4x+152x^2-4x+15.

Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

  • Isolate y in a conic equation.
  • Complete the square while preserving equality.
  • Verify vertex form by expansion.

Instructional sequence

  1. 1

    Engage

    5 min

    Expand a perfect-square binomial and identify its middle term.

  2. 2

    Direct instruction

    10 min

    Isolate y and factor the leading coefficient from the x terms.

  3. 3

    Guided practice

    13 min

    Complete the square and balance the scaled added amount.

  4. 4

    Compare and verify

    15 min

    Expand proposed vertex forms to reject nonequivalent equations.

  5. 5

    Closure

    7 min

    Name the vertex and complete the exit ticket.

Misconceptions to address

“Add the square inside parentheses without balancing it.”

Teacher move: The outside coefficient multiplies the added amount; subtract that full amount outside.

“Move the outside coefficient inside without taking account of squaring.”

Teacher move: For example, 3(x−2)23(x-2)^2 has leading term 3x23x^2, not 9x29x^2.