Skip to main content
QLDMath MethodsSyllabus dot point

Topic 1: Exponential functions

Graph and analyse exponential functions of the form y=abx+cy = 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=bxy = b^x for b>1b > 1 and 0<b<10 < b < 1, identifies the y-intercept, horizontal asymptote, domain and range, and works the QCAA-style transformation problem y=32x14y = 3 \cdot 2^{x-1} - 4.

Generated by Claude Opus 4.87 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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

Jump to a section
  1. What this dot point is asking
  2. The parent function y=bxy = b^x
  3. Transformations of y=abxh+ky = a \cdot b^{x - h} + k
  4. Reading the graph's features
  5. Why the asymptote sits at y=ky = k
  6. Solving graphically and algebraically

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 y=bxy = b^x

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 xx \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 multiplies by the base bb over each unit step in xx, so y=2xy = 2^x doubles whenever xx increases by 11. This constant-ratio behaviour is what distinguishes exponential growth from the constant-difference behaviour of a linear function, and it is the reason exponential models eventually outgrow any polynomial.

Transformations of y=abxh+ky = a \cdot b^{x - h} + k

Term Effect
aa Vertical dilation by factor aa (and reflection if a<0a < 0)
hh Horizontal translation: graph moves hh units right
kk Vertical translation: graph moves kk units up; new horizontal asymptote is y=ky = k
y-intercept
Substitute x=0x = 0: y=abh+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).

Reading the graph's features

The four numbers in y=abxh+ky = a \cdot b^{x - h} + k each control a visible feature, which is why a sketch can be built directly from the equation. The base bb sets the direction (growing if b>1b > 1, decaying if 0<b<10 < b < 1) and the steepness. The constant kk sets the horizontal asymptote and the limiting value, hh slides the curve sideways, and aa scales and possibly flips it. Because the exponential never reaches its asymptote, the range is open on that side: y>ky > k when a>0a > 0 and y<ky < k when a<0a < 0.

Why the asymptote sits at y=ky = k

As the exponent runs to the end of the domain that drives bxhb^{x - h} toward zero (negative infinity for b>1b > 1), the term abxha \cdot b^{x - h} vanishes and only the constant kk remains, so the curve flattens toward the line y=ky = k. The asymptote is therefore read straight off the constant term, and forgetting to shift it when k0k \neq 0 is the most common sketching error.

Solving graphically and algebraically

To solve abxh+k=ma \cdot b^{x - h} + k = m, isolate the exponential first, then either rewrite both sides with the same base and equate exponents, or take logarithms (the next dot point). Graphically, the solution is the xx-coordinate where the curve crosses the horizontal line y=my = m, which exists only when mm lies on the correct side of the asymptote.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20224 marksPaper 1 (technique). For f(x)=32x14f(x) = 3 \cdot 2^{x - 1} - 4, determine (a) the yy-intercept, (b) the horizontal asymptote, (c) the value of xx for which f(x)=8f(x) = 8.
Show worked answer →

(a) f(0)=3214=1.54=2.5f(0) = 3 \cdot 2^{-1} - 4 = 1.5 - 4 = -2.5, so the yy-intercept is (0,2.5)(0, -2.5).

(b) As xx \to -\infty, 2x102^{x - 1} \to 0, so f(x)4f(x) \to -4; the asymptote is y=4y = -4.

(c) 32x14=83 \cdot 2^{x - 1} - 4 = 8 gives 2x1=4=222^{x - 1} = 4 = 2^2, so x1=2x - 1 = 2 and x=3x = 3.

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

QCAA 20234 marksPaper 2 (complex familiar). The graph of y=a2x+cy = a \cdot 2^x + c has horizontal asymptote y=5y = 5 and passes through (0,8)(0, 8). (a) Determine aa and cc. (b) Determine the value of xx where y=21y = 21.
Show worked answer →

(a) The asymptote is y=cy = c, so c=5c = 5. At (0,8)(0, 8): a20+5=8a \cdot 2^0 + 5 = 8, so a=3a = 3. Thus y=32x+5y = 3 \cdot 2^x + 5.

(b) 21=32x+521 = 3 \cdot 2^x + 5 gives 2x=1632^x = \dfrac{16}{3}... solving, 2x=1632^x = \dfrac{16}{3}, so x=log2 ⁣(163)=4log232.42x = \log_2\!\left(\dfrac{16}{3}\right) = 4 - \log_2 3 \approx 2.42.

Markers reward reading cc from the asymptote, solving for aa at the intercept, and isolating the exponential before taking a logarithm.

Related dot points