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Exponential Graphs Extended

Grade 10 · Functions & Graphs · Cambridge IGCSE 0580 · Age 14–16

Welcome to Exponential & Logarithmic Graphs!

Exponential functions appear everywhere in the real world — from compound interest and population growth to radioactive decay and cooling rates. This topic is central to Cambridge IGCSE 0580 Extended and bridges algebra, graphs, and real-life modelling.

y = abx  |  A = P(1 + r/100)n  |  Half-life: N = N₀ × (½)t/T  |  loga(ax) = x

Learning Objectives

  • Recognise and sketch exponential functions y = aˣ (growth and decay)
  • Identify key features: passes through (0,1), always positive, asymptote y = 0
  • Apply transformations to y = aˣ (shifts, reflections, stretches)
  • Use compound interest formula A = P(1 + r/100)n
  • Model growth and decay using y = abx; find a and b from data
  • Sketch and interpret reciprocal graphs y = k/x (hyperbola)
  • Recognise graph types (exponential, reciprocal, quadratic, cubic) from equations
  • Solve exponential equations graphically and use logarithms as inverses

Exponential Functions

y = aˣ, shape, asymptote, growth vs decay

Transformations

Shifts, reflections of y = aˣ

Compound Interest

A = P(1 + r/100)n

Growth & Decay

y = abx, half-life, finding a and b

Reciprocal Graphs

y = k/x, hyperbola, asymptotes

Logarithms

loga(x) as inverse of aˣ, basic log rules

Learn 1 — Exponential Functions y = aˣ

The Basic Shape of y = aˣ

An exponential function has the form y = aˣ where a is a positive constant (a > 0, a ≠ 1). There are three key features every exponential graph shares:

Key Features of y = aˣ
1. The graph always passes through (0, 1) because a⁰ = 1 for any value of a.
2. The graph is always positive — it never goes below the x-axis. (aˣ > 0 for all x.)
3. The x-axis (y = 0) is a horizontal asymptote — the graph gets closer and closer to it but never touches it.

Growth: a > 1 (e.g. y = 2ˣ)

When the base a > 1, the function increases as x increases. As x → −∞, y → 0 (approaches the asymptote). As x → +∞, y → +∞.

Sketch of y = 2ˣ — key points to plot:
x = −3: y = 2⁻³ = 1/8 = 0.125
x = −2: y = 2⁻² = 1/4 = 0.25
x = −1: y = 2⁻¹ = 1/2 = 0.5
x = 0: y = 2⁰ = 1
x = 1: y = 2¹ = 2
x = 2: y = 2² = 4
x = 3: y = 2³ = 8
The graph rises steeply on the right and hugs the x-axis on the left.

Decay: 0 < a < 1 (e.g. y = (½)ˣ)

When 0 < a < 1, the function decreases as x increases. It falls steeply on the left and approaches y = 0 on the right.

Sketch of y = (½)ˣ — key points:
x = −3: y = (½)⁻³ = 2³ = 8
x = −2: y = (½)⁻² = 4
x = 0: y = 1
x = 1: y = 0.5
x = 2: y = 0.25
x = 3: y = 0.125
Note: y = (½)ˣ is the same as y = 2⁻ˣ — it is a reflection of y = 2ˣ in the y-axis.
Quick sketch rule: For growth (a>1), the left tail hugs y=0 and the right tail shoots up. For decay (0<a<1), it is the mirror image — right tail hugs y=0 and left tail shoots up. Both pass through (0,1).

