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CIE A-Level Maths Study Notes

1.1.1 Completing the Square

Completing the square is an essential technique in quadratic algebra for transforming quadratic polynomials of the form ax2+bx+cax^2 + bx + c into vertex form. This method not only aids in locating the vertex of the parabola but also in sketching its graph and solving optimisation problems.

Understanding the Process

Completing the square rewrites a quadratic equation from y=ax2+bx+cy = ax^2 + bx + c to y=p(x+q)2+ry = p(x + q)^2 + r, which is vital for identifying the vertex of the parabola.

Key Concepts:

  • The vertex of the parabola is given by (q,r)(-q, r).
  • The coefficient 'a' affects the parabola's width and direction.
  • This technique is useful for solving quadratic equations and graphing quadratic functions.

Sketching the Graph

After transforming the quadratic equation into the vertex form, we can sketch its graph. The vertex form gives a clear picture of the parabola's shape and position.

Steps to Sketch the Graph:

1. Identify the Vertex:(q,r)(-q, r) give the vertex. This point is the peak or the lowest point of the parabola, depending on the direction of opening

2. Determine the Direction:(p(x+q)2+r)(p(x + q)^2 + r) indicates the direction the parabola opens. If 'p' is positive, the parabola opens upwards; if negative, it opens downwards.

3. Plot Key Points: Plot the vertex and a few points on either side of the vertex. Use the symmetry of the parabola about the vertex.

4. Draw the Parabola: Join these points with a smooth curve to represent the parabola.

graph of parabola

Image courtesy of Studiosguy

Examples: Transforming, Locating the Vertex, and Sketching the Graph

Example 1:

Express (3x2+9x+5)(3x^2 + 9x + 5) in complete square form, find its vertex, and sketch the graph.

Solution:

3x2+9x+5=3(x2+3x)+53x^2 + 9x + 5 = 3(x^2 + 3x) + 5 =3(x+32)2(32)2+5= 3 \left(x + \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 5=3(32)274= 3 \left(\frac{3}{2}\right)^2 - \frac{7}{4}

Vertex: (32,74)\left(-\frac{3}{2}, -\frac{7}{4}\right)

Graph Sketching:

1. Plot the vertex (32,74)\left(-\frac{3}{2}, -\frac{7}{4}\right).

2. Since a=3a = 3, the parabola opens upwards and is narrower than a standard parabola y=x2y = x^2.

3. Sketch the parabola passing through the vertex.

parabola in vertex form

Example 2:

Convert 2x24x+12x^2 - 4x + 1 to vertex form, find the vertex, and sketch the graph.

Solution:

2x24x+1=2(x22x)+12x^2 - 4x + 1 = 2(x^2 - 2x) + 1=2((x1)212)+1= 2\left(\left(x - 1\right)^2 - 1^2\right) + 1=2(x1)21= 2\left(x - 1\right)^2 - 1

Vertex: (1,1)(1, -1)

Graph Sketching:

1. Plot the vertex (1,1)(1, -1).

2. With a=2a = 2, the parabola opens upwards and is narrower.

3. Draw the parabola through the vertex.

parabola in vertex form

Example 3:

Transform x2+6x+8x^2 + 6x + 8 into vertex form, determine the vertex, and sketch the graph.

Solution:

x2+6x+8=(x2+6x)+8x^2 + 6x + 8 = (x^2 + 6x) + 8=((x+3)232)+8= \left(\left(x + 3\right)^2 - 3^2\right) + 8=(x+3)21= \left(x + 3\right)^2 - 1

Vertex: (3,1)(-3, -1)

Graph Sketching:

1. Mark the vertex (3,1)(-3, -1).

2. As a=1a = 1, the parabola is standard in width and opens upwards.

3. Sketch the parabola passing through the vertex.

parabola in vertex form
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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