Finding the Rule of an Exponential Function

When finding the rule of an exponential function from a graph or a table of values, avoid the form |y=a_1(c_1)^{b(x-h)}| since it is equivalent to |y=a_2(c_2)^x|.

However, the role of the parameter |b| of the form |y=a(c)^{bx}| is useful when representing an actual situation, since it corresponds to how the percent change is calculated.

Here is an algebraic proof demonstrating the equations |y=a_1(c_1)^{b(x-h)}| and |y=a_2(c_2)^x| are perfectly equivalent. ||\begin{align}
y &= a_1(c_1)^{b(x-h)}\\
&= a_1\left(\color{blue}{(c_1)^{b}}\right)^{x-h} &&\text{since } a^{mn} = (a^m)^{^{\Large n}}\\
&= a_1\, \large{(\color{blue}{c_2})}\normalsize^{x-h} &&\text{substituting }\color{blue}{(c_1)^{b}} \ \text{with a new variable : }\large\color{blue}{c_2}\\
&= a_1 \dfrac{(c_2)^{x}}{(c_2)^h} &&\text{since } a^{m-n}=\dfrac{a^m}{a^n}\\
&= \color{red}{\dfrac{a_1}{(c_2)^h}}(c_2)^{x} &&\text{by the associative property: }a\times \frac{b}{c}=\dfrac{a}{c} \times b\\
&= \ \ \large \color{red}{a_2}(c_2)^x &&\text{substituting }\color{red}{\dfrac{a_1}{(c_2)^h}} \ \text{with a new variable: }\large\color{red}{a_2}
\end{align}||

For example, the functions |y=24(2)^{3(x-1)}| and |y=3(8)^x| are perfectly equivalent. Here is the proof. ||\begin{align}
y &= 24(2)^{3(x-1)}\\
&= 24\left(2^3\right)^{x-1} &&\text{since } a^{mn} = (a^m)^{^{\Large n}}\\
&= 24(8)^{x-1} &&\text{since } 2^3=8\\
&= 24\dfrac{(8)^x}{(8)^1}&&\text{since } a^{m-n}=\dfrac{a^m}{a^n}\\
&= \dfrac{24}{8}(8)^x &&\text{since }8^1=8\\
&= 3(8)^x &&\text{since } \dfrac{24}{8}=3
\end{align}||

In conclusion, if the graph of the function exists, it is much easier to find the equation |y=3(8)^x| than the other equation. There are only |2| parameters to determine |(a=3| and |c=8)| instead of |4|, which is half the work!

1. Find the value of parameter |a| by substituting the coordinates of the |y|-intercept into the basic equation |y=a(c)^x|.

2. Find the value of the base |c| using the coordinates of the other point.

What is the rule of the curve shown below?

1. Substitute each point into the equation to create a system of equations.

2. Determine the value of the base |c| by the comparison method.

3. Use one of the two points provided to find the value of the parameter |a.|

Determine the rule of the curve passing through the points |\left(2,\dfrac{-9}{2}\right)| and |\left(-2,\dfrac{-8}{9}\right).|

It is more convenient to use the simplified form |(y=a(c)^x+k)| than using the standard form |(y=a(c)^{b(x-h)}+k).| This form is obtained using laws of exponents.

Note: However, the value of parameter |a| in the simplified form is not the same as in the standard form. The same goes for the value of the base |c|.

1. Replace |k| with the value of the asymptote.

2. Substitute in each of the points to create a system of equations.

3. Determine the value of the base |c| using the comparison method.

4. Use one of the two points provided to find the value of the parameter |a.|

In the last example, it is possible to apply another technique for solving a system of equations other than the comparison method. This is an elimination method using division. Look at the following example.

