Express $y = 2x^2 - 4x + 7$ in the form $a(x - b)^2 + c$, where $a$, $b$ and $c$ are constants. Hence, explain why the curve lies completely above the $x$-axis.

1. Factorise the coefficient of $x^2$ from the first two terms

$y = 2(x^2 - 2x) + 7$

2. Complete the square inside the bracket

$y = 2\left[ (x - 1)^2 - (-1)^2 \right] + 7$
$y = 2\left[ (x - 1)^2 - 1 \right] + 7$

3. Expand and simplify to reach the final form

$y = 2(x - 1)^2 - 2 + 7$
$y = 2(x - 1)^2 + 5$

1. State the completed square form from part (a)

From part (a), $y = 2(x - 1)^2 + 5$.

2. Explain why the function is always positive

For all real values of $x$, $(x - 1)^2 \geqslant 0$.
Therefore, $2(x - 1)^2 \geqslant 0$.
Adding 5 to both sides, $2(x - 1)^2 + 5 \geqslant 5$.
Since $y \geqslant 5$, $y$ is always strictly greater than $0$. Thus, the curve lies completely above the $x$-axis.