Hard GCSE Turning points by completing square Questions

Challenging, exam-style GCSE Turning points by completing square questions with worked solutions. Stretch yourself on the hardest completing the square, non-monic quadratic, turning point, negative coefficient problems.

completing the squarenon-monic quadraticturning pointnegative coefficientmaximum pointminimum value
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
Write −3x2+12x−7-3x^2 + 12x - 7 in the form a(x+p)2+qa(x + p)^2 + q and hence write down the coordinates of the maximum point of the curve y=−3x2+12x−7y = -3x^2 + 12x - 7.
Show worked solution

Worked solution

  1. Identify the coefficients

    a=−3,b=12,c=−7a = -3,\quad b = 12,\quad c = -7

    Compare −3x2+12x−7-3x^2 + 12x - 7 with ax2+bx+cax^2 + bx + c.

  2. Take the factor of -3 out of the x-terms

    −3x2+12x−7=−3(x2−4x)−7-3x^2 + 12x - 7 = -3\left(x^2 - 4x\right) - 7

    Only the x2x^2 and x terms go inside the bracket; 12 ÷−3=−4\div -3 = -4.

  3. Halve the coefficient of x

    −42=−2\frac{-4}{2} = -2

    Half of -4 is -2. This is the number that goes inside the bracket.

  4. Expand the squared bracket to see what it gives

    (x−2)2=x2−4x+4(x - 2)^2 = x^2 - 4x + 4

    Squaring the bracket produces the x2x^2 and -4x terms, but it also adds 4.

  5. Rearrange to get the x-terms on their own

    x2−4x=(x−2)2−4x^2 - 4x = (x - 2)^2 - 4

    Subtract the unwanted 4 from both sides.

  6. Substitute back inside the bracket

    =−3((x−2)2−4)−7= -3\left((x - 2)^2 - 4\right) - 7

    The bracket that was multiplied by -3 is now written as a completed square.

  7. Multiply the -3 through the outer bracket

    =−3(x−2)2+12−7= -3(x - 2)^2 + 12 - 7

    -3 ×−4=12\times -4 = 12.

  8. Write the completed square

    −3x2+12x−7=−3(x−2)2+5-3x^2 + 12x - 7 = -3(x - 2)^2 + 5

    12−7=512 - 7 = 5, giving −3(x−2)2+5-3(x - 2)^2 + 5.

  9. Decide on the shape of the curve

    a=−3<0a = -3 < 0

    The coefficient of x2x^2 is negative, so the parabola is n-shaped and the turning point is a maximum.

  10. Use the fact that a square is never negative

    (x−2)2≥0 so −3(x−2)2≤0(x - 2)^2 \geq 0 \text{ so } -3(x - 2)^2 \leq 0

    Multiplying by the negative number -3 makes the term at most 0, and it equals 0 when x=2x = 2.

  11. Find the value of x at the turning point

    x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2

    The bracket is zero when x=2x = 2.

  12. Find the greatest value of y

    y=−3×0+5=5y = -3 \times 0 + 5 = 5

    With the squared term equal to 0, y takes its greatest value 5.

  13. Check by substituting back into the original

    −3(2)2+12(2)−7=5-3(2)^2 + 12(2) - 7 = 5

    Substituting x=2x = 2 into the original equation gives y=5y = 5, which confirms the turning point.

  14. State the line of symmetry

    x=2x = 2

    A parabola is symmetrical about the vertical line through its turning point, so the line of symmetry is x=2x = 2.

  15. State the maximum point

    Maximum=(2, 5)\text{Maximum} = (2,\ 5)

    The maximum point of the curve is (2,\ 5).

Answer
(2, 5)(2,\ 5)
Question 2
6 markschallenging
Write 4x2+12x+54x^2 + 12x + 5 in the form a(x+p)2+qa(x + p)^2 + q and hence solve 4x2+12x+5=04x^2 + 12x + 5 = 0. Write down the smaller solution.
Show worked solution

Worked solution

  1. Identify the coefficients

    a=4,b=12,c=5a = 4,\quad b = 12,\quad c = 5

    Compare 4x2+12x+54x^2 + 12x + 5 with ax2+bx+cax^2 + bx + c.

