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Year 10 Maths Advanced Non-Linear Relationships, Functions and Graphs

Exponential Functions

20 practice questions 2 video lessons Theory + worked examples

An exponential function has the variable in the index, as in \(y=2\times3^x-5\). Its graph has a horizontal asymptote, here \(y=-5\), and either grows or decays quickly. In NSW Year 10 Advanced maths you graph exponentials of the form \(y=k\times a^x+c\), describe their features and transformations, and use them as models (NESA outcome MA5‑NLI‑P‑01).

You will find intercepts and asymptotes, describe reflections and translations, and find \(k\), \(a\) and \(c\) from points on a curve or from a graph. You will solve equations by writing both sides with the same base and find where two exponential curves cross. Models include a ute losing value, a medical isotope's half-life, algae spreading on a dam, koala numbers in two reserves and a cooling cup of coffee. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.

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Theory

An exponential function has the pronumeral in the index, as in \(y = 2^x\). Its graph rises (growth) or falls (decay) ever more steeply or gently and approaches a horizontal asymptote. This Year 10 Advanced maths page covers the graphs of \(y = k \times a^x + c\), exponential equations, finding the rule from points, and real growth and decay models.

An exponential function has the form \(y = a^x\), where the base \(a \gt 0\) and \(a \ne 1\). If \(a \gt 1\) the graph shows exponential growth (increasing); if \(0 \lt a \lt 1\) it shows exponential decay (decreasing).

Every graph \(y = a^x\) passes through \((0,\ 1)\), because \(a^0 = 1\). Since \(a^x \gt 0\) for every \(x\), the graph never touches the \(x\)-axis. The line \(y = 0\) is a horizontal asymptote: the curve gets as close to it as you like without reaching it.

\(y = a^{-x}\) is the reflection of \(y = a^x\) in the \(y\)-axis, and \(a^{-x} = \left(\dfrac{1}{a}\right)^x\). Multiplying by \(k\) dilates the graph vertically; adding \(c\) moves it, and its asymptote, \(c\) units up.

In a model \(A = A_0 \times b^t\), \(A_0\) is the starting amount and \(b\) is the growth factor per time period: \(b = 1.05\) is \(5\%\) growth, \(b = 0.92\) is \(8\%\) decay.

This page covers the transformed forms \(y = k \times a^x + c\) and \(y = k \times a^{-x} + c\). It builds on the exponential graphs met earlier.

Exponential growth and decayThe curves y equals 2 to the x, increasing, and y equals 2 to the minus x, decreasing, are reflections of each other in the y-axis. Both pass through (0, 1) and approach the x-axis, their horizontal asymptote, without touching it. x y βˆ’2 2 1 4 O y = 2Λ£ y = 2⁻ˣ
\(y = a^x\) passes through \((0,\ 1)\) and has asymptote \(y = 0\). Replacing \(x\) by \(-x\) reflects it in the \(y\)-axis.
Graph of y = 4 times 2 to the x, minus 3The curve y equals 4 times 2 to the x minus 3 crosses the y-axis at 1, which is k plus c, and approaches the dashed horizontal asymptote y equals minus 3 to the left. x y O y-intercept k + c asymptote y = c y = 4 Γ— 2Λ£ βˆ’ 3
For \(y = k \times a^x + c\): the asymptote is \(y = c\) and the \(y\)-intercept is \(k + c\).

The transformed exponential and its key features:

\[y = k \times a^x + c \qquad \text{asymptote } y = c \qquad y\text{-intercept } k + c\]
y=k×ax+c

Index laws used to match bases:

\[(a^m)^n = a^{mn} \qquad a^{-n} = \dfrac{1}{a^n} \qquad a^{m + n} = a^m \times a^n\]
(am)n=amn
Equal bases, equal indices. If \(a^m = a^n\) (with \(a \gt 0\), \(a \ne 1\)) then \(m = n\). So \(4^x = 32\) becomes \(2^{2x} = 2^5\), giving \(x = \dfrac{5}{2}\).
Range. \(y = k \times a^x + c\) with \(k \gt 0\) has range \(y \gt c\); with \(k \lt 0\) the graph is reflected in its asymptote and the range is \(y \lt c\). A horizontal line \(y = d\) meets the graph once if \(d\) is in the range and never otherwise.
Half-life. \(A = A_0 \times \left(\dfrac{1}{2}\right)^{t/h}\) halves every \(h\) time units. After \(3h\) it is \(\dfrac{1}{8}\) of the start.