Transformations of y = aˣ

The standard transformation rules apply to exponential graphs just as they do to all functions.

y = aˣ + c — Vertical translation by c. Asymptote shifts from y = 0 to y = c. Still passes through (0, 1+c).

y = aˣ⁺ᶜ = ax+c — Horizontal translation: shifts left by c (or right if c is negative). Still passes through (−c, 1).

y = −aˣ — Reflection in the x-axis. Graph is now always negative. Asymptote remains y = 0 but curve is below it. Passes through (0, −1).

y = a⁻ˣ — Reflection in the y-axis. Growth becomes decay and vice versa.

y = k · aˣ — Vertical stretch by factor k. Passes through (0, k) instead of (0, 1). Asymptote stays at y = 0.
Example: Describe the transformation from y = 3ˣ to y = 3ˣ − 5
Vertical translation 5 units downward.
New asymptote: y = −5.
y-intercept: (0, 1 − 5) = (0, −4).
Example: Describe the transformation from y = 2ˣ to y = 2ˣ⁺³
Horizontal translation 3 units to the left.
Asymptote unchanged: y = 0.
New y-intercept: x=0 → y = 2³ = 8. So passes through (0, 8).
Do NOT confuse y = 2ˣ⁺³ (horizontal shift) with y = 2ˣ + 3 (vertical shift). In 2ˣ⁺³ the +3 is in the exponent; in 2ˣ + 3 it is added afterwards.

Learn 2 — Exponential Applications

Compound Interest

Money invested at a compound interest rate grows exponentially. The formula is:

A = P(1 + r/100)n

where A = final amount, P = principal (starting amount), r = annual percentage rate, n = number of years.

Example: £2000 invested at 5% per annum compound interest for 3 years.
A = 2000 × (1 + 5/100)³
A = 2000 × (1.05)³
A = 2000 × 1.157625
A = £2315.25
Example: Find how many years to double £500 at 6% compound interest.
1000 = 500 × (1.06)ⁿ
2 = (1.06)ⁿ — solve graphically or by trial: (1.06)¹¹ ≈ 1.898, (1.06)¹² ≈ 2.012
So n = 12 years (first integer value where amount exceeds double).

Population Growth

A population that grows by a fixed percentage each year follows an exponential model:

P = P₀ × (1 + r/100)t
Example: A town has 20 000 people and grows at 3% per year. Find the population after 10 years.
P = 20000 × (1.03)¹⁰ = 20000 × 1.3439 ≈ 26 878

Radioactive Decay and Half-Life

Radioactive material decays exponentially. The half-life T is the time for the amount to halve. The model is:

N = N₀ × (½)t/T

where N₀ is the initial amount, t is elapsed time, T is the half-life.

Example: A sample starts with 80 g and has a half-life of 4 hours. Find the amount remaining after 12 hours.
N = 80 × (½)^(12/4)
N = 80 × (½)³
N = 80 × 1/8 = 10 g
Example: How long until only 5 g remains from an original 160 g sample with half-life 6 years?
5 = 160 × (½)^(t/6)
5/160 = (½)^(t/6) → 1/32 = (½)^(t/6) → (½)⁵ = (½)^(t/6)
So t/6 = 5 → t = 30 years

The General Model y = abx

Any real-world exponential is written as y = abx where a is the y-intercept (value when x=0) and b is the growth/decay factor per unit time.

y = abx  →  a = y-intercept, b = growth factor (b>1 growth, 0<b<1 decay)
Finding a and b from two data points:
Suppose the curve passes through (0, 3) and (2, 12).
From (0, 3): 3 = a × b⁰ = a → a = 3
From (2, 12): 12 = 3 × b² → b² = 4 → b = 2
Equation: y = 3 × 2ˣ
Finding a and b when neither point is (0, y):
Curve passes through (1, 6) and (3, 54).
6 = ab¹ ... (i)
54 = ab³ ... (ii)
Divide (ii) by (i): 54/6 = b² → b² = 9 → b = 3
From (i): 6 = 3a → a = 2
Equation: y = 2 × 3ˣ
When both points have x ≠ 0, divide the two equations to cancel a. This always gives you b first, then substitute back for a.

Learn 3 — Reciprocal Graphs, Graph Types & Logarithms

Reciprocal Graphs: y = k/x

The reciprocal function y = k/x (also written y = kx⁻¹) produces a hyperbola. It has two curved branches and two asymptotes.

