Write x2 + 10x + 24 in the completed-square form (x + a)2 + b.
Worked Solution & Answer
(x + 5)2 - 25 + 24 = (x + 5)2 - 1 B1
Answer: (x + 5)2 - 1
Completing the square — Notes & Worked Examples Show / Hide
Explanation:
To write x2 + bx + c in completed square form (x + p)2 + q, halve the coefficient of x to get p = b2, then subtract p2: (x + b2)2 − (b2)2 + c. The graph of y = (x + p)2 + q has its turning point (vertex) at (−p, q) and line of symmetry x = −p.
When the coefficient of x2 is not 1 (ax2 + bx + c), factorise a out of the first two terms before completing the square inside the square brackets, then multiply out by a at the end.
Example 1: Write x2 − 8x + 5 in the form (x − a)2 + b and state the turning point of y = x2 − 8x + 5.
Solution:
Halve −8 to get −4: (x − 4)2 − (−4)2 + 5
= (x − 4)2 − 16 + 5 = (x − 4)2 − 11
Minimum occurs when x − 4 = 0 ⇒ x = 4, y = −11
Answer: (x − 4)2 − 11; turning point (4, −11)
Example 2: Write 2x2 + 12x + 7 in the form a(x + b)2 + c.
Solution:
Factorise 2 from first two terms: 2[x2 + 6x] + 7
Complete square inside: 2[(x + 3)2 − 9] + 7
Multiply out by 2: 2(x + 3)2 − 18 + 7
Answer: 2(x + 3)2 − 11
Write x2 + 4x - 3 in the completed-square form (x + a)2 + b.
Worked Solution & Answer
(x + 2)2 - 4 - 3 = (x + 2)2 - 7 B1
Answer: (x + 2)2 - 7
Given that x2 - 6x + 1 = (x - a)2 - b for all values of x,
(i) find the value of a and the value of b. [2 marks]
(ii) Hence write down the coordinates of the turning point on the graph of y = x2 - 6x + 1 [1 mark]
Worked Solution & Answer
(i)
x² – 6x + 1 = (x – 3)² – 9 + 1 = (x – 3)² – 8 M1
⇒ a = 3, b = 8 A1
⇒ a = 3, b = 8 A1
Answer (i): a = 3, b = 8
(ii)
Turning point at (3, –8) B1
Answer (ii): 3, –8
(a) Express x² + 8x – 5 in the form (x + a)² – b. [2 marks]
18 (b) Write down the coordinates of the turning point of y = (x – 7)² + 8. [2 marks]
18 (b) Write down the coordinates of the turning point of y = (x – 7)² + 8. [2 marks]
(a) Express x2 + 8x - 5 in the form (x + a)2 - b where a and b are integers. [2 marks]
Answer
(b) A curve has the equation y = (x - 7)2 + 8
Write down the coordinates of the turning point of the curve. [2 marks]
Write down the coordinates of the turning point of the curve. [2 marks]
Worked Solution & Answer
(a)
(x + 4)² – 16 – 5 = (x + 4)² – 21 M1 A1
Answer (a): Answer (x + 4)2 - 21
(b)
Minimum at x = 7, y = 8 M1
⇒ (7, 8) A1
⇒ (7, 8) A1
Answer (b): (7, 8)
(a) Write x² + 12x + 50 in the form (x + a)² + b. [2 marks]
23 (b) Turning point of y = (x – 3)² – 8. [2 marks]
23 (b) Turning point of y = (x – 3)² – 8. [2 marks]
(a) Write x2 + 12x + 50 in the form (x + a)2 + b where a and b are integers. [2 marks]
Answer
(b) A curve has the equation y = (x - 3)2 - 8
Write down the coordinates of the turning point of the curve. [2 marks]
Write down the coordinates of the turning point of the curve. [2 marks]
Worked Solution & Answer
(a)
(x + 6)² = x² + 12x + 36 M1
So x² + 12x + 50 = (x + 6)² + 14 A1
So x² + 12x + 50 = (x + 6)² + 14 A1
Answer (a): Answer (x + 6)2 + 14
(b)
Minimum occurs when (x – 3) = 0 M1
⇒ x = 3, y = –8: (3, –8) A1
⇒ x = 3, y = –8: (3, –8) A1
Answer (b): (3, –8)
Sketch the graph of
y = x2 - 6px - 7\quadwhere p > 0
showing the coordinates of the turning point, in terms of p, and the coordinates of the
intercept with the y-axis.
You must show all your working.
y = x2 - 6px - 7\quadwhere p > 0
showing the coordinates of the turning point, in terms of p, and the coordinates of the
intercept with the y-axis.
