GCSE Turning points by completing square Practice Questions

Free GCSE Turning points by completing square practice questions with full step-by-step worked solutions. Covers completed square form, turning point, minimum point, minimum value. Practise exam-style problems and check your method.

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.
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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 (−3-3,\ −4-4).

Answer
(−3, −4)(-3,\ -4)
Question 2
2 markseasy
The curve y=(x+a)2−2y = (x + a)^2 - 2 has its turning point at (5, −2)(5,\ -2). Work out the value of aa.
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Worked solution

  1. Write down the turning point in terms of a

    x+a=0⇒x=−ax + a = 0 \Rightarrow x = -a

    The turning point of (x+a)2−2(x + a)^2 - 2 is at (-a, −2-2).

  2. Compare with the given turning point

    −a=5-a = 5

    The x-coordinate of the turning point is 55.

  3. Solve for a

    a=−5a = -5

    So a=−5a = -5, and the curve is y=(x−5)2−2y = (x - 5)^2 - 2.

Answer
a=−5a = -5
Question 3
2 marksintermediate
The minimum point of the curve y=x2+8x+cy = x^2 + 8x + c is (−4, 5)(-4,\ 5). Work out the value of cc.
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Worked solution

  1. Complete the square in terms of c

    x2+8x+c=(x+4)2−16+cx^2 + 8x + c = (x + 4)^2 - 16 + c

    Half of 88 is 44, and (x+4)2(x + 4)^2 gives an extra 1616 that must be subtracted.

  2. Write down the turning point

    (−4, c−16)(-4,\ c - 16)

    The minimum point is at x=−4x = -4 with y-value c - 16.

  3. Compare y-coordinates

    c−16=5c - 16 = 5

    The minimum point is given as (−4-4, 55).

  4. Solve for c

    c=21c = 21

    Add 1616 to both sides.

  5. Write the equation of the curve

    y=x2+8x+21y = x^2 + 8x + 21

    This is the curve with the given minimum point.

  6. Check the answer

    (−4)2+8(−4)+21=16−32+21=5(-4)^2 + 8(-4) + 21 = 16 - 32 + 21 = 5

    Substituting x=−4x = -4 gives y=5y = 5, as required.

Answer
c=21c = 21
Question 4
4 markshard
The graph of y=x2y = x^2 is translated to give the graph of y=x2−6x+13y = x^2 - 6x + 13. Describe the translation as a column vector.
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Worked solution

  1. Identify the coefficients

    a=1,b=−6,c=13a = 1,\quad b = -6,\quad c = 13

    Compare x2−6x+13x^2 - 6x + 13 with ax2+bx+cax^2 + bx + c.

  2. Halve the coefficient of x

    −62=−3\frac{-6}{2} = -3

    Half of −6-6 is −3-3. This is the number that goes inside the bracket.

  3. Expand the squared bracket to see what it gives

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

    Squaring the bracket produces the x2x^2 and −6x-6x terms, but it also adds 99.

  4. Put the constant term back in

    x2−6x+13=(x−3)2−9+13x^2 - 6x + 13 = (x - 3)^2 - 9 + 13

    Replace x2−6xx^2 - 6x by (x−3)2−9(x - 3)^2 - 9 and keep the + 1313.

  5. Simplify the constants

    x2−6x+13=(x−3)2+4x^2 - 6x + 13 = (x - 3)^2 + 4

    −9+13=4-9 + 13 = 4, so the completed square is (x−3)2+4(x - 3)^2 + 4.

  6. Compare with y=x2y = x^2

    y=(x−3)2+4y = (x - 3)^2 + 4

    The curve is y=x2y = x^2 with x replaced by (x - 3), then 4 added.

  7. Interpret the bracket

    x→x−3 : a shift of 3 to the rightx \to x - 3 \text{ : a shift of } 3 \text{ to the right}

    Replacing x by x - 3 moves the graph 3 units in the positive x-direction.

  8. Interpret the constant

    +4 : a shift of 4 upwards+4 \text{ : a shift of } 4 \text{ upwards}

    Adding 4 to y moves the graph 4 units up.

  9. Check with the turning point

    (0, 0)→(3, 4)(0,\ 0) \to (3,\ 4)

    The turning point of y=x2y = x^2 is (00, 00); the new turning point is (33, 44).

  10. Write the translation as a column vector

    (34)\begin{pmatrix} 3 \\ 4 \end{pmatrix}

    The translation is 33 right and 44 up.

Answer
(34)\begin{pmatrix} 3 \\ 4 \end{pmatrix}
Question 5
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.
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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 −3-3 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 −4-4 is −2-2. 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-4x terms, but it also adds 44.

  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 44 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 −3-3 is now written as a completed square.

  7. Multiply the −3-3 through the outer bracket

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

    −3-3 ×−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 −3-3 makes the term at most 00, and it equals 00 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 (22,\ 55).

Answer
(2, 5)(2,\ 5)

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