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GCSE · Algebra - Graphs

5.9D Drawing and Using Quadratic Graphs: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculators not allowed · about 85 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Complete the table of values for y = x² - 1, for -2 ≤ x ≤ 2.
x : -2 , -1 , 0 , 1 , 2
y : _ , _ , _ , _ , _

(Total for Question 1 is 2 marks)

Write down the coordinates of the point where the graph of y = x² + 5x - 4 crosses the y-axis.

(Total for Question 2 is 1 mark)

State whether the graph of y = 8 - 3x² is u-shaped or n-shaped.

(Total for Question 3 is 1 mark)

Write down the coordinates of the turning point of the graph of y = x^2.

(Total for Question 4 is 1 mark)

Write down the coordinates of the minimum point of the graph of y = (x - 2)² + 5.

(Total for Question 5 is 1 mark)

Four students each drew a table of values for y = x² + 2x, for -3 ≤ x ≤ 3. Only one table is correct.

(a)Which table is correct?(1)

A) x: -3,-2,-1,0,1,2,3 -> y: 3,0,-1,0,3,8,15
B) x: -3,-2,-1,0,1,2,3 -> y: 15,8,3,0,-1,0,3
C) x: -3,-2,-1,0,1,2,3 -> y: 9,4,1,0,1,4,9
D) x: -3,-2,-1,0,1,2,3 -> y: 11,6,3,2,3,6,11

(b)Give a reason for your answer.(1)

(Total for Question 6 is 2 marks)

For each equation, state whether its graph is u-shaped or n-shaped.

(i)y = 4 - 7x²(1)

(ii)y = 6x² - x - 1(1)

(iii)y = -x² + 3x - 2(1)

(Total for Question 7 is 3 marks)

The graph of y = x² + 6x + 5 crosses the x-axis at two points.

(a)Find the coordinates of the two points where the graph crosses the x-axis.(2)

(b)Using the symmetry of the graph, write down the coordinates of the turning point.(2)

(c)Write down the coordinates of the y-intercept of the graph.(1)

(Total for Question 8 is 5 marks)

The graph of a quadratic function crosses the x-axis at x = -3 and x = 5.
Find the equation of the line of symmetry of the graph.

(Total for Question 9 is 2 marks)

The graph of y = x² - 3x is to be drawn for 0 ≤ x ≤ 4.

(a)Complete the table of values for y = x² - 3x.
x : 0 , 1 , 2 , 3 , 4
y : _ , _ , _ , _ , _(2)

(b)Using your table, write down the equation of the line of symmetry of the graph.(1)

(c)Hence write down the coordinates of the turning point of the graph.(1)

(Total for Question 10 is 4 marks)

The grid shows an accurately drawn graph of y = x² - 4x + 3 for 0 ≤ x ≤ 4.

(a)Write down the solutions of x² - 4x + 3 = 0.(2)

(b)Write down the coordinates of the minimum point of the graph.(1)

(c)By drawing the line y = 3 on the grid, find the solutions of x² - 4x + 3 = 3.(3)

(Total for Question 11 is 6 marks)

The graph of y = x² - 10x + 21 crosses the x-axis at two points.

(Total for Question 12 is 2 marks)

A rectangular lawn has width x metres and length (x + 5) metres. The area of the lawn, A m², is given by A = x(x + 5).

(a)Complete the table of values for A = x(x + 5), for 1 ≤ x ≤ 3.
x : 1 , 2 , 3
A : _ , _ , _(2)

(b)Explain why, in this context, x cannot take a negative value.(1)

(Total for Question 13 is 3 marks)

A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given by h = 24t - 6t^2.

(a)Complete the table of values for h = 24t - 6t^2.
t : 0 , 1 , 2 , 3 , 4
h : _ , _ , _ , _ , _(2)

(b)Using the symmetry of your table, write down the maximum height reached by the ball and the time at which it occurs.(1)

(Total for Question 14 is 3 marks)

y = x² - 8x + 10

(a)Express y in the form (x - a)² + b, where a and b are integers.(2)

(b)Hence write down the coordinates of the minimum point of the graph of y = x² - 8x + 10.(1)

(Total for Question 15 is 3 marks)

y = 3 - 2x - x²

(a)Write down the coordinates of the y-intercept of the graph.(1)

(b)State, giving a reason, whether the graph is u-shaped or n-shaped.(1)

(c)By writing y in the form -(x + p)² + q, find the coordinates of the turning point of the graph.(2)

(Total for Question 16 is 4 marks)

A market trader models her weekly profit, P pounds, using P = -2x² + 40x, where x is the price increase in pounds applied to each item, for 0 ≤ x ≤ 20.

