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Drawing quadratic graphs Quiz

Grade 5 • 5J • 10 Questions (30 Marks)

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Drawing quadratic graphs — Notes & Worked Examples Show / Hide

Explanation:

A quadratic graph y = ax2 + bx + c is a smooth, symmetrical U-shaped curve called a parabola (or ∩-shaped if a is negative). When substituting negative values into x2, always use brackets: squaring a negative number gives a positive answer, e.g. (−2)2 = +4.

Plot each (x, y) point from your table and join them with a single smooth curve (never connect points with straight ruler lines). The bottom of the curve is the minimum turning point, and the points where it crosses the x-axis (y = 0) are the roots.

Example 1: Complete the table for y = x2 − 2x − 3.

Solution:

x−2−101234
y50−3−4−305
For x = −2: (−2)2 − 2(−2) − 3 = 4 + 4 − 3 = 5
Notice the symmetrical y-values around (1, −4).

Example 2: From the graph of y = x2 − 2x − 3, find (a) the turning point and (b) the roots of x2 − 2x − 3 = 0.

Solution:

(a) Minimum point on curve: (1, −4)
(b) Crosses y = 0 at x = −1 and 3
Ans: (1, −4); x = −1, 3
x y (−1,0) (3,0) (1,−4)
Question 1
[2 marks]
x–3–2–10123y5–3–3–2–1123–5–4–3–2–11234567891011120xy
Complete the table of values for:
y = x2 - 2x - 3
On the grid, draw the graph of y = x2 - 2x - 3 for values of x from -3 to 3.
Worked Solution & Answer
Substitute each x-value into the equation (use brackets around negative numbers).
y = 12, 5, 0, -3, -4, -3, 0
x–3–2–10123y1250–3–4–30–3–2–1123–5–4–3–2–11234567891011120xy
Answer: y = 12, 5, 0, -3, -4, -3, 0
Question 2
[2 marks]
x–3–2–10123y–35–3–2–1123–5–4–3–2–11234567891011120xy
Complete the table of values for:
y = x2 + 2x - 3
On the grid, draw the graph of y = x2 + 2x - 3 for values of x from -3 to 3.
Worked Solution & Answer
Substitute each x-value into the equation (use brackets around negative numbers).
y = 0, -3, -4, -3, 0, 5, 12
x–3–2–10123y0–3–4–30512–3–2–1123–5–4–3–2–11234567891011120xy
Answer: y = 0, -3, -4, -3, 0, 5, 12
Question 3
[2 marks]
Here is the graph of y = x2 - 2x - 2
xy–2–11234–4–3–2–11234O
(a) Write down the coordinates of the turning point on the graph of y = x2 - 2x - 2 [1 mark]
(b) Write down an estimate for one of the roots of x2 - 2x - 2 = 0 [1 mark]
Worked Solution & Answer
(a)
Minimum at (1, –3) B1
Answer (a): (1, –3)
(b)
x-intercepts at x ≈ –0.7 and x ≈ 2.7 B1
Answer (b): –0.7 or 2.7
Question Paper (PDF) • Mark Scheme (PDF)
Question 4 AQA Nov 2020 2F Q22 / 2H Q7
[3 marks]
Here is the graph of y = x2 - 7x + 10 for values of x from 0 to 7
0xy1234567–4–2246810
(a) Write down the roots of x2 - 7x + 10 = 0 [2 marks]
Answer
(b) Write down the x-coordinate of the turning point of the curve. [1 mark]
Answer
Worked Solution & Answer
(a)
The roots occur where the curve crosses the x-axis: x = 2 and x = 5. M1 A1
Answer (a): Answer x = 2 and x = 5
(b)
The line of symmetry of the parabola is halfway between the roots: x = 2 + 52 = 3.5. B1
Answer (b): Answer 3.5
Question Paper (PDF) • Mark Scheme (PDF)
Question 5 AQA Jun 2022 3F Q23 / 3H Q6
[3 marks]
Here is a sketch of the curve y = x2 - 4x - 5
xyO–15–5T
(a) Write down the two roots of x2 - 4x - 5 = 0 [1 mark]
(b) Work out the coordinates of T, the turning point of the curve. [2 marks]
Worked Solution & Answer
(a)
The roots are the x-intercepts where y = 0:
From the sketch, x = -1 and x = 5. B1
Answer (a): -1, 5
(b)
The turning point lies on the line of symmetry: M1
x = -1 + 52 = 2   (or x = -b2a = --42 = 2)
Substitute x = 2 into the quadratic equation:
y = 22 - 4(2) - 5 = 4 - 8 - 5 = -9
Coordinates of T are (2, -9). A1
Answer (b): (2, -9)
Question Paper (PDF) • Mark Scheme (PDF)
Question 6
[3 marks]
Here is the graph of y = x2 - 2x - 3
xy–2–11234–4–2246O
(a) Write down the coordinates of the turning point on the graph of y = x2 - 2x - 3 [1 mark]
(b) Use the graph to find the roots of the equation x2 - 2x - 3 = 0 [2 marks]
