Completing the Square and Turning Points Revision Notes for VCE SSCE Mathematical Methods

Completing the Square and Turning Points

What is completing the square?

Completing the square is a powerful algebraic technique that allows us to rewrite a quadratic function from its polynomial form into its turning point form. This transformation makes it much easier to identify key features of a parabola, particularly its vertex (turning point).

Polynomial form: $y = ax^2 + bx + c$

Turning point form: $y = a(x - h)^2 + k$

The vertex of the parabola is at the point $(h, k)$ .

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The turning point form immediately reveals the vertex of the parabola, making it invaluable for graphing and understanding quadratic functions. This is why completing the square is such an important technique in algebra.

The completing the square process

Understanding perfect squares

First, we need to recognise the structure of a perfect square. When we expand $(x + a)^2$ , we get:

(x + a)^2 = x^2 + 2ax + a^2

Notice that the constant term $a^2$ is the square of half the coefficient of the middle term ( $2a$ ). This pattern is key to completing the square.

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Key Pattern Recognition:

In the expansion $(x + a)^2 = x^2 + 2ax + a^2$ , the relationship between the middle coefficient and the constant term is crucial:

Middle coefficient: $2a$
Half of middle coefficient: $a$
Constant term: $a^2$ (the square of half the middle coefficient)

This pattern is the foundation of the completing the square technique.

When the coefficient of $x^2$ is 1

Let's work through an example: $x^2 + 2x - 3$

This expression is not a perfect square, but we can make it one by carefully adding and subtracting a new term.

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Worked Example: Completing the Square with $a = 1$

Complete the square for $x^2 + 2x - 3$

Step 1: Find half the coefficient of $x$ and square it

Half of $2$ is $1$ , and $1^2 = 1$

Step 2: Add and subtract this value

x^2 + 2x - 3 = (x^2 + 2x + 1) - 1 - 3

Step 3: Recognise the perfect square and simplify

So the quadratic $y = x^2 + 2x - 3$ can be written as $y = (x + 1)^2 - 4$ , revealing that the vertex is at (-1, -4).

When the coefficient of $x^2$ is not 1

If the coefficient of $x^2$ is not $1$ , we must first factor it out before completing the square.

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Worked Example: Completing the Square with $a \neq 1$

Express $y = -2x^2 + 6x - 8$ in turning point form.

Solution:

Factor out $-2$ from the first two terms:

y = -2(x^2 - 3x) - 8

Now complete the square inside the brackets. Half of $-3$ is $-\frac{3}{2}$ , and $\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$

y = -2\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) - 8

Therefore, the vertex is at $\left(\frac{3}{2}, -\frac{7}{2}\right)$ and the axis of symmetry is $x = \frac{3}{2}$ .

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Common Mistake Alert:

When factoring out the coefficient of $x^2$ , remember to only factor it from the $x^2$ and $x$ terms, NOT from the constant term. Many students incorrectly factor it from all three terms, which leads to errors in the final answer.

Geometric representation

The process of completing the square can be visualised geometrically. For $x^2 + 2x$ , we can think of this as the area of rectangles that can be rearranged to form a square with a small missing piece.

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Visual Understanding:

The diagrams show how $x^2 + 2x + 1 = (x + 1)^2$ by rearranging rectangular areas to form a complete square. The red square represents the unit area we need to add to complete the square.

This geometric interpretation helps explain why we "add and subtract" the same value - we're literally completing a square shape by adding the missing piece, then compensating by subtracting it.

Using technology

Modern calculators can perform completing the square operations automatically.

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Calculator Tip:

On the TI-Nspire calculator, use menu > Algebra > Complete the Square to rearrange expressions like $x^2 - 5x + 2$ into completed square form. While technology can help verify your work, it's essential to understand the process manually for exams and deeper comprehension.

Solving equations by completing the square

Completing the square is not just useful for graphing - it's also a method for solving quadratic equations. This technique is particularly valuable when factorising is difficult or impossible.

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Worked Example: Solving by Completing the Square

Solve $2x^2 - 3x - 1 = 0$ by completing the square.

Solution:

First, divide both sides by $2$ to make the coefficient of $x^2$ equal to $1$ :

x^2 - \frac{3}{2}x - \frac{1}{2} = 0

Half of $-\frac{3}{2}$ is $-\frac{3}{4}$ , so we add and subtract $\left(-\frac{3}{4}\right)^2 = \frac{9}{16}$ :

x^2 - \frac{3}{2}x + \frac{9}{16} - \frac{9}{16} - \frac{1}{2} = 0

\left(x - \frac{3}{4}\right)^2 = \frac{9}{16} + \frac{1}{2}

\left(x - \frac{3}{4}\right)^2 = \frac{17}{16}

Take the square root of both sides:

x - \frac{3}{4} = \pm\frac{\sqrt{17}}{4}

Therefore:

x = \frac{3 \pm \sqrt{17}}{4}

The axis of symmetry formula

There's a shortcut formula for finding the axis of symmetry of any parabola without completing the square.

For a quadratic function $y = ax^2 + bx + c$ , the axis of symmetry has equation:

x = -\frac{b}{2a}

This gives us the $x$ -coordinate of the turning point directly. We can then substitute this value back into the original equation to find the $y$ -coordinate.

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The Axis of Symmetry Shortcut:

The formula $x = -\frac{b}{2a}$ comes directly from the completing the square process. It saves time when you only need the vertex coordinates and don't need the full turning point form of the equation.

Remember: This formula gives you the $x$ -coordinate of the vertex, but you still need to substitute back to find the $y$ -coordinate!

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Worked Example: Using the Axis of Symmetry Formula

Find the turning point of $y = -2x^2 + 12x - 7$ and express it in turning point form.

Solution:

Here, $a = -2$ and $b = 12$ , so:

x = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3

When $x = 3$ :

y = -2(3)^2 + 12(3) - 7 = -18 + 36 - 7 = 11

The turning point is (3, 11).

In turning point form: $y = -2(x - 3)^2 + 11$

Sketching graphs using completed square form

Once a quadratic is in turning point form, sketching its graph becomes straightforward. The completed square form reveals all the key features needed for an accurate sketch.

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Worked Example: Sketching from Completed Square Form

Sketch the graph of $y = -2x^2 + 6x - 8$ using the completed square form found earlier: $y = -2\left(x - \frac{3}{2}\right)^2 - \frac{7}{2}$

Solution:

From the turning point form, we can identify:

Vertex: $\left(\frac{3}{2}, -\frac{7}{2}\right)$
The coefficient $a = -2$ is negative, so the parabola opens downward
The $y$ -intercept occurs when $x = 0$ : $y = -8$
Since the maximum value is $-\frac{7}{2} < 0$ , there are no x-intercepts

The graph is a downward-opening parabola with its maximum point at $\left(\frac{3}{2}, -\frac{7}{2}\right)$ .

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Key Points to Remember:

Completing the square converts polynomial form to turning point form: $y = a(x - h)^2 + k$
The key step is to add and subtract $\left(\frac{b}{2}\right)^2$ where $b$ is the coefficient of $x$
If $a \neq 1$ , factor out a first before completing the square (only from the $x^2$ and $x$ terms)
The axis of symmetry formula provides a shortcut: $x = -\frac{b}{2a}$
Completing the square can be used to solve quadratic equations, even when factorising is difficult
The vertex (turning point) is at $(h, k)$ when the equation is in the form $y = a(x - h)^2 + k$
In turning point form, the sign of $a$ tells you whether the parabola opens upward ( $a > 0$ ) or downward ( $a < 0$ )

Completing the Square and Turning Points (VCE SSCE Mathematical Methods): Revision Notes