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Year 12 Maths - Methods (Unit 3 & Unit 4) Power, exponential and logarithmic functions

Exponential functions

20 practice questions 0 video lessons Theory + worked examples
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Theory

The exponential function \(y=r^x\) (with base \(r>0\)) passes through \((0,1)\), has the horizontal asymptote \(y=0\), and either grows (\(r>1\)) or decays (\(0<r<1\)) — revised here alongside its transformations \(y=r^{x-h}+k\), solving equations by equating indices, and growth-and-decay models, ready for the calculus of Unit 3.

An exponential function has the form \(y=r^x\), where the base \(r\) is a fixed positive number (\(r>0\), \(r\neq 1\)) and the variable \(x\) sits in the exponent. Because \(r^0=1\), every such graph passes through the \(y\)-intercept \((0,1)\).

As \(x\) moves in one direction the curve races upward, and in the other it flattens towards the \(x\)-axis without ever reaching it — the line \(y=0\) is a horizontal asymptote. The domain is all real \(x\) and the range is \(y>0\).

The base decides the shape: when \(r>1\) the function is increasing (exponential growth); when \(0<r<1\) it is decreasing (exponential decay). Shifting the graph with \(y=r^{x-h}+k\) moves the asymptote to \(y=k\). (The natural base \(e\) and its calculus come later, in Unit 3.)

Key idea. \(y=r^x\) (\(r>0\)): \(y\)-intercept \((0,1)\), asymptote \(y=0\), range \(y>0\). Increasing when \(r>1\), decreasing when \(0<r<1\).
Exponential growth and decay curvesTwo exponential curves on the same axes: y=2^x increases while y=(1/2)^x decreases; both pass through (0,1) and approach the horizontal asymptote y=0. x y y=2ˣ y=(½)ˣ
\(y=2^x\) (\(r>1\), growth) and \(y=\left(\dfrac12\right)^x\) (\(0<r<1\), decay): both through \((0,1)\), asymptote \(y=0\)
Transformed exponential y=2^(x+2)-1The curve y=2 to the power (x plus 2) minus 1 rises from left to right, has y-intercept (0,3) and the horizontal asymptote y=-1 shown as a dashed line. x y y=−1 (0,3)
\(y=2^{x+2}-1\): shifted left and down, asymptote \(y=-1\), \(y\)-intercept \((0,3)\)

The basic exponential function (base \(r>0\)):

\[y=r^x \qquad r>0\]
y=rx

The transformed exponential (horizontal shift \(h\), vertical shift \(k\)):

\[y=r^{x-h}+k \qquad \text{asymptote } y=k\]
y=rx-h+k

The growth-and-decay model over time \(t\), with initial amount \(N_0\):

\[N=N_0\,r^{t} \qquad r>1:\ \text{growth} \quad 0<r<1:\ \text{decay}\]
N=N0rt
Reading the base. A factor \(r=0.85\) keeps \(85\%\) each period — a \(15\%\) decrease; a factor \(r=1.05\) is \(5\%\) growth. Compare the base with \(1\).

How to sketch \(y=r^{x-h}+k\)

  1. Asymptote first. The \(+k\) lifts the whole curve, so the horizontal asymptote is \(y=k\) (not \(y=0\)).
  2. Find the \(y\)-intercept. Substitute \(x=0\) into \(y=r^{x-h}+k\) and simplify.
  3. Set the shape. Increasing when \(r>1\), decreasing when \(0<r<1\); the range is \(y>k\).
Solving exponential equations. Write both sides as a power of the same base, then equate the indices — e.g. \(2^x=32=2^5\) gives \(x=5\). If equal bases are not possible, solve with technology.
Example 1 — Features of \(y=r^x\)
State the \(y\)-intercept, horizontal asymptote, range, and whether \(y=\left(\dfrac14\right)^x\) is increasing or decreasing.
Solution
\(y\)-intercept — substitute \(x=0\):
\(y(0)\)\(=\)\(\left(\dfrac14\right)^{0}\)
\(=\)\(1\)

\(y\)-intercept \((0,1)\).

Asymptote and range — read from the basic shape:

The curve flattens towards the \(x\)-axis, so the asymptote is \(y=0\) and the range is \(y>0\).

Increasing or decreasing — compare the base with \(1\):

Here \(r=\dfrac14\), and \(0<r<1\), so the graph is decreasing (decay).