Key features of y = k/x:
• Vertical asymptote: x = 0 (the graph never crosses the y-axis)
• Horizontal asymptote: y = 0 (the graph never crosses the x-axis)
• If k > 0: branches in quadrants 1 and 3 (both positive or both negative)
• If k < 0: branches in quadrants 2 and 4
• The graph has rotational symmetry of order 2 about the origin.
Example: Sketch y = 6/x
x = 1: y = 6  |  x = 2: y = 3  |  x = 3: y = 2  |  x = 6: y = 1
x = −1: y = −6  |  x = −2: y = −3  |  x = −6: y = −1
Two smooth curves — one in Q1, one in Q3 — approaching both axes but never touching.
Note: y = k/x is the SAME type of graph as inverse proportion (from the Ratio & Proportion topic). The hyperbola and the inverse proportion graph are identical.

Recognising Graph Types from Equations

In the exam you may be given equations and asked to match them to graphs, or given graphs and asked to identify equations. Learn these shapes:

y = ax + b — Straight line (linear). Constant gradient.

y = ax² + bx + c — Parabola (quadratic). U-shape (a>0) or n-shape (a<0).

y = ax³ + ... — Cubic. S-shaped, can have two turning points.

y = k/x — Hyperbola (reciprocal). Two branches, asymptotes at axes.

y = abˣ or y = aˣ — Exponential. Passes through (0,a), asymptote y=0, always positive (if a>0).

y = a√x — Square root. Starts at origin, increasing curve that flattens out.
Common exam trap: An exponential y = 2ˣ and a power function y = x² look similar for small positive x but behave very differently. The exponential has a horizontal asymptote; the power function does not. The exponential passes through (0,1); the square passes through (0,0).

Solving Exponential Equations Graphically

To solve an equation like 2ˣ = 3x − 1, draw both y = 2ˣ and y = 3x − 1 on the same axes. The x-coordinates of intersection points are the solutions.

Worked approach:
Rearrange so one side is a familiar graph. For 2ˣ = 5:
Draw y = 2ˣ. Draw the horizontal line y = 5. Where they intersect gives x ≈ 2.32.
(Exact answer: x = log₂ 5, which the Extended curriculum introduces.)

Introduction to Logarithms (Extended)

The logarithm is the inverse operation of an exponential. If y = aˣ, then x = loga(y).

loga(aˣ) = x   and   aloga(x) = x
Reading loga(x): "the power you raise a to, in order to get x."
log₂(8) = 3 because 2³ = 8.
log₃(81) = 4 because 3⁴ = 81.
log₁₀(1000) = 3 because 10³ = 1000.
log₅(1) = 0 because 5⁰ = 1 (always true for any base).
Basic log rules (Extended IGCSE level):
loga(xy) = loga(x) + loga(y) — product rule
loga(x/y) = loga(x) − loga(y) — quotient rule
loga(xⁿ) = n · loga(x) — power rule
loga(1) = 0 and loga(a) = 1 for any valid base a.
Graph of y = loga(x):
It is the reflection of y = aˣ in the line y = x.
Passes through (1, 0) — because loga(1) = 0.
Passes through (a, 1) — because loga(a) = 1.
Vertical asymptote at x = 0. Defined only for x > 0.
Using logs to solve 2ˣ = 50: Take log of both sides → x log 2 = log 50 → x = log 50 / log 2 ≈ 5.64. This is the change-of-base formula in action.

Example 1 — Sketching y = 3ˣ and identifying features

Q: Sketch y = 3ˣ for −2 ≤ x ≤ 3. State the asymptote and y-intercept.
M1 — Table of values:
x = −2: 3⁻² = 1/9 ≈ 0.11
x = −1: 3⁻¹ = 1/3 ≈ 0.33
x = 0: 3⁰ = 1
x = 1: 3¹ = 3
x = 2: 3² = 9
x = 3: 3³ = 27
A1: y-intercept: (0, 1). Horizontal asymptote: y = 0. Graph is always positive, increasing left to right, steepening.