You must show all your working.
Worked Solution & Answer
y = (x – 3p)² – 9p² – 7 ⇒ turning point at (3p, –9p² – 7) in 4th quadrant since p > 0 M4
y-intercept at (0, –7) A1
y-intercept at (0, –7) A1
Answer: Parabola sketched with y-intercept at (0, –7) and minimum turning point at (3p, –9p² – 7) in the 4th quadrant
(a) Write 28 + 24x – 6x² in the form a – b(x – c)². (3) (b) Sketch the graph of y = 28 + 24x – 6x², showing the turning point and y-intercept. (3)
(a) Write 28 + 24x - 6x2 in the form a - b(x - c)2 where a, b and c are integers. [3 marks]
(b) On the axes below, sketch the graph of y = 28 + 24x - 6x2
Show clearly the coordinates of the turning point and the coordinates of the point of intersection of the graph with the y-axis. [3 marks]
Show clearly the coordinates of the turning point and the coordinates of the point of intersection of the graph with the y-axis. [3 marks]
Worked Solution & Answer
(a)
28 – 6(x² – 4x) = 28 – 6[(x – 2)² – 4] = 52 – 6(x – 2)² M2 A1
Answer (a): 52 - 6(x - 2)2
(b)
Max at (2, 52), y-intercept at (0, 28) M2 A1
Answer (b): Inverted parabola with maximum at (2, 52) and y-intercept at (0, 28)
(a) Write x2 - 6x + 11 in the form (x - a)2 + b.
(b) Sketch the graph of y = x2 - 6x + 11.
Show clearly the coordinates of any turning points.
(b) Sketch the graph of y = x2 - 6x + 11.
Show clearly the coordinates of any turning points.
(a) Write x2 - 6x + 11 in the form (x - a)2 + b. [3 marks]
(b) Sketch the graph of y = x2 - 6x + 11.
Show clearly the coordinates of any turning points. [3 marks]
Show clearly the coordinates of any turning points. [3 marks]
Worked Solution & Answer
(a)
x² − 6x + 11 = (x − 3)² − 9 + 11 = (x − 3)² + 2. M2 A1
Answer (a): (x - 3)2 + 2
(b)
Minimum turning point at (3, 2) in the first quadrant, crossing the positive y-axis at (0, 11). M2 A1
Answer (b): U-shaped parabola in the upper half-plane with minimum turning point labelled (3, 2)
(a) Write x2 + 8x + 3 in the form (x + a)2 - b.
(b) Sketch the graph of y = x2 + 8x + 3.
Show clearly the coordinates of any turning points and the y-intercept.
(b) Sketch the graph of y = x2 + 8x + 3.
Show clearly the coordinates of any turning points and the y-intercept.
(a) Write x2 + 8x + 3 in the form (x + a)2 - b. [3 marks]
(b) Sketch the graph of y = x2 + 8x + 3.
Show clearly the coordinates of any turning points and the y-intercept. [4 marks]
Show clearly the coordinates of any turning points and the y-intercept. [4 marks]
Worked Solution & Answer
(a)
(x + 4)² − 16 + 3 = (x + 4)² − 13. M2 A1
Answer (a): (x + 4)2 - 13
(b)
Minimum at (−4, −13), y-intercept at (0, 3) M3 A1
Answer (b): U-shaped parabola with minimum at (−4, −13) and y-intercept at (0, 3)
(a) Write x2 + 4x - 16 in the form (x + a)2 - b. [3 marks]
(b) Solve the equation x2 + 4x - 16 = 0.
Give your answers in surd form as simply as possible. [4 marks]
Give your answers in surd form as simply as possible. [4 marks]
x =
(b) Sketch the graph of y = x2 + 4x - 16, showing clearly the coordinates of any turning points. [3 marks]
Worked Solution & Answer
(a)
(x + 2)² − 4 − 16 = (x + 2)² − 20. M2 A1
Answer (a): (x + 2)2 - 20
(b)
(x + 2)² = 20 M2
⇒ x + 2 = ±√20 = ±2√5 M1
⇒ x = −2 ± 2√5. A1
⇒ x + 2 = ±√20 = ±2√5 M1
⇒ x = −2 ± 2√5. A1
Answer (b): x = -2 + 2√(5) or -2 - 2√(5)
(b)
Draw a U-shaped parabola crossing the negative y-axis and having its minimum turning point at (−2, −20). M2 A1
Answer (b): U-shaped parabola with minimum turning point in the 3rd quadrant labelled (−2, −20)