How to sketch \(y = k \times a^x + c\)

  1. Asymptote: draw the dashed line \(y = c\).
  2. \(y\)-intercept: substitute \(x = 0\) to get \(k + c\).
  3. Direction: increasing if \(a \gt 1\) and \(k \gt 0\); a negative index (\(a^{-x}\)) or a base below \(1\) reverses it; a negative \(k\) flips it to the other side of the asymptote.
  4. \(x\)-intercept (if \(0\) is in the range): solve \(k \times a^x + c = 0\) using a common base.

Problem types

  1. Solving equations: rewrite both sides with the same base (\(625 = 5^4\), \(\dfrac{1}{16} = 2^{-4}\), \(8^{x + 1} = 2^{3x + 3}\)) and equate indices.
  2. Finding the rule from points: \((0,\ y_0)\) gives \(k + c\) at once. For \(y = k \times a^x\) with two other points, divide one equation by the other to remove \(k\). For \(y = a^x + c\), use \((0,\ y_0)\) to find \(c\) first. If the asymptote is shown, it gives \(c\).
  3. Matching graphs and equations: check the asymptote, the \(y\)-intercept and whether it increases or decreases.
  4. Transformations: reflect in the \(y\)-axis by replacing \(x\) with \(-x\); translate up by adding to the whole rule; translate the asymptote with it.
  5. Two exponentials: to find where they meet, match the bases. If one is \(2^{x + 1}\) and the other contains \(2^x\), write \(2^{x + 1} = 2 \times 2^x\) and solve for \(2^x\) first.
  6. Models: substitute to find a value; use a table of whole-number values to find when a model first passes a target. For two models, tabulate both side by side. A model like \(T = 20 + 70 \times 0.95^t\) approaches \(20\) (its asymptote) but never reaches it.
Example 1 β€” Features of an exponential graph
For \(y = 2^{x + 1} - 8\), find the \(y\)-intercept, the \(x\)-intercept, the horizontal asymptote and the range, then sketch the graph.
Solution

\(y\)-intercept, \(x = 0\): \(y = 2^1 - 8 = -6\).

\(x\)-intercept, \(y = 0\): write \(8\) as a power of \(2\) and equate the indices.

\(2^{x + 1}\)\(=\)\(8 = 2^3\)
\(x + 1\)\(=\)\(3\)
\(x\)\(=\)\(2\)
2x+1=23

\(2^{x + 1} \gt 0\) for every \(x\) and gets close to \(0\) as \(x\) becomes very negative, so \(y\) approaches \(-8\): asymptote \(\boldsymbol{y = -8}\), range \(\boldsymbol{y \gt -8}\). The base \(2 \gt 1\), so the graph increases.

Graph of y = 2 to the (x + 1), minus 8Increasing exponential curve with y-intercept minus 6, x-intercept 2, and dashed horizontal asymptote y equals minus 8. x y 2 O βˆ’6 y = βˆ’8
Example 2 β€” Same base, equal indices
Solve
(i) \(5^{2x - 1} = 125\)
(ii) \(8^x = \dfrac{1}{4}\).
(iii) Find where the curves \(y = 9^x\) and \(y = 3^{x + 3}\) meet.
Solution

Write both sides as powers of the same base, then set the indices equal.

