Revision Library

revisionlibrary.co.uk

FOUNDATION

5.9F Drawing Quadratic Graphs: Foundation Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the value of y when x = 0 for the quadratic function y = x² + 3x + 2.

(Total for Question 1 is 1 mark)

Complete the table of values for y = x² - 2x.
Fill in values for x = 0 and x = 2.

(a)Find y when x = 0.(1)

(b)Find y when x = 2.(1)

(Total for Question 2 is 2 marks)

Plot the points for the function y = x² - 1 for x = -1, 0 and 1. Give the three pairs as (x, y).

(a)Find y when x = -1.(1)

(b)Write down the three coordinate pairs for x = -1, 0, 1.(1)

(Total for Question 3 is 2 marks)

Draw a simple sketch of the graph y = x². On the sketch, mark and label the points when x = -2, -1, 0, 1, 2.

(a)Find the y values for x = -2, -1, 0, 1, 2.(2)

(b)Mark and label these five points on the sketch and draw the curve shape.(1)

(Total for Question 4 is 3 marks)

The graph of y = (x - 1)² - 2 is the graph of y = x² shifted. State the coordinates of the vertex (turning point) of this parabola.

(a)Explain how the graph of y = (x - 1)² - 2 is shifted from y = x².(1)

(b)State the coordinates of the vertex.(3)

(Total for Question 5 is 4 marks)

Sketch the graph of y = x² - 4. On your sketch, show where the graph meets the x-axis.

(a)Find the x-intercepts (solutions) of x² - 4 = 0.(2)

(b)Sketch the parabola and mark the x-intercepts on the x-axis and the vertex.(3)

(Total for Question 6 is 5 marks)

The graph of y = 2x² is narrower than y = x². Explain why and then find y when x = 1 and x = 2.

(a)Explain in one sentence why y = 2x² is narrower than y = x².(1)

(b)Find y when x = 1.(2)

(c)Find y when x = 2.(2)

(Total for Question 7 is 5 marks)

A student is given the graph of y = x² and is asked to draw the graph of y = (x + 2)². Describe how to transform the graph and give the vertex of the new graph.

(a)Describe the transformation from y = x² to y = (x + 2)².(2)

(b)Give the vertex (turning point) of y = (x + 2)².(4)

(Total for Question 8 is 6 marks)

A simple real-life context: A small ball is thrown up and its height h metres after t seconds is modelled by h = -t² + 4t + 1. Find the height after t = 0, t = 1 and t = 4. Then state whether the ball is above the ground at t = 4 seconds.

(a)Find h when t = 0.(1)

(b)Find h when t = 1.(2)

(c)Find h when t = 4.(2)

(d)Is the ball above the ground at t = 4 seconds? Explain briefly.(1)

(Total for Question 9 is 6 marks)

Stretch question for higher foundation: Sketch the graph of y = -x² + 2x + 3. Give the vertex and the x-intercepts (to the nearest whole number if needed).

(a)Find the vertex (turning point) of y = -x² + 2x + 3.(3)

(b)Find the x-intercepts (solutions) of -x² + 2x + 3 = 0. Give answers to the nearest whole number.(3)

(Total for Question 10 is 6 marks)

Mark scheme · 5.9F Drawing Quadratic Graphs: Foundation Tier Practice

Question 1

B1 y = 2 cao
Answer: 2 [y = 0² + 3(0) + 2 = 2 ]

Question 2

(a) B1 0 cao
(a) Answer: 0 [0² - 2(0) = 0]
(b) B1 0 cao
(b) Answer: 0 [2² - 2(2) = 4 - 4 = 0]

Question 3

(a) B1 0 cao
(a) Answer: 0 [(-1)² - 1 = 1 - 1 = 0]
(b) B1 (-1, 0), (0, -1), (1, 0) cao
(b) Answer: (-1, 0), (0, -1), (1, 0)

Question 4

(a) M1 correct method to square the x values shown or listed
(a) A1 y values 4, 1, 0, 1, 4 cao
(a) Answer: y = 4, 1, 0, 1, 4
(b) A1 curve sketched through the five labelled points with correct U shape
(b) Answer: Sketch with points plotted at (-2,4), (-1,1), (0,0), (1,1), (2,4) and a U-shaped curve through them.

Question 5

(a) B1 shift right 1 and down 2 oe
(a) Answer: Shifted right 1 and down 2.
(b) M1 identify h = 1 and k = -2 or show (1, -2)
(b) A1 vertex (1, -2) cao
(b) A1 reason: vertex from completing square/form sign applied oe
(b) Answer: (1, -2)

Question 6

(a) M1 method: factorise or take square root, show x² = 4
(a) A1 x = 2 and x = -2 cao
(a) Answer: x = 2 and x = -2
(b) M1 correct U-shape sketched with symmetry and intercepts marked at -2 and 2
(b) A1 vertex at (0, -4) indicated
(b) A1 curve passes through intercepts and has correct vertical position
(b) Answer: Sketch with x-intercepts at (-2,0) and (2,0) and vertex at (0,-4).

Question 7

(a) B1 because the coefficient 2 makes y grow faster for a given x oe
(a) Answer: Because the factor 2 multiplies x² so y increases faster for each x, making the parabola steeper.
(b) M1 substitute x = 1 into 2x²
(b) A1 y = 2 cao
(b) Answer: 2 [2(1)² = 2]
(c) M1 substitute x = 2 into 2x²
(c) A1 y = 8 cao
(c) Answer: 8 [2(2)² = 2*4 = 8]

Question 8

(a) M1 state shift left or right with sign identified
(a) A1 shift left 2 oe
(a) Answer: The graph is shifted 2 units to the left.
(b) M1 identify h = -2 and k = 0 or state (-2,0)
(b) A1 vertex (-2, 0) cao
(b) A1 brief reason: because (x + 2)² has minimum at x = -2
(b) A1 correct coordinates and reason given
(b) Answer: (-2, 0)

Question 9

(a) B1 h = 1 cao
(a) Answer: 1 [h = 0 + 0 + 1 = 1]
(b) M1 substitute t = 1 into -t² + 4t + 1
(b) A1 h = 4 cao
(b) Answer: 4 [-(1)² + 4(1) + 1 = -1 + 4 + 1 = 4]
(c) M1 substitute t = 4 into -t² + 4t + 1
(c) A1 h = 1 cao
(c) Answer: 1 [-(4)² + 4(4) + 1 = -16 + 16 + 1 = 1]
(d) B1 yes, because h = 1 > 0 oe
(d) Answer: Yes. At t = 4, h = 1 metre, so the ball is 1 m above the ground.

Question 10

(a) M1 complete the square or use formula x = -b/(2a)
(a) A1 x-coordinate 1 cao
(a) A1 y-coordinate 4 cao
(a) Answer: (1, 4)
(b) M1 use quadratic formula or factorise correctly
(b) A1 x = -1 and x = 3 shown to 1 d.p or whole numbers given
(b) A1 both intercepts stated (-1, 0) and (3, 0) cao
(b) Answer: x-intercepts at (-1, 0) and (3, 0)

Drawing Quadratic Graphs: Foundation Tier Practice - Worksheets, Questions and Revision