Example 2 — Compound Interest

Q: Sarah invests £4500 at 3.5% per annum compound interest. Find the amount after 6 years, correct to the nearest penny.
M1: A = P(1 + r/100)ⁿ = 4500 × (1.035)⁶
M1: (1.035)⁶ = 1.22926...
A1: A = 4500 × 1.22926 = £5531.68

Example 3 — Finding a and b from a graph

Q: A curve of the form y = abx passes through (0, 5) and (3, 40). Find a and b.
B1: Substitute (0, 5): 5 = a × b⁰ = a  →  a = 5
M1: Substitute (3, 40): 40 = 5 × b³  →  b³ = 8  →  b = 2
A1: Equation: y = 5 × 2ˣ

Example 4 — Radioactive Decay / Half-Life

Q: A radioactive substance has a half-life of 5 years. It starts with 200 g. How much remains after 20 years?
M1: N = N₀ × (½)^(t/T) = 200 × (½)^(20/5) = 200 × (½)⁴
M1: (½)⁴ = 1/16
A1: N = 200 × 1/16 = 12.5 g

Example 5 — Reciprocal Graph y = k/x

Q: The graph of y = k/x passes through the point (3, 8). Find k. State the equations of both asymptotes.
M1: Substitute (3, 8): 8 = k/3  →  k = 24
A1: Asymptotes: x = 0 (vertical) and y = 0 (horizontal)

Example 6 — Evaluating Logarithms

Q: Without a calculator, find: (a) log₂(32)   (b) log₃(1/9)   (c) log₅(5)
(a) 2? = 32 → 2⁵ = 32  →  log₂(32) = 5
(b) 3? = 1/9 = 3⁻²  →  log₃(1/9) = −2
(c) 5¹ = 5  →  log₅(5) = 1   (log of the base always = 1) A1

Common Mistakes in Exponential Graphs

These are the errors examiners see most often. Understanding WHY they're wrong helps you avoid them.

Mistake 1 — Thinking the exponential graph touches the x-axis

✗ Wrong: "y = 2ˣ crosses the x-axis somewhere for negative x"
✓ Correct: y = 2ˣ is ALWAYS positive. It gets very close to y = 0 but NEVER touches or crosses it. The x-axis is a horizontal asymptote.

2ˣ > 0 for all real x because a positive number raised to any power is still positive. Even 2⁻¹⁰⁰⁰ is a tiny positive number, not zero.

Mistake 2 — Confusing y = 2ˣ (exponential) with y = x² (quadratic)

✗ Wrong: Both pass through the origin (0, 0) and look similar — treating them as the same type.
✓ Correct: y = 2ˣ passes through (0, 1), not (0, 0). y = x² passes through (0, 0). Their shapes, asymptotes and growth rates are completely different.

The x is the exponent in an exponential (base is fixed, power varies). In y = x², x is the base (base varies, power is fixed). These are completely different functions.

Mistake 3 — Wrong base: confusing growth and decay

✗ Wrong: "y = (0.5)ˣ grows because it has a positive coefficient"
✓ Correct: Base 0 < a < 1 gives DECAY. As x increases, (0.5)ˣ gets smaller. Only base a > 1 gives growth.

Check: (0.5)¹ = 0.5, (0.5)² = 0.25, (0.5)³ = 0.125 — clearly decreasing. Growth requires a > 1 (e.g. base 1.1, 2, 10).

Mistake 4 — Using simple interest instead of compound interest

✗ Wrong: £1000 at 5% for 3 years → A = 1000 + 3 × 50 = £1150
✓ Correct: A = 1000 × (1.05)³ = 1000 × 1.157625 = £1157.63

Simple interest adds the same amount each year. Compound interest multiplies by the factor each year — the interest earns interest. Always use the formula A = P(1 + r/100)ⁿ for compound.

