← Unit 2: Functions, calculus and probability

VICMath MethodsSyllabus dot point

How are exponential functions analysed?

Sketch and analyse exponential functions of the form $y = a \cdot b^{x - h} + k$, identifying key features (intercepts, asymptote, domain, range) and applying transformations

A focused answer to the VCE Maths Methods Unit 2 dot point on exponential functions. Sketches $y = b^x$ for $b > 1$ and $0 < b < 1$, identifies the y-intercept, horizontal asymptote, domain and range, and works the VCAA SAC-style transformation problem.

Generated by Claude OpusReviewed by Better Tuition Academy4 min answer

Have a quick question? Jump to the Q&A page

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 IMATH_0

For b>1b > 1:

  • Domain: all real xx. Range: y>0y > 0.
  • y-intercept: (0,1)(0, 1).
  • Horizontal asymptote: y=0y = 0 as xβ†’βˆ’βˆžx \to -\infty.
  • Increasing, concave up.

For 0<b<10 < b < 1:

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

Transformed form IMATH_8

  • IMATH_9 vertical dilation by factor ∣a∣|a| (reflection in xx-axis if a<0a < 0).
  • IMATH_13 horizontal shift right by hh units.
  • IMATH_15 vertical shift up by kk units; new asymptote y=ky = k.

Key features

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

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

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

Solving exponential equations

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

Worked example

Sketch y=βˆ’2β‹…3xβˆ’2+6y = -2 \cdot 3^{x - 2} + 6.

Parameters: a=βˆ’2a = -2, h=2h = 2, k=6k = 6.

  • Asymptote: y=6y = 6, approached from below.
  • y-intercept: βˆ’2β‹…3βˆ’2+6=βˆ’2/9+6=5.78-2 \cdot 3^{-2} + 6 = -2/9 + 6 = 5.78.
  • As xβ†’+∞x \to +\infty, 3xβˆ’2β†’βˆž3^{x-2} \to \infty, so βˆ’2β‹…3xβˆ’2β†’βˆ’βˆž-2 \cdot 3^{x-2} \to -\infty and the graph descends.
  • Decreasing throughout.

Common traps

Direction of horizontal translation. y=bxβˆ’2y = b^{x-2} shifts right by 22, not left.

Forgetting to update the asymptote. When adding kk, the asymptote moves with it.

Mixing the order of transformations. Apply horizontal shift inside the exponent, then dilation, then vertical shift.

In one sentence

The exponential function y=abxβˆ’h+ky = ab^{x-h} + k has horizontal asymptote y=ky = k, y-intercept abβˆ’h+kab^{-h} + k, domain all real xx, and range y>ky > k if a>0a > 0 (or y<ky < k if a<0a < 0); the parent y=bxy = b^x is increasing for b>1b > 1 and decreasing for 0<b<10 < b < 1.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksFor $f(x) = 2 \cdot 3^{x-1} + 5$, find (a) the y-intercept, (b) the horizontal asymptote, (c) the value of $x$ for which $f(x) = 23$.
Show worked answer β†’

(a) y-intercept. f(0)=2β‹…3βˆ’1+5=2/3+5=17/3β‰ˆ5.67f(0) = 2 \cdot 3^{-1} + 5 = 2/3 + 5 = 17/3 \approx 5.67.

(b) Asymptote. As xβ†’βˆ’βˆžx \to -\infty, 3xβˆ’1β†’03^{x-1} \to 0, so f(x)β†’5f(x) \to 5. Asymptote: y=5y = 5.

(c) Solve f(x)=23f(x) = 23.

2β‹…3xβˆ’1+5=23β‡’2β‹…3xβˆ’1=18β‡’3xβˆ’1=9=322 \cdot 3^{x-1} + 5 = 23 \Rightarrow 2 \cdot 3^{x-1} = 18 \Rightarrow 3^{x-1} = 9 = 3^2.

xβˆ’1=2x - 1 = 2, so x=3x = 3.

Markers reward substitution, identification of the asymptote from the constant, and equating powers.

Related dot points