Worked Solution & Answer
(a)
Minimum point on parabola is at (1, –4) B1
Answer (a): (1, -4)
(b)
Graph crosses x-axis at x = –1 and x = 3 M1 A1
Answer (b): –1 and 3
Question Paper (PDF) • Mark Scheme (PDF)
Question 7 AQA Nov 2021 3F Q15
[3 marks]
y = x2 - 2
(a) Complete the table of values for y = x2 - 2
@@TABULAR0@@ [1 mark]
(b) Draw the graph of y = x2 - 2 for values of x from -3 to 3 [2 marks]
–3–2–1123–3–2–112345678910xyO
Worked Solution & Answer
(a)
For x = -3: (-3)2 - 2 = 9 - 2 = 7
For x = 2: 22 - 2 = 4 - 2 = 2
For x = 3: 32 - 2 = 9 - 2 = 7 B1
Answer (a): When x = -3, y = 7; when x = 2, y = 2; when x = 3, y = 7
(b)
Plot the 7 coordinates from the table and join with a smooth symmetrical U-shaped curve. M1 A1
–3–2–1123–3–2–112345678910xyO
Answer (b): Answer Smooth parabolic curve plotted through (-3,7), (-2,2), (-1,-1), (0,-2), (1,-1), (2,2), (3,7)
Question Paper (PDF) • Mark Scheme (PDF)
Question 8
[3 marks]
The graph of y = f(x) is drawn on the grid.
–2–11234–4–3–2–1123456Oyx
(a) Write down the coordinates of the turning point of the graph. [1 mark]
(b) Write down estimates for the roots of f(x) = 0 [1 mark]
(c) Use the graph to find an estimate for f(1.5) [1 mark]
Worked Solution & Answer
(a)
Minimum point is at (1, –3) B1
Answer (a): 1, -3
(b)
Graph crosses y = 0 at x ≈ –0.75 and x ≈ 2.75 B1
Answer (b): -0.75, 2.75
(c)
At x = 1.5, y ≈ –2.8 B1
Answer (c): -2.8
Question Paper (PDF) • Mark Scheme (PDF)
Question 9 IGCSE Jun 2023 2H Q4
[4 marks]
20 (a) Complete table of values for y = x² – x – 4. (2) (b) Draw the graph of y = x² – x – 4 for –3 ≤ x ≤ 3. (2)
(a) Complete the table of values for y = x2 - x - 4 [2 marks]
xy–3–22–101–423
(b) On the grid below, draw the graph of y = x2 - x - 4 for values of x from -3 to 3 [2 marks]
–3–2–1123–5–4–3–2–112345678910Oxy
Worked Solution & Answer
(a)
At x = –3: 9 – (–3) – 4 = 8 M1
x = –1: 1 – (–1) – 4 = –2
x = 0: –4
x = 2: 4 – 2 – 4 = –2
x = 3: 9 – 3 – 4 = 2. A1
xy–38–22–1–20–41–42–232
Answer (a): 8, (2), –2, –4, (–4), –2, 2
(b)
At x = –3: 9 – (–3) – 4 = 8 M1
x = –1: 1 – (–1) – 4 = –2
x = 0: –4
x = 2: 4 – 2 – 4 = –2
x = 3: 9 – 3 – 4 = 2. A1
–3–2–1123–5–4–3–2–112345678910Oxy
Answer (b): Smooth parabola plotted through (–3, 8), (–2, 2), (–1, –2), (0, –4), (1, –4), (2, –2), (3, 2)
Question Paper (PDF) • Mark Scheme (PDF)
Question 10
[5 marks]
y = x² + x – 4
(a) Complete the table of values for x from –3 to 2. (2)
(b) Draw the graph on the grid. (2)
(c) Write down the coordinates of the turning point of the graph. (1)
(a) Complete the table of values for y = x2 + x - 4
@@TABULAR0@@ [2 marks]
(b) On the grid, draw the graph of y = x2 + x - 4 for values of x from –3 to 2 [2 marks]
xyO–3–2–112–5–4–3–2–1123
(c) Write down the coordinates of the turning point of the graph of y = x2 + x - 4 [1 mark]
Worked Solution & Answer
(a)
x = –2: 4 – 2 – 4 = –2 M1
x = 0: –4
x = 1: 1 + 1 – 4 = –2
x = 2: 4 + 2 – 4 = 2 A1
Answer (a): (2), –2, (–4), –4, –2, 2
(b)
Graph drawn M1 A1
xyO–3–2–112–5–4–3–2–1123
Answer (b): Graph drawn
(c)
Turning point is at x = –0.5, y = (–0.5)² + (–0.5) – 4 = 0.25 – 0.5 – 4 = –4.25 B1
Answer (c): (–0.5, –4.25)
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 30 marks)
Drawing quadratic graphs Worksheets Next: Cubic/reciprocal graphs →

Grade 5 Quizzes

25 Topics
5A - Combining ratios5B - Direct and inverse proportion
5C - Reverse percentages5D - Standard form
5E - Speed5F - Changing the subject
5G - Quadratic expressions5H - Solving quadratic equations
5J - Drawing quadratic graphs5K - Cubic/reciprocal graphs
5L - Simultaneous equations5M - Sector areas and arc lengths
5N - Sim. equ. graphically5P - Midpoint of a line segment
5Q - Gradient of a line5R - Equation of a line
5S - Spheres and cones5T - Similar shapes (lengths)
5U - Trigonometry (SOHCAHTOA)5V - Exact trig. values
5W - Probability trees5X - Vectors
5Y - Venn diagrams5Z - Density
5AA - Pressure