Year 11 Methods (Unit 1 & 2) Quadratics

Completing The Square And Turning Points

Q: What does completing the square do?

It rewrites a x squared plus b x plus c as a times x minus h all squared plus k, which shows the turning point h comma k and the axis of symmetry x equals h directly.

Q: What number do I add and subtract?

Half the coefficient of x, then squared. Add it to form the square and subtract it to keep the value the same.

Q: How do I complete the square when a is not one?

Factor a out of the x squared and x terms first, complete the square inside the bracket, then expand the a back through.

Q: How do I read the turning point?

From y equals a times x minus h all squared plus k the turning point is h comma k: h makes the bracket zero and k is the constant.

Q: Is the turning point a maximum or a minimum?

A minimum if a is greater than zero, where the parabola opens up, and a maximum if a is less than zero, where the parabola opens down.

Q: Do I need calculus to find the turning point?

No. In Year 11 you read the turning point straight from completed-square form; differentiation is not required here.

20 practice questions 1 video lesson Theory + worked examples

Master completing the square for Queensland Year 11 Mathematical Methods (QCAA). Completing the square rewrites a quadratic in turning-point form, revealing its turning point at a glance.

You will learn to complete the square on monic and non-monic quadratics, read the turning point and axis of symmetry directly, and state the maximum or minimum value — no calculus needed, and a stepping stone to solving and sketching parabolas.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Completing The Square And Turning Points - Video - Finding Turning Points using Completing the Square Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), completing the square rewrites $y=ax^2+bx+c$ in the form $y=a(x-h)^2+k$. From that form you can read the turning point $(h,k)$, the axis of symmetry $x=h$, and the maximum or minimum value $k$ directly — without calculus.

Completing the square turns $x^2+bx$ into a perfect square by adding and subtracting $\left(\dfrac{b}{2}\right)^2$, giving $\left(x+\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^2$. For a non-monic quadratic, factor the leading coefficient out of the $x^2$ and $x$ terms first.

In the completed form $y=a(x-h)^2+k$, the turning point (vertex) is $(h,k)$ and the axis of symmetry is $x=h$. The turning point is a minimum when $a>0$ and a maximum when $a<0$; the value there is $k$.

Half the coefficient of \(x\), then square it. That single number \(\left(\dfrac{b}{2}\right)^2\) is what you add and subtract to build the perfect square.

$y=x^2+6x+5=(x+3)^2-4$: minimum turning point $(-3,-4)$, axis $x=-3$.

When $a<0$, $k$ is a maximum value — here the vertex $(6,32)$.

The completing-the-square identity and what it reveals:

\[x^2+bx=\left(x+\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^2\]

x^{2} + b x = {(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2}

\[y=a(x-h)^2+k \quad\Rightarrow\quad \text{vertex } (h,\,k),\ \text{value } k\]

y = a {(x - h)}^{2} + k

Minimum or maximum. \(k\) is the minimum value if \(a>0\) and the maximum value if \(a<0\); either way it occurs at \(x=h\).

How to complete the square

Factor out $a$ from the $x^2$ and $x$ terms (skip this if $a=1$).
Halve and square the coefficient of $x$ inside the bracket, adding and subtracting $\left(\dfrac{b}{2}\right)^2$ to form a perfect square.
Write $a(x-h)^2+k$ and read off the turning point $(h,k)$, the axis $x=h$, and whether $k$ is a minimum or maximum.

Example 1 — Monic completing the square

Write $y=x^2+6x+5$ in the form $y=(x-h)^2+k$ and state the turning point.

Solution

Half the coefficient of $x$ is $3$; square it to get $9$. Add and subtract $9$:

$y$

$=$

$x^2+6x+9-9+5$

Group the first three terms as a perfect square:

$y$	$=$	$(x+3)^2-9+5$
	$=$	$(x+3)^2-4$

$y=(x+3)^2-4$; turning point $(-3,-4)$, a minimum, axis $x=-3$.

(- 3, - 4)

Example 2 — Turning point and minimum value

Find the turning point, axis of symmetry and minimum value of $y=x^2-8x+11$.

Solution

Half of $-8$ is $-4$; its square is $16$. Add and subtract $16$:

$y$

$=$

$x^2-8x+16-16+11$

Form the perfect square and simplify:

$y$	$=$	$(x-4)^2-16+11$
	$=$	$(x-4)^2-5$

Turning point $(4,-5)$; axis $x=4$; minimum value $-5$.

