Applications (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Bacterial Population Growth

To sketch this graph, we first identify the key features using the form $y = k(a^x)$:

For $N = 200(3^t)$, we have $k = 200$ and $a = 3$.

Finding key features:

• Asymptote: The horizontal asymptote is at $N = 0$ (the number of cells can only approach zero as time goes backwards towards negative infinity, but time cannot actually be negative in this context)

• y-intercept: When $t = 0$, we calculate: $N = 200(3^0) = 200(1) = 200$

• Domain: Since time cannot be negative in this context, $t \geq 0$

• Range: Since the population starts at $200$ and grows without bound, $N > 0$ (or more specifically, $N \geq 200$ for the practical domain)

The graph shows exponential growth starting from $200$ cells at $t = 0$ and increasing rapidly as time progresses.

Worked Example: Sketching a Compound Interest Graph

Strategy:

To sketch an exponential graph, we need to:

• Find the y-intercept by setting $t = 0$
• Identify the horizontal asymptote for a function of the form $y = k(a^x)$, which is $y = 0$
• Calculate an additional point by substituting a value for $t$ (such as $t = 3$)
• Plot these points and the asymptote, then draw the curve

Solution:

The y-intercept occurs when $t = 0$:

$A = 5000 \times 1.03^0 = 5000 \times 1 = 5000$

So the point is $(0, 5000)$.

The horizontal asymptote is $A = 0$.

For an additional point, we substitute $t = 3$:

$A = 5000(1.03^3) \approx 5463.63$

The point is approximately $(3, 5464)$.

The graph shows a positive y-intercept and increases over time, which makes sense for a savings account earning interest. All key features are clearly labelled on the graph.

Worked Example: Analyzing End Behaviour

Strategy:

To analyze end behaviour, we examine what happens to the function $A = 5000(1.03^t)$ as $t$ becomes very large. Since the base $1.03$ is greater than $1$, this represents exponential growth.

Solution:

As $t \to \infty$, the term $1.03^t$ increases without bound. Since we're multiplying this ever-increasing value by $5000$, the account balance $A$ also tends to infinity.

The end behaviour is: as $t \to \infty$, $A \to \infty$.

This makes sense in the real world, as a savings account with compound interest will continue to grow indefinitely over time (assuming no withdrawals are made).

Worked Example: Calculating Account Balance

Strategy:

We substitute $t = 3$ into the formula $A = 5000(1.03^t)$ and evaluate.

Solution:

Checking our answer:

The balance of $5464 is greater than the initial $5000, which makes sense because the account earns interest. This value also matches the point we calculated when sketching the graph.