Problem 14 - Algebra 2

Question

Write an exponential function \(y=a b^{x}\) for a graph that includes the given points. $$ (-3,24),(-2,12) $$

Step-by-Step Solution

Answer

The exponential function that includes the given points (-3,24) and (-2,12) is \(y = 192 \cdot 2^{x}\).

Step 1: Plug the given points into the exponential function

Plugging the values from the point (-3,24) into the equation results in \(24 = a \cdot b^{-3}\). Similarly, for the point (-2,12), the equation derived will be \(12 = a \cdot b^{-2}\). These give us two equations from which the values of 'a' and 'b' can be determined.

Step 2: Solve the system of equations

We have a system of two equations as \[ \begin{align*} 24 = a \cdot b^{-3} \ 12 = a \cdot b^{-2} \end{align*} \] Divide the second equation by the first to find the value of 'b'. The 'a' variable will cancel out, leaving \[ \begin{align*} \frac{12}{24} = \frac{b^{-2}}{b^{-3}} \end{align*} \] Solving for 'b', the result is 'b = 2'. Once 'b' is known, it can be substituted back in any of the equations to find 'a'. Using the first equation: \[ \begin{align*} 24 = a \cdot 2^{-3} \end{align*} \] gives 'a = 192'

Step 3: Write out the exponential function

Having found the values of 'a' and 'b', the exponential function can now be written as \(y = 192 \cdot 2^{x}\).

Key Concepts

Exponential Equations

Exponential equations are mathematical expressions where variables appear as exponents. This means they follow the general form of \( y = a \cdot b^x \). These equations are powerful in modeling growth and decay in various real-life scenarios, like finance or biology.
Understanding exponential functions begins with recognizing the base \( b \) and the coefficient \( a \). The base \( b \) determines the rate of growth or decay:

• If \( b > 1 \), the function represents exponential growth.
• If \( b < 1 \), it signifies exponential decay.

The coefficient \( a \) reflects the initial or starting value of the function – think of it as the function's baseline when \( x = 0 \). Importantly, exponential equations differ from polynomial ones because of the variable in the exponent, which makes them grow or shrink at increasing rates.
Working with exponential equations often involves solving for unknowns by using logarithmic properties or converting the equations into manageable algebraic expressions. This turns potentially complex problems into more solvable ones.

Solving Systems of Equations

When dealing with exponential functions involving two points, as in our exercise, a common method is to set up a system of equations. This technique allows us to extract values for unknown constants from given points.
In our exercise, plugging the points \((-3, 24)\) and \((-2, 12)\) into the function form \(y = a \cdot b^x\), results in two equations:

• \(24 = a \cdot b^{-3}\)
• \(12 = a \cdot b^{-2}\)

Solving begins by manipulating these equations to isolate or eliminate variables. A strategic approach is dividing one equation by the other:

• \(\frac{12}{24} = \frac{b^{-2}}{b^{-3}}\)

This division allows for cancellation of the \(a\) variable, simplifying the process to finding \(b\). After determining \(b\), substitute back into one of the original equations to find \(a\). Always remember, solving systems can be less straightforward than solving individual equations, but with practice, it becomes a manageable and systematic process.

Algebra Problem Solving

Algebra is a fundamental branch of mathematics, essential for problem solving. In the context of exponential functions, it involves manipulating expressions to find the unknown variables.
Key strategies in solving algebraic problems include:

• Understanding the type of equation (e.g., exponential) and its features.
• Breaking down complex problems into smaller, more manageable parts.
• Applying algebraic rules, such as division and power rules, to simplify expressions.

In our example, the initial challenge was to determine the constants \(a\) and \(b\) in the exponential function. By creating a system of equations and applying basic algebraic manipulations—like dividing equations to cancel out terms—we found these constants.
Successful algebra problem solving often involves recognizing which techniques or transformations will most directly lead to a solution. With exponential functions, understanding the interplay between coefficients and exponents is critical. This comprehension aids in rewriting and solving expressions efficiently, leading to clearer and more accessible solutions.