Mistake 5 — Asymptotes of y = k/x: saying "the graph approaches the origin"

✗ Wrong: "The asymptotes of y = 12/x are at the origin (0, 0)"
✓ Correct: The asymptotes are the LINES x = 0 (the y-axis) and y = 0 (the x-axis). They are axes, not a point.

An asymptote is a LINE that the curve approaches. For y = k/x, as x → ∞, y → 0; as x → 0, y → ±∞. The two asymptotes are the coordinate axes themselves.

Mistake 6 — Forgetting that loga(1) = 0, not 1

✗ Wrong: log₂(1) = 1
✓ Correct: log₂(1) = 0 because 2⁰ = 1. loga(a) = 1 (the base itself gives log = 1).

Memorise: loga(1) = 0 always, loga(a) = 1 always. Confusing these two is one of the most common errors on log questions.

Key Formulas — Exponential & Logarithmic Graphs

Formula / FactNotes / When to Use
y = abxGeneral exponential model. a = y-intercept, b = growth/decay factor. b>1: growth; 0<b<1: decay.
A = P(1 + r/100)nCompound interest. P = principal, r = % rate per period, n = number of periods.
N = N₀ × (½)t/THalf-life / radioactive decay. T = half-life, t = elapsed time.
loga(ax) = xInverse relationship. Log undoes the exponential.
aloga(x) = xInverse relationship. Exponential undoes the log.
loga(xy) = loga(x) + loga(y)Product rule. Split log of a product.
loga(x/y) = loga(x) − loga(y)Quotient rule. Split log of a fraction.
loga(xn) = n loga(x)Power rule. Bring down the exponent.
loga(1) = 0Any base: log of 1 is always 0.
loga(a) = 1Any base: log of the base itself is 1.
Key graph: y = aˣ passes through (0,1), always positive, asymptote y = 0
Key graph: y = k/x — hyperbola, asymptotes x = 0 and y = 0, branches in Q1 & Q3 (k>0)
Finding b from two points: divide equations to eliminate a, then solve for b

Graph Shape Summary

y = aˣ (a>1): Exponential growth — rises steeply right, approaches y=0 left.
y = aˣ (0<a<1): Exponential decay — falls steeply right, approaches y=0 right.
y = k/x: Hyperbola — two branches, asymptotes along both axes.
y = loga(x): Reflection of y = aˣ in line y = x. Passes through (1,0), asymptote x = 0.
y = abx: Same shape as y = bˣ but vertically scaled by factor a (y-intercept is a, not 1).

Exponential Graph Plotter — y = abx

Enter values of a and b to plot the curve y = abx. Try a=1, b=2 for y=2ˣ; a=1, b=0.5 for decay; a=3, b=2 for y=3×2ˣ.

Enter a and b, then click "Plot Curve" to begin.

Tip: For y = 2ˣ use a=1, b=2. For y = (½)ˣ use a=1, b=0.5. For compound interest model A = P(1.05)ⁿ use a=P, b=1.05.

Reciprocal Graph Plotter — y = k/x

Enter k to plot the hyperbola y = k/x.

Enter k and click "Plot y = k/x".

Exercise 1 — Evaluating Exponential Expressions

1. Evaluate 2⁵

2. Evaluate 3⁻² (enter as a decimal or fraction — answer to 4 d.p. if needed)

3. Evaluate (½)⁴

4. Evaluate (0.1)³

5. What is the y-intercept of y = 5 × 3ˣ?

6. For y = 4ˣ, what is y when x = 0?

7. For y = 2 × 5ˣ, find y when x = 3.

8. For y = 100 × (0.5)ˣ, find y when x = 4.

Exercise 2 — Compound Interest A = P(1 + r/100)ⁿ

1. £1000 invested at 4% p.a. compound interest for 2 years. Find A (£, to nearest penny).