After obtaining the 2 equations |7.5=a(c)^{-1}| and |0.06=a(c)^2,| divide them like so: ||\begin{align} \frac{7.5}{0.06} &= \frac{a(c)^{-1}}{a(c)^2} \\ 125 &= a^{1-1}(c)^{-1-2}\end{align}|| Since |a^{1-1}=a^0=1,| |a| disappears. ||\begin{align}125 &=\frac{1}{(c)^3}\\ (c)^3 &= \frac{1}{125}\\ \sqrt[3]{(c)^3} &= \sqrt[3]{\frac{1}{125}} \\ c &= \frac{\sqrt[3]{1}}{\sqrt[3]{125}} = \frac{1}{5} = 0.2 \end{align}|| Obviously, we get the same result with this strategy.

With the exponential function, there is a multiplying factor between the changes in the dependent variable every time the independent variable increases by |1|. The multiplying factor corresponds to the base of the function.

The following is the table of values of the function |y=2(3)^x-1|.

Note the multiplying factor is |3| and that it corresponds to the base |c| of the exponential function.

1. Find the variations of the dependent variable (the changes in the independent variable must be |1|) to determine the multiplying factor |(c).|

2. Substitute two |(x,y)| coordinate pairs into the equation  |y=a(c)^x+k.|

3. Isolate |k| in both equations.

4. Solve the system of equations algebraically to find the value of the parameter |a.|

5. Replace |a| in either of the two equations to find the value of the parameter |k.|

6. Write the rule of the exponential function.

The rule of the exponential function of the form |y=a(c)^x+k| is shown in the following table of values.

Why does the method presented in the finding the equation of an exponential function of the form |y=a(c)^x+k| work?

The following is the proof.

Without loss of generality, take three consecutive coordinate pairs whose |x|-coordinates increase by |1|, such as |(x_1,y_1),(x_2,y_2),(x_3,y_3)|.

Thus: ||\begin{align} y_1 &= a(c)^{x_1}+k \\ y_2 &= a(c)^{x_2}+k \\ y_3 &= a(c)^{x_3}+k \end{align}||

Calculate the changes in the independent variable. ||\begin{align} y_2-y_1 &= \big(a(c)^{x_2}+k\big)-\big(a(c)^{x_1}+k\big) \\ \Rightarrow\ y_2-y_1 &=a(c)^{x_2}-a(c)^{x_1} \\\\ y_3-y_2 &= \big(a(c)^{x_3}+k\big) - \big(a(c)^{x_2}+k\big) \\ \Rightarrow\ y_3-y_2 &= a(c)^{x_3}-a(c)^{x_2} \end{align}||

At this stage, factor out a common factor from each equation. ||\begin{align} y_2-y_1 &= a\big((c)^{x_2}-(c)^{x_1}\big) \\ y_3-y_2 &= a\big((c)^{x_3}-(c)^{x_2}\big) \end{align}||

Note that |x_2=x_1+1| and |x_3=x_2+1|.

Make substitutions in the two calculated equations. ||\begin{align} y_2-y_1 &= a\big((c)^{x_1+1}-(c)^{x_1}\big) \\ y_3-y_2 &= a\big((c)^{x_2+1}-(c)^{x_2}\big) \end{align}||

Again, factor out a common factor from each equation. ||\begin{align} y_2-y_1 &= a(c)^{x_1}\big(c-1\big) \\ y_3-y_2 &= a(c)^{x_2}\big(c-1\big) \end{align}||

The only thing remaining is to divide the two equations. ||\dfrac{y_3-y_2}{y_2-y_1} = \dfrac{a(c)^{x_2}(c-1)}{a(c)^{x_1}(c-1)}||

||\dfrac{y_3-y_2}{y_2-y_1} = \frac{(c)^{x_2}}{(c)^{x_1}} = \frac{(c)^{x_1+1}}{(c)^{x_1}}||

Note, |\dfrac{(c)^{x_1+1}}{(c)^{x_1}} = \dfrac{c}{1}.|

Therefore: ||\dfrac{y_3-y_2}{y_2-y_1} = c||

This explains the variation in the values in the dependant variable; when the |x|-values are consecutive and differ by |1|, it is possible to find the base |c| of the exponential function in the form |y=a(c)^x+k|.