Turning Points by Completing the Square [H]

Turning Points by Completing the Square

Completing the square rewrites \( y = ax^2 + bx + c \) in the form \( y = a(x + p)^2 + q \). This form directly reveals the turning point: since a square term \( (x+p)^2 \geq 0 \), the minimum (or maximum if \( a < 0 \)) occurs when the squared term is zero, i.e. when \( x = -p \). The turning point is therefore \( (-p, q) \).

Method for \( y = x^2 + bx + c \)

  1. Halve the coefficient of \( x \): take \( \left(\frac{b}{2}\right)^2 \) as the correction term.
  2. Write \( y = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c \).
  3. Simplify the constant. The turning point is \( \left(-\frac{b}{2},\ c - \left(\frac{b}{2}\right)^2\right) \).

Method when \( a eq 1 \)

Factor \( a \) out of the \( x^2 \) and \( x \) terms first, complete the square inside the bracket, then expand the factor of \( a \) back.

\[ y = 2x^2 + 8x - 3 = 2(x^2 + 4x) - 3 = 2\left[(x+2)^2 - 4\right] - 3 = 2(x+2)^2 - 8 - 3 = 2(x+2)^2 - 11 \]

Turning point: \( (-2,\ -11) \) — minimum, since \( a = 2 > 0 \).

Worked Examples

Write \( y = x^2 - 8x + 3 \) in completed square form and state the turning point.
\[ y = (x-4)^2 - 16 + 3 = (x-4)^2 - 13 \]

Turning point: \( (4,\ -13) \) — minimum (\(a=1>0\)).

Write \( y = x^2 + 6x + 11 \) in completed square form.
\[ y = (x+3)^2 - 9 + 11 = (x+3)^2 + 2 \]

Turning point: \( (-3,\ 2) \) — this lies above the x-axis, confirming no real roots (\( \Delta = 36-44=-8<0 \)).

Write \( y = 3x^2 - 12x + 7 \) in the form \( a(x-p)^2 + q \). State the vertex.
\[ y = 3(x^2 - 4x) + 7 = 3\left[(x-2)^2 - 4\right] + 7 = 3(x-2)^2 - 12 + 7 = 3(x-2)^2 - 5 \]

Vertex: \( (2,\ -5) \) — minimum (\(a=3>0\)).

Write \( y = -x^2 + 4x + 1 \). Find the maximum value.
\[ y = -(x^2 - 4x) + 1 = -\left[(x-2)^2 - 4\right] + 1 = -(x-2)^2 + 5 \]

Maximum value is 5, occurring at \( x = 2 \). Vertex: \( (2, 5) \).

Solve \( x^2 - 6x + 7 = 0 \) by completing the square, leaving the answer in surd form.
\[ (x-3)^2 - 9 + 7 = 0 \implies (x-3)^2 = 2 \implies x = 3 \pm \sqrt{2} \]
Show that \( y = x^2 + 4x + 7 \) has no real roots by completing the square.
\[ y = (x+2)^2 - 4 + 7 = (x+2)^2 + 3 \]

Since \( (x+2)^2 \geq 0 \) for all real \( x \), we have \( y \geq 3 > 0 \). The function is always positive — it never equals zero, so there are no real roots. ✓

 Key Takeaways

  • \( y = a(x+p)^2 + q \): turning point is at \( (-p, q) \). Minimum if \( a > 0 \), maximum if \( a < 0 \).
  • For \( y = x^2+bx+c \): complete the square by writing \( (x+b/2)^2 - (b/2)^2 + c \).
  • For \( a \neq 1 \): factor \( a \) out of the squared and linear terms first, then complete the square inside the bracket.
  • Once in completed square form, the minimum (or maximum) value of \( y \) is immediately visible as the constant \( q \).
  • If \( a > 0 \) and \( q > 0 \), the function is always positive — no real roots. If \( a > 0 \) and \( q < 0 \), two real roots exist.