Problem 59 - Algebra 2

Question

Write an exponential equation \(y=a b^{x}\) for a graph that includes the given points. $$ (-1,16),(3,1) $$

Step-by-Step Solution

Answer

The exponential equation that fits the given points (-1,16) and (3,1) is \(y = 2\cdot(0.5)^x\).

Step 1: Substitute the points into the exponential equation

Substitute the pairs of x-y values into the form \(y = ab^x\) for both (-1,16) and (3,1). This gives us two separate equations.\n Equation 1 is obtained from (-1,16), yielding \(16 = ab^{-1}\). \n Equation 2 is obtained from (3,1), yielding \(1 = ab^{3}\).

Step 2: Solve the equations simultaneously to find 'a' and 'b'

Divide equation 1 by equation 2 to solve for 'b'. This gives \(16 = b^{-4}\). Solving this yields \(b = 0.5\).\n Then substitute \(b = 0.5\) into equation 2 to solve for 'a'. This gives \(a = 2\).

Step 3: Substitute 'a' and 'b' into the original formula

The initial exponential equation is \(y = ab^x\). Plug in the values of 'a' and 'b' that we have found, which transforms this equation into \(y = 2\cdot(0.5)^x\).

Key Concepts

Graphing

Graphing exponential equations can often give you a beautiful, smooth curve that helps visualize the behavior of the equation. This makes understanding the relationship between the variables a lot easier.
For an exponential equation like \( y = a b^x \), the graph will show how rapidly or slowly the output \( y \) changes with regard to the input \( x \). Generally, if \( b > 1 \), the graph will depict an increasing function, showing exponential growth. Conversely, if \( 0 < b < 1 \), as is the case in our example where \( b = 0.5 \), the graph will display an exponential decay.
When plotting exponential equations, consider these key points:

• The y-intercept occurs at \( y = a \) when \( x = 0 \).
• The graph will never touch the x-axis as it approaches it infinitely.
• The sign of \( a \) influences the graph's direction; a positive \( a \) keeps the graph above the x-axis, while a negative \( a \) would place it below.

Visualizing the graph with these considerations helps make abstract equations more tangible.

Solving Equations

Solving equations, particularly exponential ones, can sometimes seem tricky, but breaking them down into steps makes it manageable. In our example, we need to find the constants \( a \) and \( b \) by using the points \((-1,16)\) and \((3,1)\).
First, substitute these points into the general form \( y = ab^x \). This gives us two equations: \( 16 = ab^{-1} \) and \( 1 = ab^{3} \). From these, we can solve for \( b \) and \( a \).
The approach involves:

• Using algebraic manipulation. Start by dividing one equation by the other to eliminate one of the variables.
• Plugging the value of \( b \) back into one of the equations to solve for \( a \).

This method allows us to find the unique solution that matches the points provided, which demonstrates how solving such equations can model real-life situations accurately.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, especially when dealing with exponential equations. The idea is to substitute one equation into another to isolate and eventually solve for the unknowns.
In our example, after forming the equations \( 16 = ab^{-1} \) and \( 1 = ab^3 \), we use substitution to solve for \( b \).
Here's how it unfolds:

• Divide the two equations to simplify and solve for \( b \), which results in \( 16 = b^{-4} \). Solve this to find \( b = 0.5 \).
• Then, substitute \( b = 0.5 \) back into the easier equation to solve for \( a \). This gives \( a = 2 \).

This technique neatly isolates each variable by sequential substitution, allowing us to solve the system efficiently and achieve the solution \( y = 2(0.5)^x \). Understanding and practicing this method grants flexibility in tackling different types of equations.