Completing the square and turning points

(a) $(x + 3)^2 - 7$

1. Halve the coefficient of $x$: $a = 3$.
   $$(x + 3)^2 = x^2 + 6x + 9$$
2. Compensate for the extra 9.
   $$x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7$$

• M1 Writing $(x + 3)^2$.
• A1 $(x + 3)^2 - 7$.

(b) $(-3, -7)$

1. $(x + 3)^2 \ge 0$ and equals 0 when $x = -3$, so the minimum is $(-3, -7)$.

• B1 The correct answer, $(-3, -7)$.

(a) $(-5,-8)$

1. $$(x+5)^{2}-8$$
2. Complete the square: $x^{2}$ + 10x + 17 = (x + 5)$^{2}$ $-$ 8.
3. A square is at least zero, with its minimum at x = $-5.$
4. Then y = $-8$, so the turning point is ($-5$, $-8$).

• P1 Establishing $(x+5)^{2}-8$ or an equivalent valid method.
• A1 Correct answer: $(-5,-8)$

(a) $450$ m²

1. $$x(60-2x)$$
2. $$-2(x-15)^{2}+450$$
3. Let each side perpendicular to the wall be x metres. The third fenced side is 60 $-$ 2x metres.
4. The area is $A = x(60 - 2x) = -2x^{2} + 60x$.
5. Complete the square: A = $-2$(x $-$ 15)$^{2}$ + 450.
6. A square is non-negative, so the greatest area is 450 $m^{2}$, achieved with sides 15 m and 30 m.

• P1 Establishing $x(60-2x)$ or an equivalent valid method.
• P1 Establishing $-2(x-15)^{2}+450$ or an equivalent valid method.
• A1 Correct answer: $450$ m²

(a) $(x-3)^{2}+1$

1. Halve the linear coefficient to form the square and adjust its constant.
   $$x^{2}-6x+9+1$$
2. Therefore $(x-3)^{2}+1$.

• M1 Halve the linear coefficient to form the square and adjust its constant.
• A1 Correct answer: $(x-3)^{2}+1$

(b) $(3,1)$

1. The square is zero at x=a and cannot be negative.
   $$(x-3)^{2}=0$$
2. Therefore $(3,1)$.

• M1 The square is zero at x=a and cannot be negative.
• A1 Correct answer: $(3,1)$

(a) $-2(x-6)^{2}+72$

1. The remaining fenced side is total minus twice x.
   $$x(24-2x)$$
2. Complete the square in the resulting quadratic.
   $$-2(x-6)^{2}+72$$
3. Therefore $-2(x-6)^{2}+72$.

• P1 The remaining fenced side is total minus twice x.
• P1 Complete the square in the resulting quadratic.
• A1 Correct answer: $-2(x-6)^{2}+72$

(b) $72$ m²

1. The negative square term is greatest when it is zero.
   $$x=6$$
2. Therefore $72$ m².

• P1 The negative square term is greatest when it is zero.
• A1 Correct answer: $72$ m²

(a) $-5\pm\sqrt7$

1. Complete the square and isolate it.
   $$(x+5)^{2}=7$$
2. Use both signs of the square root.
3. Therefore $-5\pm\sqrt7$.

• M1 Complete the square and isolate it.
• M1 Use both signs of the square root.
• A1 Correct answer: $-5\pm\sqrt7$

(b) $-10$

1. The opposite surd terms cancel when the roots are added.
   $$(-5-\sqrt{7})+(-5+\sqrt{7})$$
2. Therefore $-10$.

• M1 The opposite surd terms cancel when the roots are added.
• A1 Correct answer: $-10$

(a) $2(x - 3)^2 - 13$

1. Take out the factor 2 from the $x$ terms.
   $$2(x^2 - 6x) + 5$$
2. Complete the square inside.
   $$2\left[(x - 3)^2 - 9\right] + 5 = 2(x - 3)^2 - 18 + 5$$
3. $$= 2(x - 3)^2 - 13$$

• M1 Taking out 2: $2(x^2 - 6x) + 5$.
• M1 Writing $2(x - 3)^2$ with a compensating term.
• A1 $2(x - 3)^2 - 13$.

(b) $x = 3 \pm \sqrt{\frac{13}{2}}$

1. $$2(x - 3)^2 = 13 \Rightarrow (x - 3)^2 = \frac{13}{2}$$
2. $$x = 3 \pm \sqrt{\tfrac{13}{2}} \;\left(= 3 \pm \tfrac{\sqrt{26}}{2}\right)$$

• M1 Rearranging to $(x - 3)^2 = \frac{13}{2}$.
• A1 $x = 3 \pm \sqrt{\frac{13}{2}}$ (or $3 \pm \frac{\sqrt{26}}{2}$).

(a) $E = (x - 4)^2 + 4 \ge 4 > 0$

1. $$x^2 - 8x + 20 = (x - 4)^2 - 16 + 20 = (x - 4)^2 + 4$$
2. $(x - 4)^2 \ge 0$ for all $x$, so $E \ge 4 > 0$.

• M1 Writing $(x - 4)^2 + 4$.
• C1 Stating that a square is never negative, so $E$ is at least 4 and so positive.

(b) $\frac{1}{4}$

1. The fraction is greatest when the denominator is smallest.
2. The smallest value of $(x - 4)^2 + 4$ is 4 (when $x = 4$), so the greatest value is $\frac{1}{4}$.

• P1 Recognising the fraction is greatest when $E$ is least, and finding the least value 4.
• A1 The correct answer, $\frac{1}{4}$.