Question

Find an exponential function whose graph has the horizontal asymptote \(y=2\) and contains the points (0,3) and \((1,5) .\)

Short answer

The exponential function is \( y = 3^x + 2 \).

Step-by-step solution

Identify the general form of the exponential function

An exponential function can generally be written as \( y = a \times b^x + c \), where \(c\) is the horizontal asymptote. Given the horizontal asymptote \( y = 2 \), we can refine this to \( y = a \times b^x + 2 \).

Use the point (0,3) in the function

Substitute \(x = 0\) and \(y = 3\) into the refined function. This gives us: \[ 3 = a \times b^0 + 2 \] Since \( b^0 = 1 \), this simplifies to: \[ 3 = a + 2 \] Solving for \(a\): \[ a = 1 \]

Substitute \(a\) back into the function

Now, using \( a = 1 \), the equation becomes: \[ y = 1 \times b^x + 2 \] or more simply: \[ y = b^x + 2 \]

Use the point (1,5) to find \(b\)

Substitute \(x = 1\) and \(y = 5\) into the equation: \[ 5 = b^1 + 2 \] Simplifying: \[ 5 = b + 2 \] Solving for \(b\): \[ b = 3 \]

Write the final exponential function

Substitute \(b = 3\) back into the equation to get the final form of the exponential function: \[ y = 3^x + 2 \]

Key concepts

horizontal asymptote

The horizontal asymptote of a function is a horizontal line that the graph of the function approaches but never actually touches. For exponential functions, the horizontal asymptote helps determine the end behavior of the function. In our given problem, the horizontal asymptote is at \(y=2\). This tells us that as \(x\) becomes very large or very small, the value of \(y\) will get closer and closer to 2, but it will never quite reach it. So, the function values gravitate around \(y = 2\) for extreme values of \(x\).

exponential function form

An exponential function can generally be written in the form \(y = a \times b^x + c\). Here, \(a\) is a coefficient that stretches or compresses the graph vertically. \(b\) is the base of the exponential and it determines the rate of growth or decay. \(c\) is the horizontal asymptote value.
In this exercise, the provided horizontal asymptote \(y = 2\) tells us that \(c = 2\). When substituting the given points and solving for \(a\) and \(b\), we found the function to be \(y = 3^x + 2\).

solving exponential equations

Solving exponential equations involves manipulating the equation to isolate the variable, often using properties of exponents. Let's break down the solution from our original exercise:

First, we use the point (0, 3) to find \(a\):
The equation becomes \(3 = a \times 1 + 2\) so \(a = 1\).
With \(a = 1\), substituting in the second point (1, 5):
The modified equation becomes \(5 = b + 2\), which simplifies to \(b = 3\).
Thus, substituting these findings back into the form, our final function is \(y = 3^x + 2\).

substitution method in algebra

The substitution method in algebra is a technique used to solve systems of equations or verify points against a given function.
Here, we used this method to plug in known values into the general exponential function form to find unknowns like \(a\) and \(b\).
Substituting \(x = 0\) and \(y = 3\) helped us determine \(a\), while using \(x = 1\) and \(y = 5\) enabled us to find \(b\).
This step-by-step substitution ensured our values of \(a\) and \(b\) precisely fit the points provided, resulting in the accurate exponential function \(y = 3^x + 2\).