Turning points by completing square Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Turning points by completing square questions. See exactly how to solve problems on completed square form, turning point, minimum point, minimum value.

completed square formturning pointminimum pointminimum valueline of symmetrycompleting the square
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
The curve y=(x+3)2−4y = (x + 3)^2 - 4 has a turning point. Write down the coordinates of the turning point.

Worked solution

  1. Compare with the completed square form

    y=(x+p)2+q with p=3, q=−4y = (x + p)^2 + q \text{ with } p = 3,\ q = -4

    In the form (x+p)2+q(x + p)^2 + q the turning point is at (-p, q).

  2. Find the value of x at the turning point

    x+3=0⇒x=−3x + 3 = 0 \Rightarrow x = -3

    The bracket is zero when x=−3x = -3.

  3. State the minimum point

    Minimum=(−3, −4)\text{Minimum} = (-3,\ -4)

    The minimum point of the curve is (-3,\ -4).

Answer
(−3, −4)(-3,\ -4)
Question 2
1 markeasy
The curve y=(x−5)2+2y = (x - 5)^2 + 2 has a minimum point. Write down the coordinates of the minimum point.

Worked solution

  1. Compare with the completed square form

    y=(x+p)2+q with p=−5, q=2y = (x + p)^2 + q \text{ with } p = -5,\ q = 2

    The turning point of (x+p)2+q(x + p)^2 + q is at (-p, q).

  2. Find the value of x at the turning point

    x−5=0⇒x=5x - 5 = 0 \Rightarrow x = 5

    The bracket is zero when x=5x = 5.

  3. State the minimum point

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

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

Answer
(5, 2)(5,\ 2)
Question 3
1 markeasy
Write down the minimum value of y=(x−4)2+7y = (x - 4)^2 + 7.

Worked solution

  1. Look at the squared bracket

    (x−4)2≥0(x - 4)^2 \geq 0

    A square is never negative, so the smallest value of (x−4)2(x - 4)^2 is 0.

  2. Make the bracket as small as possible

    x=4⇒(x−4)2=0x = 4 \Rightarrow (x - 4)^2 = 0

    When x=4x = 4 the bracket is zero.

  3. Read the minimum value of y

    y=0+7=7y = 0 + 7 = 7

    The least value y can take is 7.

Answer
77
Question 4
1 markeasy
Write down the value of xx at which y=(x+6)2−1y = (x + 6)^2 - 1 takes its minimum value.

Worked solution

  1. Look at the squared bracket

    (x+6)2≥0(x + 6)^2 \geq 0

    The smallest a square can be is 0.

  2. Make the bracket zero

    x+6=0x + 6 = 0

    The bracket is zero when x+6=0x + 6 = 0.

  3. Solve for x

    x=−6x = -6

    The minimum value of y occurs when x=−6x = -6.

Answer
x=−6x = -6
Question 5
1 markeasy
Write down the equation of the line of symmetry of the curve y=(x−2)2+5y = (x - 2)^2 + 5.

Worked solution

  1. Find the turning point

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

    The turning point is where the bracket is zero, at x=2x = 2.

  2. Recall the symmetry of a parabola

    symmetry about the turning point\text{symmetry about the turning point}

    A parabola is symmetrical about the vertical line through its turning point.

  3. Write the equation of the line

    x=2x = 2

    The line of symmetry is the vertical line x=2x = 2.

Answer
x=2x = 2

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