(a)Complete the table of values for P = -2x² + 40x.
x : 0 , 5 , 10 , 15 , 20
P : 0 , _ , 200 , _ , 0(2)

(b)Using the symmetry of the table, find the maximum weekly profit and the value of x at which it occurs.(2)

(c)Interpret what the value P = 0 represents at both x = 0 and x = 20.(1)

(Total for Question 17 is 5 marks)

The graph of y = x² + px + q has a minimum point at (3, -5), where p and q are constants.

(Total for Question 18 is 3 marks)

y = x² - 6x + 11

(a)Show that y = (x - 3)² + 2.(2)

(b)Hence explain why the equation x² - 6x + 11 = 0 has no real solutions.(2)

(Total for Question 19 is 4 marks)

Without solving the equation, determine how many times the graph of y = 2x² - 5x + 4 crosses the x-axis.

(Total for Question 20 is 3 marks)

The graphs of y = x² - 2x - 3 and y = x + 1 intersect at two points.

(Total for Question 21 is 5 marks)

f(x) = x² - 4x + 7. The graph of y = f(x) - c has its minimum point on the x-axis, where c is a constant.

(Total for Question 22 is 3 marks)

The graph of y = x² + kx + 9, where k is a non-zero constant, touches the x-axis at exactly one point.

(Total for Question 23 is 5 marks)

The line y = 2x + c is a tangent to the curve y = x² - 4x + 9 (it touches the curve at exactly one point), where c is a constant.

(Total for Question 24 is 5 marks)

Mark scheme · 5.9D Drawing and Using Quadratic Graphs: Fluency and Exam Drill

Question 1

B1 at least 3 of the 5 y-values correct
B1 all 5 y-values correct
Answer: 3, 0, -1, 0, 3

Question 2

B1 (0, -4) oe
Answer: (0, -4)

Question 3

B1 n-shaped
Answer: n-shaped

Question 4

B1 (0, 0)
Answer: (0, 0)

Question 5

B1 (2, 5)
Answer: (2, 5)

Question 6

(a) B1 A
(a) Answer: A
(b) B1 correct substitution check shown for at least one value, e.g. at x=1, 1²+2(1)=1+2=3 oe
(b) Answer: At x = 1: 1² + 2(1) = 1 + 2 = 3, which only matches table A

Question 7

(i) B1 n-shaped
(i) Answer: n-shaped
(ii) B1 u-shaped
(ii) Answer: u-shaped
(iii) B1 n-shaped
(iii) Answer: n-shaped

Question 8

(a) M1 correct factorisation (x + 1)(x + 5) = 0 oe
(a) A1 (-1, 0) and (-5, 0) both cao
(a) Answer: (-1, 0) and (-5, 0)
(b) M1 finds the midpoint of the roots, x = (-1 + -5)/2 = -3, ft from part (a)
(b) A1 (-3, -4) cao
(b) Answer: (-3, -4)
(c) B1 (0, 5)
(c) Answer: (0, 5)

Question 9

M1 uses the midpoint of the two roots, (-3 + 5) / 2
A1 x = 1 cao
Answer: x = 1

Question 10

(a) B1 at least 3 of the 5 y-values correct
(a) B1 all 5 y-values correct
(a) Answer: 0, -2, -2, 0, 4
(b) B1 x = 1.5 oe, ft from candidate's table
(b) Answer: x = 1.5
(c) B1 (1.5, -2.25) ft from part (b)
(c) Answer: (1.5, -2.25)

Question 11

(a) B1 x = 1
(a) B1 x = 3
(a) Answer: x = 1 and x = 3
(b) B1 (2, -1)
(b) Answer: (2, -1)
(c) M1 line y = 3 drawn on the grid
(c) A1 x = 0
(c) A1 x = 4
(c) Answer: x = 0 and x = 4

Question 12

M1 correct factorisation (x - 3)(x - 7) = 0 oe
A1 (3, 0) and (7, 0) both cao
Answer: (3, 0) and (7, 0)

Question 13

(a) B1 at least 2 of the 3 A-values correct
(a) B1 all 3 A-values correct
(a) Answer: 6, 14, 24
(b) B1 x represents a width/length, and a physical length cannot be negative oe
(b) Answer: x is a width (a length), and a length cannot be negative