(4, - 5)

Example 3 — Non-monic completing the square

Write $y=2x^2+12x+7$ in completed-square form and state the minimum value.

Solution

Factor $2$ from the $x^2$ and $x$ terms:

$y$

$=$

$2(x^2+6x)+7$

Inside the bracket, half of $6$ is $3$, squared is $9$; add and subtract $9$:

$y$	$=$	$2(x^2+6x+9-9)+7$
	$=$	$2\big((x+3)^2-9\big)+7$

Expand the $2$ back through and simplify:

$y$	$=$	$2(x+3)^2-18+7$
	$=$	$2(x+3)^2-11$

$y=2(x+3)^2-11$; turning point $(-3,-11)$; minimum value $-11$.

(- 3, - 11)

Example 4 — Maximum in a context

A small stall's daily profit is $P=-2x^2+24x-40$ dollars, where $x$ is the price rise in dollars. Find the price rise that maximises profit and the maximum profit.

Solution

Factor $-2$ from the $x^2$ and $x$ terms:

$P$

$=$

$-2(x^2-12x)-40$

Half of $-12$ is $-6$, squared is $36$; add and subtract $36$ inside:

$P$	$=$	$-2(x^2-12x+36-36)-40$
	$=$	$-2\big((x-6)^2-36\big)-40$

Expand the $-2$ back through:

$P$	$=$	$-2(x-6)^2+72-40$
	$=$	$-2(x-6)^2+32$

Since $a=-2<0$, the vertex is a maximum. The greatest value of $P$ occurs when the square is zero, at $x=6$.

A price rise of $\$6$ gives the maximum profit of $\$32$.

(6, 32)

Common pitfalls

Forgetting to subtract the square back. You must add $\left(\dfrac{b}{2}\right)^2$ and subtract it, so the expression is unchanged.

Not factoring $a$ first. For $2x^2+12x+7$, take the $2$ out of the $x$-terms before halving and squaring; halving $12$ directly gives the wrong result.

Sign of the turning point. $y=(x+3)^2-4$ has vertex $(-3,-4)$: the $x$-coordinate is the value that makes the bracket zero, so $-3$, not $+3$.

Mislabelling max and min. $k$ is a minimum only when $a>0$; if $a<0$ the parabola opens down and $k$ is a maximum.

Frequently asked questions

What does completing the square do?

It rewrites $ax^2+bx+c$ as $a(x-h)^2+k$, which shows the turning point $(h,k)$ and the axis of symmetry $x=h$ directly.

What number do I add and subtract?

Half the coefficient of $x$, then squared: $\left(\dfrac{b}{2}\right)^2$. Add it to form the square and subtract it to keep the value the same.

How do I complete the square when $a\ne1$?

Factor $a$ out of the $x^2$ and $x$ terms first, complete the square inside the bracket, then expand the $a$ back through.

How do I read the turning point?

From $y=a(x-h)^2+k$ the turning point is $(h,k)$: $h$ makes the bracket zero and $k$ is the constant.

Is the turning point a maximum or a minimum?

A minimum if $a>0$ (parabola opens up) and a maximum if $a<0$ (parabola opens down).

Do I need calculus to find the turning point?

No. In Year 11 you read the turning point straight from completed-square form; differentiation is not required here.

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Graphing Quadratics In Polynomial Form

\(y\)	\(=\)	\((x+3)^2-9+5\)
	\(=\)	\((x+3)^2-4\)

\(y\)	\(=\)	\((x-4)^2-16+11\)
	\(=\)	\((x-4)^2-5\)

\(y\)	\(=\)	\(2(x^2+6x+9-9)+7\)
	\(=\)	\(2\big((x+3)^2-9\big)+7\)

\(y\)	\(=\)	\(2(x+3)^2-18+7\)
	\(=\)	\(2(x+3)^2-11\)

\(P\)	\(=\)	\(-2(x^2-12x+36-36)-40\)
	\(=\)	\(-2\big((x-6)^2-36\big)-40\)

Completing The Square And Turning Points

📖 Theory

How to complete the square

Common pitfalls

Frequently asked questions

Theory