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QLDMath MethodsSyllabus dot point

Topic 1: Exponential functions

Graph and analyse exponential functions of the form $y = a \cdot b^x + c$, identifying key features (intercepts, asymptote, domain, range) and applying transformations

A focused answer to the QCE Math 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 QCAA-style transformation problem $y = 3 \cdot 2^{x-1} - 4$.

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What this dot point is asking

QCAA wants you to graph exponential functions, identify their key features, and apply standard transformations (vertical and horizontal translations, vertical dilations, reflections) to the parent function y=bxy = b^x.

The parent function IMATH_1

For base b>1b > 1 (for example y=2xy = 2^x):

  • 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 function. Concave up.

For 0<b<10 < b < 1 (for example y=(1/2)xy = (1/2)^x):

  • Same domain, range, intercept, asymptote.
  • Decreasing function. Concave up.

The function value doubles every fixed step in xx (for y=2xy = 2^x, between any two values of xx differing by 11).

Transformations of IMATH_15

Term Effect
IMATH_16 Vertical dilation by factor aa (and reflection if a<0a < 0)
IMATH_19 Horizontal translation: graph moves hh units right
IMATH_21 Vertical translation: graph moves kk units up; new horizontal asymptote is IMATH_23

y-intercept. Substitute x=0x = 0: y=abβˆ’h+ky = a b^{-h} + k.

Horizontal asymptote. Always y=ky = k (the value the function approaches as the exponent goes to negative infinity for b>1b > 1, or positive infinity for b<1b < 1).

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

Solving graphically and algebraically

To solve aβ‹…bxβˆ’h+k=ma \cdot b^{x-h} + k = m, isolate the exponential and either rewrite both sides with the same base or take logarithms (next dot point). Graphically, the solution is the xx-coordinate where the curve crosses y=my = m.

Worked example

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

  • Reflection: a=βˆ’2a = -2 flips the graph vertically.
  • Horizontal shift: h=2h = 2 shifts the curve 22 units right.
  • Vertical shift: k=6k = 6 shifts the curve 66 units up.
  • Horizontal asymptote: y=6y = 6, approached from below.
  • y-intercept: y=βˆ’2β‹…3βˆ’2+6=βˆ’2/9+6=5.778y = -2 \cdot 3^{-2} + 6 = -2/9 + 6 = 5.778.
  • As xβ†’+∞x \to +\infty, 3xβˆ’2β†’+∞3^{x-2} \to +\infty and βˆ’2β‹…3xβˆ’2β†’βˆ’βˆž-2 \cdot 3^{x-2} \to -\infty, so yβ†’βˆ’βˆžy \to -\infty.
  • Decreasing throughout.

Common traps

Confusing the direction of horizontal translation. y=bxβˆ’2y = b^{x-2} moves the graph right 22 units (not left). The graph shifts in the same direction as hh.

Forgetting to update the asymptote. When you add a constant, the asymptote moves. Failing to update the asymptote in a sketch loses marks.

Mixing the order of transformations. Apply horizontal shift inside the exponent first, then dilations, then vertical shift. Order matters when combining.

Treating the asymptote as a point on the graph. The graph approaches the asymptote but never touches it.

In one sentence

The exponential function y=abxβˆ’h+ky = a b^{x-h} + k has horizontal asymptote y=ky = k, y-intercept abβˆ’h+ka b^{-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, and standard transformations apply.

Past exam questions, worked

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

Year 11 SAC4 marksFor the function $f(x) = 3 \cdot 2^{x-1} - 4$, find (a) the y-intercept, (b) the horizontal asymptote, (c) the value of $x$ when $f(x) = 8$.
Show worked answer β†’

(a) y-intercept. Substitute x=0x = 0.

f(0)=3β‹…2βˆ’1βˆ’4=1.5βˆ’4=βˆ’2.5f(0) = 3 \cdot 2^{-1} - 4 = 1.5 - 4 = -2.5.

(b) Horizontal asymptote. As xβ†’βˆ’βˆžx \to -\infty, 2xβˆ’1β†’02^{x-1} \to 0, so f(x)β†’βˆ’4f(x) \to -4.

Asymptote: y=βˆ’4y = -4.

(c) Solve f(x)=8f(x) = 8.

3β‹…2xβˆ’1βˆ’4=83 \cdot 2^{x-1} - 4 = 8

3β‹…2xβˆ’1=123 \cdot 2^{x-1} = 12

2xβˆ’1=4=222^{x-1} = 4 = 2^2

xβˆ’1=2x - 1 = 2, so x=3x = 3.

Markers reward correct substitution, identification of the asymptote from the constant term, and equating powers of 22.

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