Question 14

(a) B1 at least 3 of the 5 h-values correct
(a) B1 all 5 h-values correct
(a) Answer: 0, 18, 24, 18, 0
(b) B1 maximum height = 24 m at t = 2 seconds, ft from candidate's table
(b) Answer: 24 metres at t = 2 seconds

Question 15

(a) M1 (x - 4)² seen (correct value of a)
(a) A1 (x - 4)² - 6 oe cao (correct value of b)
(a) Answer: (x - 4)² - 6
(b) B1 (4, -6) ft from part (a)
(b) Answer: (4, -6)

Question 16

(a) B1 (0, 3)
(a) Answer: (0, 3)
(b) B1 n-shaped, because the coefficient of x² is negative (-1) oe
(b) Answer: n-shaped, because the coefficient of x² is -1, which is negative
(c) M1 correctly completes the square, e.g. y = -((x + 1)² - 4) oe, or -(x+1)²+4
(c) A1 (-1, 4) cao
(c) Answer: (-1, 4)

Question 17

(a) B1 P = 150 at x = 5
(a) B1 P = 150 at x = 15
(a) Answer: x=5: P=150, x=15: P=150
(b) M1 identifies that the maximum lies midway between x = 0 and x = 20 (or between the two equal P-values), x = 10
(b) A1 maximum profit = 200 pounds at x = 10
(b) Answer: Maximum profit = 200 pounds, at x = 10
(c) B1 a sensible interpretation, e.g. the price increase is either too small or too large to generate any profit oe
(c) Answer: At x = 0 there is no price increase, and at x = 20 the price increase is so large that no profit is made; both give zero profit in this model

Question 18

M1 uses -p/2 = 3 to find p
A1 p = -6
B1 q = 4, ft from candidate's p, using 9 + 3p + q = -5
Answer: p = -6, q = 4

Question 19

(a) M1 (x - 3)² seen
(a) A1 correctly shows (x - 3)² + 2 = x² - 6x + 9 + 2 = x² - 6x + 11 (cso)
(a) Answer: y = (x - 3)² + 2
(b) M1 states or uses that (x - 3)² ≥ 0 for all real x
(b) A1 correct conclusion: since (x - 3)² ≥ 0, y = (x - 3)² + 2 ≥ 2, so y is never 0, so the graph never crosses the x-axis and there are no real solutions (cso)
(b) Answer: Since (x - 3)² ≥ 0 for all real x, y = (x - 3)² + 2 ≥ 2 always, so y can never equal 0, meaning the graph never crosses the x-axis and the equation has no real solutions

Question 20

M1 attempts b² - 4ac with a = 2, b = -5, c = 4
A1 discriminant = -7 cao
B1 ft conclusion: since the discriminant is negative, the graph does not cross the x-axis (0 times)
Answer: The graph does not cross the x-axis (0 times), since the discriminant is negative

Question 21

M1 equates x² - 2x - 3 = x + 1 oe
dM1 rearranges to a correct three-term quadratic x² - 3x - 4 = 0 oe, dependent on the previous M1
M1 correct factorisation (x - 4)(x + 1) = 0, or correct use of the quadratic formula
A1 x = 4 and x = -1
A1 y = 5 and y = 0 respectively, cao
Answer: (4, 5) and (-1, 0)

Question 22

M1 completes the square correctly, f(x) = (x - 2)² + 3
M1 sets the minimum value of f(x) - c equal to 0, i.e. 3 - c = 0, dependent on the previous M1
A1 c = 3
Answer: c = 3

Question 23

M1 uses the condition for equal roots, b² - 4ac = 0, with a = 1, b = k, c = 9
M1 forms k² - 36 = 0 and rearranges to k² = 36, dependent on the previous M1
A1 k = 6
A1 k = -6
B1 touch points ft: (-3, 0) when k = 6, and (3, 0) when k = -6
Answer: k = 6, touching at (-3, 0); or k = -6, touching at (3, 0)

Question 24

M1 equates x² - 4x + 9 = 2x + c oe
M1 rearranges to a correct three-term quadratic x² - 6x + (9 - c) = 0, dependent on the previous M1
M1 sets the discriminant of this quadratic equal to 0, (-6)² - 4(1)(9 - c) = 0
A1 simplifies to 4c = 0 oe
A1 c = 0 cao
Answer: c = 0

Drawing and Using Quadratic Graphs: Fluency and Exam Drill - Worksheets, Questions and Revision