2. £5000 invested at 3% p.a. compound interest for 5 years. Find A (£, to 2 d.p.).

3. £800 at 6% p.a. compound for 3 years. Find A (£, to 2 d.p.).

4. £2000 at 2.5% p.a. compound for 4 years. Find A (£, to 2 d.p.).

5. A car is bought for £12000. It depreciates at 15% per year. Find its value after 3 years (£, to 2 d.p.).

6. £3000 at 5% p.a. compound for 10 years. Find A (£, to 2 d.p.).

7. Find the compound interest EARNED (not the total amount) on £6000 at 4% for 2 years (£, to 2 d.p.).

8. A population of 50000 grows at 2% per year. Find the population after 5 years (nearest whole number).

Exercise 3 — Finding a and b in y = abx

1. Curve y = abˣ passes through (0, 4) and (1, 12). Find b.

2. Curve y = abˣ passes through (0, 7) and (2, 63). Find b.

3. Curve y = abˣ passes through (0, 10) and (3, 80). Find a and then find b. Enter b.

4. Curve y = abˣ passes through (1, 6) and (2, 18). Find a.

5. Curve y = abˣ passes through (2, 12) and (4, 108). Find b.

6. y = abˣ passes through (0, 5) and (4, 80). Find y when x = 6.

7. The model y = 3 × bˣ passes through (5, 96). Find b.

8. y = abˣ passes through (1, 10) and (3, 250). Find a.

Exercise 4 — Half-Life and Decay

1. Half-life = 10 years, start = 160 g. Amount after 30 years (g)?

2. Half-life = 4 hours, start = 512 g. Amount after 20 hours (g)?

3. Half-life = 2 years. Original sample = 640 g. How many years until 40 g remain?

4. A sample decays by 20% each year. Start = 500 g. Amount after 3 years (to 2 d.p.)?

5. A model is N = 200 × (0.75)ᵗ. Find N when t = 4 (to 2 d.p.).

6. Half-life = 6 hours. What fraction of the original remains after 18 hours?

7. Half-life = 5 days, start = 800 g. Amount after 25 days (g)?

8. A car depreciates at 12% per year. It is worth £9000 now. Find its value in 4 years (£, to 2 d.p.).

Exercise 5 — Reciprocal Graphs & Logarithms

1. y = 12/x. Find y when x = 3.

2. y = k/x passes through (4, 5). Find k.

3. log₂(16) = ?

4. log₃(27) = ?

5. log₅(1) = ?

6. log₄(64) = ?

7. log₂(1/8) = ?

8. y = 30/x. Find x when y = 5.

Practice — 25 Mixed Questions

NC = Non-calculator    C = Calculator allowed

NC 1. What is the y-intercept of y = 7 × 2ˣ?

NC 2. Evaluate (½)⁵.

NC 3. log₂(64) = ?

C 4. £2000 at 5% compound for 3 years. Find A (£, to 2 d.p.).

NC 5. Half-life = 8 years, start = 400 g. Amount after 24 years (g)?

NC 6. y = abˣ passes through (0, 6) and (1, 18). Find b.

C 7. y = 3 × (0.8)ˣ. Find y when x = 5 (to 3 s.f.).

NC 8. y = 20/x. Find y when x = 5.

NC 9. What is the horizontal asymptote of y = 5 × 3ˣ?

C 10. Population 80000 grows at 1.5% per year. Find population after 10 years (nearest whole number).

NC 11. log₁₀(10000) = ?

C 12. £4500 depreciates at 8% per year for 5 years. Find value (£, to 2 d.p.).

NC 13. y = abˣ passes through (2, 20) and (3, 100). Find b.

NC 14. log₃(1/9) = ?

NC 15. Half-life = 3 years, start = 96 g. Amount after 12 years (g)?

