Constructing and comparing linear and exponential models: tables of values, graphs, predictions and the limits of each model (Year 12 Algebraic relationships, 2024 Mathematics Standard 1 syllabus)

Substitute each value of $x$.

Shape. The graph rises slowly at first and then more and more steeply. It crosses the $y$-axis at $1$. This is exponential growth, a curve rather than a straight line.

Marker's note: one mark for the correct table, one for describing the curve as increasingly steep.

(a) Look at the differences. $7 - 4 = 3$, $10 - 7 = 3$, $13 - 10 = 3$, $16 - 13 = 3$. The difference is constant, so the relationship is linear, with $y = 3x + 4$.

(b) Look at the ratios. $\frac{6}{3} = 2$, $\frac{12}{6} = 2$, $\frac{24}{12} = 2$, $\frac{48}{24} = 2$. The ratio is constant, so the relationship is exponential, with $y = 3 \times 2^x$.

Marker's note: one mark for each classification with the reason (constant difference, constant ratio), one for a correct rule.

(a) Substitute.

(Exact values are $2315.25$ and $2431.0125$.)

(b) Compare after $10$ years. Exponential: $2000 \times 1.05^{10} \approx 3258$. Linear: $2000 + 100 \times 10 = 3000$.

The exponential model gives about $258$ more people. The linear model adds the same amount each year, whereas the exponential model adds $5\%$ of an ever larger population, so the gap widens with time.

Marker's note: one mark for the table, one for each value after $10$ years, one for the comparison.

(a) After 7 hours

$N = 100 \times 2^7 = 100 \times 128 = 12\,800$ cells.

(b) First above $10\,000$

Test whole hours: $t = 6$ gives $100 \times 64 = 6400$, and $t = 7$ gives $12\,800$. So the culture first exceeds $10\,000$ after $7$ hours.

(c) Limitation

The culture cannot keep doubling forever, because food and space run out, so the model is only reliable for a limited time.

Marker's note: one mark for $12\,800$, one for checking $t = 6$ and $t = 7$, one for a valid limitation.

(a) Values.

(b) Comparison. Both graphs start at $1000 and are equal after $1$ year ($1060). Plan A is a straight line, growing by $60 each year. Plan B is a curve that gets steeper because interest is earned on the interest as well. From year $2$ onwards Plan B is higher, and the difference grows.

Marker's note: one mark for the two correct Plan A values, one for the two correct Plan B values, one for describing the line against the curve, and one for the reason.

(a) Declining balance

$V(1) = 30\,000 \times 0.85 = \$25\,500.00$. $V(3) = 30\,000 \times 0.85^3 = \$18\,423.75$. $V(5) = 30\,000 \times 0.85^5 \approx \$13\,311.16$.

(b) After $7$ years

Exponential: $30\,000 \times 0.85^7 \approx \$9617.31$. Straight line: $30\,000 - 4500 \times 7 = 30\,000 - 31\,500 = -\$1500$.

(c) Which is sensible

The straight-line model gives a negative value, which is impossible, because it removes the same amount every year regardless of how little the car is worth. The exponential model keeps the value positive and falls by a smaller dollar amount each year, which matches how cars lose value. The exponential model is more sensible over a long period, although it never reaches $0, so it is also only an approximation.

Marker's note: one mark each for the three values in (a), one for both values in (b), and one for a reasoned conclusion in (c).

Test values of $n$. The population has doubled when $1.05^n \ge 2$. Trying values: $1.05^{14} \approx 1.980$, which is less than $2$, and $1.05^{15} \approx 2.079$, which is at least $2$. So it takes $15$ years.

Straight line comparison. A straight-line model adds a fixed amount ($5\%$ of the starting value) every year. It would reach double the starting value after exactly $20$ years, since $20 \times 5\% = 100\%$. The exponential model doubles sooner because each year's $5\%$ is taken from a larger and larger total.

Marker's note: one mark for testing values around the answer, one for $15$, one for the explanation.