  2. Take the factor of 4 out of the x-terms

    4x2+12x+5=4(x2+3x)+54x^2 + 12x + 5 = 4\left(x^2 + 3x\right) + 5

    Only the x2x^2 and x terms go inside the bracket; 12÷4=312 \div 4 = 3.

  3. Halve the coefficient of x

    32=32\frac{3}{2} = \frac{3}{2}

    Half of 3 is 3/2. This is the number that goes inside the bracket.

  4. Expand the squared bracket to see what it gives

    (x+32)2=x2+3x+94(x + \frac{3}{2})^2 = x^2 + 3x + \frac{9}{4}

    Squaring the bracket produces the x2x^2 and 3x terms, but it also adds 94\frac{9}{4}.

  5. Rearrange to get the x-terms on their own

    x2+3x=(x+32)2−94x^2 + 3x = (x + \frac{3}{2})^2 - \frac{9}{4}

    Subtract the unwanted 94\frac{9}{4} from both sides.

  6. Substitute back inside the bracket

    =4((x+32)2−94)+5= 4\left((x + \frac{3}{2})^2 - \frac{9}{4}\right) + 5

    The bracket that was multiplied by 4 is now written as a completed square.

  7. Multiply the 4 through the outer bracket

    =4(x+32)2−9+5= 4(x + \frac{3}{2})^2 - 9 + 5

    4 ×−94=−9\times -\frac{9}{4} = -9.

  8. Write the completed square

    4x2+12x+5=4(x+32)2−44x^2 + 12x + 5 = 4(x + \frac{3}{2})^2 - 4

    −9+5=−4-9 + 5 = -4, giving 4(x+32)2−44(x + \frac{3}{2})^2 - 4.

  9. Set the completed square equal to zero

    4(x+32)2−4=04\left(x + \frac{3}{2}\right)^2 - 4 = 0

    The equation 4x2+12x+5=04x^2 + 12x + 5 = 0 becomes this.

  10. Add 4 to both sides

    4(x+32)2=44\left(x + \frac{3}{2}\right)^2 = 4

    Move the constant across.

  11. Divide both sides by 4

    (x+32)2=1\left(x + \frac{3}{2}\right)^2 = 1

    Isolate the squared bracket.

  12. Take the square root of both sides

    x+32=±1x + \frac{3}{2} = \pm 1

    Remember both square roots of 1.

  13. Solve for x

    x=−32+1=−12 or x=−32−1=−52x = -\frac{3}{2} + 1 = -\frac{1}{2} \text{ or } x = -\frac{3}{2} - 1 = -\frac{5}{2}

    Subtract 3/2 from each of +1 and -1.

  14. Check by factorising

    4x2+12x+5=(2x+1)(2x+5)4x^2 + 12x + 5 = (2x + 1)(2x + 5)

    Expanding (2x + 1)(2x + 5) gives 4x2+10x+2x+5=4x2+12x+54x^2 + 10x + 2x + 5 = 4x^2 + 12x + 5, and the roots agree.

  15. Write down the smaller solution

    x=−52x = -\frac{5}{2}

    Since −5/2=−2.5-5/2 = -2.5 is less than −1/2=−0.5-1/2 = -0.5, the smaller solution is -5/2.

Answer
x=−52x = -\frac{5}{2}
Question 3
6 markschallenging
A rectangle has a perimeter of 2424 cm. One side is xx cm long. Show that the area of the rectangle is greatest when the rectangle is a square, and work out this greatest area.
Show worked solution

Worked solution

  1. Write an expression for the other side

    other side=12−x\text{other side} = 12 - x

    The perimeter is 24 cm, so two adjacent sides add up to 12 cm.

  2. Write an expression for the area

    A=x(12−x)A = x(12 - x)

    Area of a rectangle = length x width.

