Completing the Square for a Maximum

When the $x^2$ coefficient is negative, completing the square gives a maximum instead of a minimum, because the squared term is now subtracted. The method is the same; only the sign of the vertex changes.

Express $3 + 4x - 2x^2$ in the form $a - b(x - c)^2$, and hence state the maximum value. [4]

The working

Step 1, reorder and factor the $-2$ from the $x$-terms: $3 + 4x - 2x^2 = -2x^2 + 4x + 3 = -2(x^2 - 2x) + 3 \quad \text{(M1)}$

Step 2, complete the square inside the bracket ($x^2 - 2x = (x-1)^2 - 1$): $= -2\big[(x - 1)^2 - 1\big] + 3 \quad \text{(M1)}$

Step 3, multiply back and combine ($-2 \times -1 = +2$, then $+2 + 3 = 5$): $= -2(x - 1)^2 + 5 = 5 - 2(x - 1)^2 \quad \text{(A1)}$

Step 4, read the maximum. Since $-2(x - 1)^2 \le 0$, the expression is largest when the bracket is zero: $\text{maximum value} = 5 \ (\text{at } x = 1) \quad \text{(A1)}$

Where the marks are won and lost

• Factor out the negative coefficient from the $x$-terms first: $-2(x^2 - 2x)$. Handling the sign here is the crux.
• Multiplying back: $-2 \times (-1) = +2$, so the constant becomes $+2 + 3 = 5$. A sign slip on this product is the classic error.
• Negative coefficient means maximum, not minimum, the vertex is the highest point.

Common mistakes

• Forgetting to multiply the inside $-1$ by the outside $-2$ (giving the wrong constant).
• Treating the turning point as a minimum.
• Sign errors when factoring $-2$ out of $+4x$ (it becomes $-2x$ inside).

Full method: Completing the Square notes. See also Maximum/Minimum & the Vertex. Topic home: Quadratic Functions pillar.