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Trigonometric & exp. Graphs Quiz

Grade 7 • 7G • 10 Questions (27 Marks)

Score: 0 / 10
Trigonometric & exp. Graphs — Notes & Worked Examples Show / Hide

Explanation:

Both y = sin x and y = cos x oscillate between −1 and 1 with a period of 360°: sin x starts at (0°, 0), peaks at (90°, 1), crosses at (180°, 0), reaches (270°, −1), and returns to (360°, 0); cos x starts at (0°, 1) and is symmetrical about x = 180°. y = tan x has period 180° with vertical asymptotes at 90° and 270°.

Exponential graphs y = abx (b > 0) always pass through (0, a) because b0 = 1, and have the x-axis (y = 0) as a horizontal asymptote. When b > 1 the graph shows exponential growth; when 0 < b < 1 it shows exponential decay.

Example 1: Solve sin x = −0.5 for 0° ≤ x ≤ 360°.

1 −0.5 −1 90° 180° 270° 360° 210° 330°

Solution:

Base acute angle: sin−1(0.5) = 30°
sin x is negative in 180° < x < 360° (by wave symmetry):
x = 180° + 30° = 210° and x = 360° − 30° = 330°
Answer: x = 210° and x = 330°

Example 2: The curve y = abx passes through (0, 5) and (3, 40). Find a and b.

(0, 5) (3, 40) y = abx 3 40

Solution:

Substitute (0, 5): 5 = a × b0 ⇒ a = 5
Substitute (3, 40) into y = 5bx: 40 = 5b3
b3 = 405 = 8 ⇒ b = ∛8 = 2
Answer: a = 5, b = 2
Question 1
[1 mark]
Sana needs to draw the graph of y = 3x for 0 ≤ x ≤ 4
She draws the graph shown on the grid.
yx12345102030405060708090O
Write down one thing Sana has done wrong.
Worked Solution & Answer
When x = 0, y = 3⁰ = 1, so the y-intercept should be (0, 1) C1
yx12345102030405060708090O
Answer: The graph should cross the y-axis at (0, 1), not (0, 0), because 3⁰ = 1
Question 2
[2 marks]
Here is a sketch of part of the graph of y = kx where k is a positive constant.
The graph passes through the point with coordinates (-1, 1.5)
xyO(–1, 1.5)
Find the value of k.
k =
Worked Solution & Answer
Substitute (–1, 1.5): 1.5 = k⁻¹ M1
⇒ 1/k = 3/2
⇒ k = 2/3. A1
Answer: k = 23
Question 3
[2 marks]
Sketch the graph of y = \tan x° for 0 ≤ x ≤ 360
yx090180270360
Worked Solution & Answer
Asymptotes at 90° and 270° M1
passes through (0, 0), (180, 0), (360, 0) with period 180°. A1
yx090180270360
Answer: Tangent curve with vertical asymptotes at x = 90° and x = 270° and x-intercepts at 0°, 180°, 360°
Question 4
[2 marks]
On the grid, sketch the curve with equation y = 2x
Give the coordinates of any points of intersection with the axes.
xyO
Worked Solution & Answer
Passes through (0, 1) on the y-axis M1
approaches y = 0 as x → –∞ and increases rapidly for x > 0. A1
xyO(0, 1)
Answer: Exponential curve increasing above the x-axis through (0, 1)
Question 5 AQA Jun 2019 3H Q23
[2 marks]
Here is a sketch of the curve y = 2x
xyO1
On the axes above, sketch the curve y = 3x
Worked Solution & Answer
y = 3x also passes through (0, 1) M1
For x > 0, 3x > 2x, so the curve is steeper and lies above y = 2x
For x < 0, 3x < 2x, so the curve is closer to the x-axis and lies below y = 2x. A1
xyO1y = 3x
Answer: Exponential curve passing through (0, 1), steeper for x > 0 and below for x < 0
Question 6 AQA Nov 2024 3H Q23
[3 marks]
The distance of a particle from a point is d metres after t seconds.
@@TABULAR0@@
012345101520253035404550td
Work out the values of a and b.
Worked Solution & Answer
At t = 0, d = 3 ⇒ a × b⁰ = 3 ⇒ a = 3. M2
At t = 1, d = 6 ⇒ 3 × b¹ = 6 ⇒ b = 2. A1
Answer: a = 3, b = 2
Question 7
[3 marks]
Here is the graph of y = \sin x° for -180 ≤ x ≤ 180
1–1O–180–9090180yx
(a) Use the graph to find estimates for the solutions of
\sin x° = 0.3 for -180 ≤ x ≤ 180
[2 marks]
(b) Write down a value of x such that
\sin(x + 20)° = 0 for -180 ≤ x ≤ 180
[1 mark]
x =
Worked Solution & Answer
(a)
Read x at y = 0.3: x ≈ 18 and x ≈ 162 M1 A1
Answer (a): 18 and 162
(b)
x + 20 = 0 or 180 ⇒ x = –20 or 160 B1
Answer (b): x = -20 or 160
Question 8 AQA Jun 2025 1H Q14
[4 marks]
(a) Complete table of values for y = 2x. [2 marks]
14 (b) Draw the graph of y = 2x for x from –1 to 3. [2 marks]
(a) Complete the table of values for y = 2x
@@TABULAR0@@
[2 marks]
(b) Draw the graph of y = 2x for values of x from -1 to 3 [2 marks]
–1123–2–112345678910Oxy
Worked Solution & Answer
(a)
2⁻¹ = 1/2 = 0.5 M1
2¹ = 2
2³ = 8 A1
Answer (a): x = –1: 0.5 (or 1/2); x = 1: 2; x = 3: 8
(b)
Plot all points and connect with a smooth increasing curve M1 A1
–1123–2–112345678910Oxy
Answer (b): Smooth curve drawn through (–1, 0.5), (0, 1), (1, 2), (2, 4), (3, 8)
Question 9
[4 marks]
(a) Sketch the graph of y = sin x° for 0 ≤ x ≤ 360. (2) (b) Solve 2 sin x° = –1 for 0 ≤ x ≤ 360. (2)
(a) Sketch the graph of y = \sin x° for 0 ≤ x ≤ 360 [2 marks]
xyO90180270360
(b) Solve the equation 2\sin x° = -1 for 0 ≤ x ≤ 360 [2 marks]
Worked Solution & Answer
(a)
Sine wave from (0,0) to peak (90,1), through (180,0), trough (270,–1), to (360,0) with 1 and –1 labelled on y-axis M1 A1
xyO901802703601–1
Answer (a): Sine wave from (0,0) to peak (90,1), through (180,0), trough (270,–1), to (360,0) with 1 and –1 labelled on y-axis
(b)
sin x° = –1/2. Since sin 30° = 1/2, in 3rd and 4th quadrants x = 180 + 30 = 210 and x = 360 – 30 = 330 M1 A1
Answer (b): 210, 330
Question 10 IGCSE Specimen 2H Q16
[4 marks]
The diagram shows part of a sketch of the curve y = \sin x°
PQOyx
(a) (i) the point P
(ii) the point Q
[2 marks]
(b) Sketch the graph of y = \tan x for 0° ≤ x ≤ 360°
Show the coordinates of any points of intersection with the coordinate axes.
[2 marks]
Oyx
Worked Solution & Answer
(a)
sin x° crosses the x-axis at P(180, 0) and reaches its minimum at Q(270, –1) M1 A1
Answer (a): (i) (180, 0), (ii) (270, –1)
(b)
tan x has vertical asymptotes at x = 90° and x = 270° and crosses the axes at (0, 0), (180, 0) and (360, 0) M1 A1
Oyx(180, 0)(360, 0)
Answer (b): y = tan x sketched with asymptotes at 90° and 270° and x-intercepts at (0, 0), (180, 0), (360, 0)
Score: 0 / 10 (0 / 27 marks)