  3. Expand the bracket

    A=12x−x2A = 12x - x^2

    Multiply x by each term inside the bracket.

  4. Write in the usual order

    A=−x2+12xA = -x^2 + 12x

    The coefficient of x2x^2 is negative, so A has a maximum.

  5. Take out the factor of -1

    A=−(x2−12x)A = -(x^2 - 12x)

    Every term inside the bracket changes sign.

  6. Halve the coefficient of x

    −122=−6\frac{-12}{2} = -6

    Half of -12 is -6, so the bracket will be (x - 6).

  7. Expand the squared bracket

    (x−6)2=x2−12x+36(x - 6)^2 = x^2 - 12x + 36

    Squaring adds an unwanted 36.

  8. Rearrange

    x2−12x=(x−6)2−36x^2 - 12x = (x - 6)^2 - 36

    Subtract the extra 36.

  9. Substitute back

    A=−[(x−6)2−36]A = -\left[(x - 6)^2 - 36\right]

    Put the completed square inside the outer bracket.

  10. Multiply out the minus sign

    A=−(x−6)2+36A = -(x - 6)^2 + 36

    Both terms inside the bracket change sign.

  11. Use the fact that a square is never negative

    (x−6)2≥0⇒−(x−6)2≤0(x - 6)^2 \geq 0 \Rightarrow -(x - 6)^2 \leq 0

    So A is never more than 36.

  12. Find the value of x giving the maximum

    x−6=0⇒x=6x - 6 = 0 \Rightarrow x = 6

    The area is greatest when x=6x = 6 cm.

  13. Find the other side

    12−6=612 - 6 = 6

    The other side is also 6 cm, so the rectangle is a square.

  14. Check the area

    6×6=366 \times 6 = 36

    A 6 cm by 6 cm square has area 36 cm236\text{ cm}^2 and perimeter 24 cm.

  15. State the greatest area

    Amax⁡=36 cm2A_{\max} = 36 \text{ cm}^2

    The area is greatest, at 36 cm236\text{ cm}^2, when the rectangle is a 6 cm square.

Answer
Amax⁡=36 cm2A_{\max} = 36 \text{ cm}^2
Question 4
5 markschallenging
Prove that the curve y=x2−10x+29y = x^2 - 10x + 29 lies entirely above the xx-axis.
Show worked solution

Worked solution

  1. Identify the coefficients

    a=1,b=−10,c=29a = 1,\quad b = -10,\quad c = 29

    Compare x2−10x+29x^2 - 10x + 29 with ax2+bx+cax^2 + bx + c.

  2. Halve the coefficient of x

    −102=−5\frac{-10}{2} = -5

    Half of -10 is -5. This is the number that goes inside the bracket.

  3. Expand the squared bracket to see what it gives

    (x−5)2=x2−10x+25(x - 5)^2 = x^2 - 10x + 25

    Squaring the bracket produces the x2x^2 and -10x terms, but it also adds 25.

  4. Rearrange to get the x-terms on their own

    x2−10x=(x−5)2−25x^2 - 10x = (x - 5)^2 - 25

    Subtract the unwanted 25 from both sides.

  5. Put the constant term back in

    x2−10x+29=(x−5)2−25+29x^2 - 10x + 29 = (x - 5)^2 - 25 + 29

    Replace x2−10xx^2 - 10x by (x−5)2−25(x - 5)^2 - 25 and keep the + 29.

  6. Simplify the constants

    x2−10x+29=(x−5)2+4x^2 - 10x + 29 = (x - 5)^2 + 4

    −25+29=4-25 + 29 = 4, so the completed square is (x−5)2+4(x - 5)^2 + 4.

  7. Use the fact that a square is never negative

    (x−5)2≥0(x - 5)^2 \geq 0

    For every real value of x the square is 0 or more.

  8. Add 4 to both sides

    (x−5)2+4≥4(x - 5)^2 + 4 \geq 4

    Adding 4 keeps the inequality true.