NC 16. y = k/x passes through (6, 4). Find k.

C 17. £8000 at 3.5% compound for 6 years. Find A (£, to 2 d.p.).

NC 18. For y = 2ˣ, does the graph pass through (0, 0), (0, 1), or (1, 0)? Enter 0 for (0,0), 1 for (0,1), 2 for (1,0).

NC 19. Evaluate log₆(36).

C 20. y = 5 × 2ˣ. Find y when x = 7.

C 21. Car bought for £15000, depreciates 10% per year. Value after 4 years (£, to 2 d.p.).

NC 22. y = abˣ passes through (0, 3) and (4, 48). Find b.

NC 23. State the two asymptotes of y = 15/x. Enter the y-value of the horizontal asymptote.

C 24. £10000 at 4% p.a. compound for 8 years. Find A (£, to 2 d.p.).

NC 25. Half-life = 10 days, start = 1024 g. After how many days does 32 g remain?

Challenge — 12 Questions (IGCSE Extended Level)

1. y = abˣ passes through (2, 18) and (5, 4374). Find a.

2. £P invested at 6% p.a. compound interest triples in value. Using trial and improvement or logs, find the number of years (nearest whole year).

3. A substance decays by 5% per hour. How many hours until it is below 50% of its original value? (nearest whole hour)

4. Solve graphically (estimate to 1 d.p.): 2ˣ = x + 3. The positive solution for x is approximately ?

5. log₂(x) + log₂(x−2) = 3. Find x. (Use log rules: log₂(x(x−2))=3 → x(x−2)=8)

6. A population doubles every 12 years. Starting at 5000, find the population after 48 years.

7. The graph of y = abx passes through (1, 6) and (4, 162). Find y when x = 6.

8. log₃(2x − 1) = 4. Find x.

9. An investment grows from £1000 to £1500 in 5 years with compound interest. Find the annual rate r (to 2 d.p., enter as a % e.g. 8.45).

10. Simplify: log₅(125) − log₅(5). Answer is a single integer.

11. The curve y = 4 × bˣ passes through (3, 500). Find b (to 3 s.f.).

12. For y = 3 × 2ˣ − 12, what is the equation of the horizontal asymptote? Enter the y-value.

Exam Style Questions

Mark-scheme style. Show working in your book. Enter final answers for self-marking.

Question 1 — Compound Interest [5 marks]

Maria invests £6000 in a bank account. The account pays compound interest at a rate of 3.2% per annum.

(a) Find the amount in the account after 4 years. Give your answer to the nearest penny.
(b) Find the total compound interest earned over the 4 years.

Question 2 — Exponential Model y = abx [6 marks]

The number of bacteria N in a culture is modelled by N = abt, where t is the time in hours.

(a) When t = 0, N = 500. When t = 3, N = 4000. Find the value of a.
(b) Find the value of b. Give your answer to 3 significant figures.
(c) Find the number of bacteria after 5 hours. Give your answer to the nearest whole number.

Question 3 — Half-Life [5 marks]

A radioactive substance has a half-life of 8 years. The initial mass is 320 grams.

(a) Find the mass remaining after 32 years.
(b) Find the mass remaining after 20 years. Give your answer to 2 decimal places.

Question 4 — Reciprocal Graph [4 marks]

The graph of y = k/x passes through the point (4, 7.5).

(a) Find the value of k.
(b) Write down the equations of the two asymptotes of the graph.
Enter the y-value of the horizontal asymptote:
(c) Find the value of x when y = 6.

Question 5 — Logarithms & Transformations [5 marks]

Consider the function y = 2ˣ.

(a) Evaluate log₂(128).
(b) The graph of y = 2ˣ is translated 3 units upward. Write down the equation of the new asymptote.
Enter the y-value:
(c) The graph of y = 2ˣ is translated 2 units to the right to give y = 2^(x−2). Find the y-intercept of the new graph (y-value when x=0).