\(\therefore\) \(y\)-intercept \((0,1)\); asymptote \(y=0\); range \(y>0\); decreasing (decay)
(14)0=1
Example 2 — Transformation
For \(y=2^{x+2}-1\), state the horizontal asymptote, the \(y\)-intercept and the range.
Solution
Rewrite in the form \(y=r^{x-h}+k\):

\(y=2^{x-(-2)}+(-1)\): shifted left \(2\) and down \(1\).

Asymptote — the vertical shift \(k\):
\(\text{asymptote}\)\(:\)\(y=k=-1\)
\(y\)-intercept — substitute \(x=0\):
\(y(0)\)\(=\)\(2^{0+2}-1\)
\(=\)\(2^{2}-1\)
\(=\)\(3\)
Range — above the asymptote (base \(>1\)):
\(\text{range}\)\(:\)\(y>-1\)
\(\therefore\) Asymptote \(y=-1\); \(y\)-intercept \((0,3)\); range \(y>-1\)
Graph of y=2^(x+2)-1Increasing exponential curve with y-intercept (0,3) and dashed horizontal asymptote y=-1. x y y=−1 (0,3)
y=22-1=3
Example 3 — Equate the indices
Solve for \(x\): (a) \(5^x=125\); (b) \(2^x=\dfrac{1}{16}\).
Solution
(a) \(5^{x}=125\)
Write \(125\) as a power of \(5\):
\(5^{x}\)\(=\)\(5^{3}\)
Equate the indices:
\(x\)\(=\)\(3\)
\(\therefore\) \(x=3\)
(b) \(2^{x}=\dfrac{1}{16}\)
Write \(\dfrac{1}{16}\) as a power of \(2\):
\(2^{x}\)\(=\)\(2^{-4}\)
Equate the indices:
\(x\)\(=\)\(-4\)
\(\therefore\) \(x=-4\)
5x=53,x=3
Example 4 — Decay model
A car bought for \(\$36\,000\) loses \(20\%\) of its value each year. Find its value after \(3\) years.
Solution
Find the factor — losing \(20\%\) keeps \(80\%\):
\(r\)\(=\)\(1-0.20=0.8\)
Model — value after \(t\) years (decay):
\(V\)\(=\)\(36\,000\times 0.8^{\,t}\)
Value after \(3\) years — substitute \(t=3\):
\(V(3)\)\(=\)\(36\,000\times 0.8^{3}\)
\(=\)\(36\,000\times 0.512\)
\(=\)\(18\,432\)
\(\therefore\) After \(3\) years the car is worth \(\$18\,432\)
36000×0.83=18432

Common pitfalls

Every \(y=r^x\) passes through \((0,1)\). Because \(r^0=1\) for any base, changing the base makes the curve steeper or flatter but does not move the \(y\)-intercept away from \((0,1)\).
The asymptote moves with \(k\), not \(h\). For \(y=r^{x-h}+k\) the horizontal asymptote is \(y=k\) (the vertical shift), so the range is \(y>k\). The \(h\) only slides the graph sideways.
Growth vs decay is about the base. \(0<r<1\) gives decay, \(r>1\) gives growth. A factor of \(0.85\) is a \(15\%\) decrease per period — not \(85\%\).

Frequently asked questions

What is an exponential function?

It has the form \(y=r^x\), where the base \(r\) is a fixed positive number and the variable \(x\) is the exponent. Every such graph passes through \((0,1)\), has asymptote \(y=0\), domain all real \(x\) and range \(y>0\).

What is the horizontal asymptote of y = r^x?

The line \(y=0\) (the \(x\)-axis): the curve approaches it but never touches it. After a vertical shift \(y=r^{x-h}+k\) the asymptote moves to \(y=k\).

How do you solve an exponential equation with the same base?

Write both sides as a power of the same base, then equate the indices. For example \(2^x=32=2^5\) gives \(x=5\). If equal bases are not possible, solve with technology.

What does y = r^(x-h)+k do to the graph of y = r^x?

\(h\) shifts it horizontally (right if \(h>0\), left if \(h<0\)) and \(k\) shifts it vertically, moving the asymptote to \(y=k\) so the range becomes \(y>k\).

When is an exponential function growth or decay?

When \(r>1\) it increases (growth); when \(0<r<1\) it decreases (decay). In \(N=N_0\,r^{t}\) the base \(r\) decides which.

What does a factor of 0.85 mean in an exponential model?

Multiplying by \(0.85\) keeps \(85\%\) each period, a \(15\%\) decrease (decay). A factor of \(1.05\) is \(5\%\) growth. Compare the base with \(1\).