(i) \(125 = 5^3\):

\(2x - 1\)\(=\)\(3\)
\(x\)\(=\)\(\mathbf{2}\)

(ii) \(8 = 2^3\) and \(\dfrac{1}{4} = 2^{-2}\):

\(2^{3x}\)\(=\)\(2^{-2}\)
\(3x\)\(=\)\(-2\)
\(x\)\(=\)\(\mathbf{-\dfrac{2}{3}}\)
23x=2−2

(iii) \(9^x = (3^2)^x = 3^{2x}\):

\(3^{2x}\)\(=\)\(3^{x + 3}\)
\(2x\)\(=\)\(x + 3\)
\(x\)\(=\)\(3\)
\(y\)\(=\)\(9^3 = 729\)

The curves meet at \(\mathbf{(3,\ 729)}\).

Example 3 β€” Finding the rule from two points
The curve \(y = k \times a^x\), where \(a \gt 0\), passes through \((1,\ 10)\) and \((3,\ 250)\).
(i) Find \(a\) and \(k\).
(ii) Find \(x\) when \(y = 1250\).
Solution

(i) Substitute both points:

\(ka\)\(=\)\(10\quad (1)\)
\(ka^3\)\(=\)\(250\quad (2)\)

Divide (2) by (1) so that \(k\) cancels:

\(a^2\)\(=\)\(25\)
\(a\)\(=\)\(5\quad (a \gt 0)\)
\(k\)\(=\)\(10 \div 5 = 2\)
a2=25,a=5,k=2

So \(y = \boldsymbol{2 \times 5^x}\).

(ii)

\(2 \times 5^x\)\(=\)\(1250\)
\(5^x\)\(=\)\(625 = 5^4\)
\(x\)\(=\)\(\mathbf{4}\)
Example 4 β€” Growth model and a table of values
A bacteria culture has \(N = 500 \times 1.3^t\) cells after \(t\) hours.
(i) By what percentage does it grow each hour?
(ii) Find \(N\) after \(3\) hours.
(iii) Use a table of values to find the first whole number of hours after which there are more than \(10\,000\) cells.
Solution

(i) \(1.3 = 1 + 0.3\), so the culture grows by \(\mathbf{30\%}\) each hour.

\(\text{(ii) } N\)\(=\)\(500 \times 1.3^3\)
\(\)\(=\)\(1098.5 \approx \mathbf{1099}\text{ cells}\)

(iii) Try whole numbers of hours near the answer:

\(t\) (hours)\(10\)\(11\)\(12\)
\(N\)\(6\,893\)\(8\,961\)\(11\,649\)

\(N\) first exceeds \(10\,000\) after \(\mathbf{12}\) hours.

Common pitfalls

\(2 \times 3^x\) is not \(6^x\). Apply the index before multiplying: at \(x = 2\), \(2 \times 3^2 = 18\), but \(6^2 = 36\).
The intercept is not always \(1\). Only \(y = a^x\) passes through \((0,\ 1)\). For \(y = k \times a^x + c\) the \(y\)-intercept is \(k + c\).
The asymptote is never reached. \(y = 3^x + 4\) gets close to \(4\) but never equals it, so the range is \(y \gt 4\), with a strict inequality.
Match the base before equating indices. \(25^x = 125\) does not give \(x = 5\). Write \(5^{2x} = 5^3\), so \(x = \dfrac{3}{2}\).

Frequently asked questions

What is an exponential function?

A function where the variable is in the index, such as y = 2 to the power x. With a base greater than 1 it shows growth; with a base between 0 and 1 it shows decay.

What is the asymptote of an exponential graph?

For y = a to the x it is the x-axis, y = 0. For y = k times a to the x plus c, the asymptote moves to y = c. The graph gets closer and closer to this line but never touches it.

How do you solve an exponential equation?

Write both sides as powers of the same base, using index laws, then set the indices equal and solve the resulting equation.

How do you find the equation of an exponential curve from two points?

Substitute each point into y = k times a to the x to get two equations, then divide one by the other so that k cancels. Solve for a, then substitute back to find k.

How do you tell exponential growth from decay?

Look at the base, or the growth factor in a model. A base greater than 1 means growth; a base between 0 and 1 means decay. A negative index such as a to the minus x reverses the direction.