Determining Rules for Graphs of Exponential and Logarithmic Functions Revision Notes for VCE SSCE Mathematical Methods

Determining Rules for Graphs of Exponential and Logarithmic Functions

When you're given a graph of an exponential or logarithmic function, you often need to find the rule that describes it. This process involves identifying the general form of the function and then using known points on the graph to determine the specific parameter values.

Understanding the general approach

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The key principle is that if you have two unknown parameters in your function rule, you need two points on the graph to find them. By substituting the coordinates of these points into the general form of the function, you create a system of two equations that you can solve simultaneously.

Exponential functions of the form $y = ae^x + b$

General form and parameters

The most common form for transformed exponential functions is:

y = ae^x + b

where:

$a$ controls the vertical stretch or compression
$b$ represents the vertical shift (and gives the horizontal asymptote at $y = b$ )

Method for finding $a$ and $b$

The following steps outline the systematic approach to finding parameters:

Step 1: Identify two points on the graph

Step 2: Substitute each point's coordinates into the general form to create two equations

Step 3: Solve the system of equations (often by subtraction to eliminate one variable)

Step 4: Find the other parameter by substitution

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Worked Example: Finding Parameters for an Exponential Function

Consider an exponential function with rule $y = ae^x + b$ that passes through the points $(0, 6)$ and $(3, 22)$ .

To find the values of $a$ and $b$ :

Substitute the first point $(0, 6)$ :

6 = ae^0 + b

6 = a + b \quad \text{(equation 1)}

Substitute the second point $(3, 22)$ :

22 = ae^3 + b \quad \text{(equation 2)}

Subtract equation (1) from equation (2):

16 = ae^3 - a

16 = a(e^3 - 1)

a = \frac{16}{e^3 - 1}

Substitute back into equation (1):

b = 6 - a

b = 6 - \frac{16}{e^3 - 1}

b = \frac{6e^3 - 22}{e^3 - 1}

Therefore, the function has rule:

y = \left(\frac{16}{e^3 - 1}\right)e^x + \frac{6e^3 - 22}{e^3 - 1}

Logarithmic functions of the form $y = a\log_e(x + b)$

General form and parameters

For transformed logarithmic functions, the common form is:

y = a\log_e(x + b)

where:

$a$ controls the vertical stretch or compression
$b$ represents the horizontal shift (the vertical asymptote occurs at $x = -b$ )

Method for finding $a$ and $b$

The approach is similar to exponential functions, but we use logarithmic properties to simplify.

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Worked Example: Finding Parameters for a Logarithmic Function

Consider a logarithmic function with rule $y = a\log_e(x + b)$ that passes through the points $(5, 0)$ and $(8, 1)$ .

To find the values of $a$ and $b$ :

Substitute the first point $(5, 0)$ :

0 = a\log_e(5 + b) \quad \text{(equation 1)}

Substitute the second point $(8, 1)$ :

1 = a\log_e(8 + b) \quad \text{(equation 2)}

From equation (1):

\log_e(5 + b) = 0

5 + b = e^0

5 + b = 1

b = -4

Substitute $b = -4$ into equation (2):

1 = a\log_e(8 - 4)

1 = a\log_e 4

a = \frac{1}{\log_e 4}

Therefore, the rule is:

y = \frac{1}{\log_e 4}\log_e(x - 4)

Exponential functions of the form $y = Ae^{bt}$

Alternative exponential form

Sometimes exponential functions are written as:

y = Ae^{bt}

This form is particularly useful for modelling growth and decay, where $t$ often represents time.

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When working with this form, division is often the most efficient method for solving the system of equations, as it eliminates the parameter $A$ and allows you to solve for $b$ directly.

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Worked Example: Finding Parameters Using Division Method

Given that $y = Ae^{bt}$ with $y = 6$ when $t = 1$ and $y = 8$ when $t = 2$ , find $A$ and $b$ .

Substitute the first point $(1, 6)$ :

6 = Ae^b \quad \text{(equation 1)}

Substitute the second point $(2, 8)$ :

8 = Ae^{2b} \quad \text{(equation 2)}

Divide equation (2) by equation (1):

\frac{8}{6} = \frac{Ae^{2b}}{Ae^b}

\frac{4}{3} = e^b

b = \log_e\frac{4}{3}

Substitute back into equation (1):

6 = Ae^{\log_e\frac{4}{3}}

Using the property $e^{\log_e a} = a$ :

6 = A \times \frac{4}{3}

A = \frac{6 \times 3}{4} = \frac{18}{4} = \frac{9}{2}

Therefore, the rule is:

y = \frac{9}{2}e^{\left(\log_e\frac{4}{3}\right)t}

Note that this can also be written as:

y = \frac{9}{2}\left(\frac{4}{3}\right)^t

Using technology to solve systems of equations

When solving systems of equations to find parameters, CAS calculators can be very helpful for checking your work and handling complex calculations efficiently.

TI-Nspire method

Navigate to: menu > Algebra > Solve System of Equations > Solve System of Equations

Enter your equations and the calculator will solve for the unknowns.

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Common Calculator Error to Avoid:

Don't use the 'e' from the alpha keys on your calculator, as it will be treated as a variable. Use the exponential function key instead to ensure correct calculations.

Casio ClassPad method

Select the simultaneous equations template, then enter your equations. Select $e$ from the Math1 keyboard and choose the parameters you're solving for from the Var keyboard.

Key tips for success

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Identifying the Correct Form

Look at the graph carefully to identify which form of function you're dealing with:

Horizontal asymptote suggests $y = ae^x + b$ (the asymptote is at $y = b$ )
Vertical asymptote suggests logarithmic function (the asymptote is at $x = -b$ )

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Using Special Points

Strategic point selection can simplify your calculations:

For exponential functions, the $y$ -intercept (where $x = 0$ ) is often helpful because $e^0 = 1$
For logarithmic functions, look for where $y = 0$ (the $x$ -intercept) because this tells you when $\log_e(\text{something}) = 0$

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Essential Logarithmic Properties

Keep these properties in mind when solving:

$\log_e 1 = 0$ (because $e^0 = 1$ )
$e^{\log_e a} = a$ for all $a > 0$

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Choosing Your Solution Method

You can solve systems by:

Subtraction (to eliminate one variable)
Division (particularly useful for exponential forms like $y = Ae^{bt}$ )
Using CAS calculator technology (for verification and complex systems)

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Key Points to Remember:

To find two unknown parameters, you need two points on the graph
For $y = ae^x + b$ , the parameter $b$ gives the horizontal asymptote
For $y = a\log_e(x + b)$ , the vertical asymptote is at $x = -b$
When solving systems, look for efficient methods like division for exponential forms
Use the property $e^{\log_e a} = a$ to simplify expressions
CAS calculators can efficiently solve systems of equations, but ensure you're using the correct exponential function (not the variable 'e')

Determining Rules for Graphs of Exponential and Logarithmic Functions (VCE SSCE Mathematical Methods): Revision Notes