Look at the coefficient of \(x\). It is 10. We halve it to get 5 and then square that. So 5² = 25.

We are going to add this 25 into the expression but then immediately subtract it out again. That might seem like a strange thing to do, but it will give us the perfect square that we need.

$$ \begin{eqnarray} &\ & x^2+10x+28 \\[6pt] &=&\ x^2+10x+25-25+28 \\[6pt] &=&\ (\underbrace{x^2+10x+25}_{\textsf{a perfect square}})-25+28 \\[6pt] &=&\ (x+5)^2+3 \end{eqnarray} $$

The coefficient of \(x\) is 6. We halve this to get 3 and then square that: 3² = 9.

We add this 9 into the expression but then immediately subtract it out again. The first three terms will be our perfect square.

$$ \begin{eqnarray} &\ & x^2+6x+1 \\[6pt] &=&\ x^2+6x+9-9-1 \\[6pt] &=&\ (\underbrace{x^2+6x+9}_{\textsf{a perfect square}})-9+1 \\[6pt] &=&\ (x+3)^2-8 \end{eqnarray} $$

The coefficient of \(x\) is -2. We halve this to get -1 and then square that: (-1)² = 1.

We add this 1 into the expression but then immediately subtract it out again.

$$ \begin{eqnarray} &\ & x^2-2x+5 \\[6pt] &=&\ x^2-2x+1-1+5 \\[6pt] &=&\ (\underbrace{x^2-2x+1}_{\textsf{a perfect square}})-1+5 \\[6pt] &=&\ (x-1)^2+4 \end{eqnarray} $$

The coefficient of \(x\) is -4. We halve this to get -2 and then square that: (-2)² = 4.

We add this 4 into the expression but then immediately subtract it out again.

$$ \begin{eqnarray} &\ & x^2-4x-9 \\[6pt] &=&\ x^2-4x+4-4-9 \\[6pt] &=&\ (\underbrace{x^2-4x+4}_{\textsf{a perfect square}})-4-9 \\[6pt] &=&\ (x-2)^2-13 \end{eqnarray} $$

This example is trickier because the coefficient of \(x\) is odd. One like this hasn't yet appeared in a National 5 exam paper, but the course specification says that the coefficient can be any integer, so be prepared!

We halve the \(5\) to get \(\frac{5}{2}\) and then square that: \( (\frac{5}{2})^2=\frac{25}{4}. \)

So we add the \(\frac{25}{4}\) into the expression and then immediately subtract it out again.

$$ \begin{eqnarray} x^2+5x-1 &=&\ x^2+5x+\small\frac{25}{4}\normalsize-\small\frac{25}{4}\normalsize-1\\[6pt] &=&\ \large(\normalsize\underbrace{x^2+5x+\small\frac{25}{4}\normalsize}_{\textsf{a perfect square}}\large)\normalsize-\small\frac{25}{4}\normalsize -1\\[6pt] &=&\ \left(x+\small\frac{5}{2}\normalsize\right)^2-\small\frac{25}{4}\normalsize-\small\frac{4}{4}\normalsize\\[6pt] &=&\ \left(x+\small\frac{5}{2}\normalsize\right)^2-\small\frac{29}{4}\normalsize \end{eqnarray} $$

We halve the \(-9\) to get \(-\frac{9}{2}\) and then square that: \( (-\frac{9}{2})^2=\frac{81}{4}. \)

So we add the \(\frac{81}{4}\) into the expression and then immediately subtract it out again.

$$ \begin{eqnarray} x^2-9x+3 &=&\ x^2-9x+\small\frac{81}{4}\normalsize-\small\frac{81}{4}\normalsize+3\\[6pt] &=&\ \large(\normalsize\underbrace{x^2-9x+\small\frac{81}{4}\normalsize}_{\textsf{a perfect square}}\large)\normalsize-\small\frac{81}{4}\normalsize +3\\[6pt] &=&\ \left(x-\small\frac{9}{2}\normalsize\right)^2-\small\frac{81}{4}\normalsize+\small\frac{12}{4}\normalsize\\[6pt] &=&\ \left(x-\small\frac{9}{2}\normalsize\right)^2-\small\frac{69}{4}\normalsize \end{eqnarray} $$

The coefficient of \(x\) is \(-8\) so we first calculate \((\frac{-8}{2})^2=(-4)^2=16\).

Then we complete the square:

$$ \begin{eqnarray} x^2-8x+5 &=&\ x^2-8x+16-16+5\\[6pt] &=&\ (x^2-8x+16)-16+5\\[6pt] &=&\ (x-4)^2-11 \end{eqnarray} $$

Now it is easy to state the coordinates of the turning point, because perfect squares like \((x-4)^2\) can never be negative. The smallest they can ever be is \(0\). In this example, \((x-4)^2=0\) when \(x=4\).

Finally, when \(x=4\), we can just read off from the formula that \(y=-11\). So the turning point is \((4,-11)\).

$$ \begin{eqnarray} &\ & x^2-14x+44 \\[6pt] &=&\ x^2-14x+49-49+44 \\[6pt] &=&\ (x^2-14x+49)-49+44 \\[6pt] &=&\ (x-7)^2-5 \end{eqnarray} $$

$$ \begin{eqnarray} &\ & x^2+8x-7 \\[6pt] &=&\ x^2+8x+16-16-7 \\[6pt] &=&\ (x^2+8x+16)-16-7 \\[6pt] &=&\ (x+4)^2-23 \end{eqnarray} $$

$$ \begin{eqnarray} &\ & x^2+10x-15 \\[6pt] &=&\ x^2+10x+25-25-15 \\[6pt] &=&\ (x^2+10x+25)-25-15 \\[6pt] &=&\ (x+5)^2-40 \end{eqnarray} $$

(a)  This part is just a straightforward 2-mark completing the square question. $$ \begin{eqnarray} &\ & x^2+8x+15 \\[6pt] &=&\ x^2+8x+16-16+15 \\[6pt] &=&\ (x^2+8x+16)-16+15 \\[6pt] &=&\ (x+4)^2-1 \end{eqnarray} $$

(b)  This part requires us to use the fact that the turning point of the graph of \(y=\left(x+a\right)^2+b\) is \(\left(-a,b\right).\) So the turning point is \(\left(-4,-1\right).\)