Exponential functions and graphs (VCE Maths Methods Unit 2)

What this dot point is asking

VCAA wants you to graph exponential functions, identify their key features (intercepts, asymptote, domain, range), and apply standard transformations.

Parent function $y = b^x$

For $b > 1$:

• Domain: all real $x$. Range: $y > 0$.
• y-intercept: $(0, 1)$.
• Horizontal asymptote: $y = 0$ as $x \to -\infty$.
• Increasing, concave up.

For $0 < b < 1$:

• Same domain, range, intercept and asymptote.
• Decreasing, concave up.

Transformed form $y = a b^{x - h} + k$

• $a$ vertical dilation by factor $|a|$ (reflection in $x$-axis if $a < 0$).
• $h$ horizontal shift right by $h$ units.
• $k$ vertical shift up by $k$ units; new asymptote $y = k$.

Key features

y-intercept: $f(0) = a b^{-h} + k$.

Horizontal asymptote: $y = k$ (always).

Range: $y > k$ (if $a > 0$) or $y < k$ (if $a < 0$).

Solving exponential equations

If both sides can be rewritten in the same base, equate exponents. Otherwise take logs (next dot point).

The base $e$ and growth and decay models

The number $e \approx 2.718$ is the natural base for continuous growth and decay, written $y = A e^{kt}$. When $k > 0$ the quantity grows; when $k < 0$ it decays. The constant $A$ is the initial value $y(0)$, and the rate constant $k$ controls how fast the quantity changes. Two staples follow: the doubling time of a growing quantity is $\dfrac{\ln 2}{k}$, and the half-life of a decaying quantity is $\dfrac{\ln 2}{|k|}$. Solving these requires taking natural logarithms of both sides, which is why exponential modelling and logarithms are taught together.

Solving exponential equations

There are two routes. If both sides can be written to the same base, equate the exponents (for example $3^{x-1} = 9 = 3^2$ gives $x - 1 = 2$). If they cannot, take logarithms of both sides and use the power law to bring the exponent down, then divide. Real modelling problems almost always need the logarithm route, since the target value is rarely an exact power of the base. Always isolate the exponential term before applying logs.

In one sentence

The exponential function $y = ab^{x-h} + k$ has horizontal asymptote $y = k$, y-intercept $ab^{-h} + k$, domain all real $x$, and range $y > k$ if $a > 0$ (or $y < k$ if $a < 0$); the parent $y = b^x$ is increasing for $b > 1$ and decreasing for $0 < b < 1$.

Examples in context

Example 1. Drug concentration decay. A medication's blood concentration is modelled by $C = 80(0.5)^{t} + 4$ mg/L, where $t$ is hours and the $0.5$ base gives a one-hour halving. The horizontal asymptote is $C = 4$ mg/L (a residual baseline), the initial concentration is $C(0) = 80 + 4 = 84$ mg/L, and the range is $C > 4$. The curve is decreasing since the base is between $0$ and $1$.

Example 2. Investment growth. A balance grows as $A = 1000(1.05)^{x}$ dollars after $x$ years. This has $y$-intercept $\$1000$ (the initial deposit), asymptote $A = 0$, and is increasing since $1.05 > 1$. To reach $\$1276$, solve $1.05^x = 1.276$; since $1.05^5 = 1.2763$, it takes about $5$ years.

Try this

Q1. State the $y$-intercept, asymptote and range of $y = 3(2)^x - 1$. [3 marks]

• Cue. $y$-intercept $3(1) - 1 = 2$; asymptote $y = -1$; range $y > -1$.

Q2. Solve $5 \cdot 2^{x-1} + 3 = 23$. [3 marks]

• Cue. $2^{x-1} = 4 = 2^2$, so $x - 1 = 2$, giving $x = 3$.

Q3. A population is $P = 200(1.5)^{t}$. (a) State the initial population. (b) Find $P$ after $3$ time units. [1+2 marks]

• Cue. (a) $P(0) = 200$. (b) $P(3) = 200(1.5)^3 = 200 \times 3.375 = 675$.