  9. Interpret the inequality

    y≥4y \geq 4

    Every point on the curve has y-coordinate at least 4.

  10. Find the turning point

    x=5, y=4x = 5,\ y = 4

    The minimum point of the curve is (5, 4).

  11. Compare with the x-axis

    y=0 on the x-axisy = 0 \text{ on the } x\text{-axis}

    A point on the x-axis has y-coordinate 0.

  12. Show y=0y = 0 is impossible

    4>04 > 0

    The smallest y-value on the curve is 4, which is greater than 0.

  13. Check with the discriminant

    b2−4ac=100−116=−16b^2 - 4ac = 100 - 116 = -16

    A negative discriminant confirms there are no real roots.

  14. Interpret the discriminant

    −16<0⇒no real roots-16 < 0 \Rightarrow \text{no real roots}

    The curve never meets the x-axis.

  15. Write the conclusion

    y=(x−5)2+4≥4>0y = (x - 5)^2 + 4 \geq 4 > 0

    Since y is always at least 4, the curve lies entirely above the x-axis, as required.

Answer
y=(x−5)2+4≥4>0y = (x - 5)^2 + 4 \geq 4 > 0
Question 5
5 markschallenging
The curve y=a(x−3)2+5y = a(x - 3)^2 + 5 passes through the point (1, 13)(1,\ 13). Work out the value of aa.
Show worked solution

Worked solution

  1. Write down the turning point

    x−3=0⇒x=3, y=5x - 3 = 0 \Rightarrow x = 3,\ y = 5

    Whatever the value of a, the turning point of a(x−3)2+5a(x - 3)^2 + 5 is at (3, 5).

  2. Use the point on the curve

    13=a(1−3)2+513 = a(1 - 3)^2 + 5

    Substitute x=1x = 1 and y=13y = 13 into the equation of the curve.

  3. Work out the bracket

    13=a(−2)2+513 = a(-2)^2 + 5

    1−3=−21 - 3 = -2.

  4. Square the bracket

    13=4a+513 = 4a + 5

    (−2)2=4(-2)^2 = 4, so a(−2)2=4aa(-2)^2 = 4a.

  5. Subtract 5 from both sides

    8=4a8 = 4a

    13−5=813 - 5 = 8.

  6. Divide by 4

    a=2a = 2

    So the curve is y=2(x−3)2+5y = 2(x - 3)^2 + 5.

  7. Check the point (1, 13)

    2(1−3)2+5=2(4)+5=132(1 - 3)^2 + 5 = 2(4) + 5 = 13

    The point (1, 13) does lie on the curve.

  8. Decide on the shape

    a=2>0a = 2 > 0

    The coefficient of the squared term is positive, so the curve is U-shaped with a minimum.

  9. Use the fact that a square is never negative

    2(x−3)2≥02(x - 3)^2 \geq 0

    The smallest the squared term can be is 0.

  10. State the turning point

    (3, 5)(3,\ 5)

    The minimum point of the curve is (3, 5).

  11. Expand to check

    2(x−3)2+5=2(x2−6x+9)+52(x - 3)^2 + 5 = 2(x^2 - 6x + 9) + 5

    Expand the bracket to write the curve in the usual form.

  12. Multiply out

    =2x2−12x+18+5= 2x^2 - 12x + 18 + 5

    Multiply each term inside the bracket by 2.

  13. Simplify

    y=2x2−12x+23y = 2x^2 - 12x + 23

    18+5=2318 + 5 = 23.

  14. Check the line of symmetry

    x=−−122×2=3x = -\frac{-12}{2 \times 2} = 3

    The line of symmetry is x=3x = 3, which agrees with the turning point.

  15. State the answer

    a=2,turning point (3, 5)a = 2,\quad \text{turning point } (3,\ 5)

    The value of a is 2 and the turning point is (3, 5).

Answer
a=2 (the turning point is (3, 5))a = 2 \text{ (the turning point is } (3,\ 5) \text{)}

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