Problem 45 - Precalculus with Limits

Question

Solving a System of Equations Graphically In Exercises \(45-48,\) use a graphing utility to solve the system of equations. Find the solution(s) accurate to two decimal places. $$\left\\{\begin{array}{r}{y=e^{x}} \\ {x-y+1=0}\end{array}\right.$$

Step-by-Step Solution

Answer

Please verify the answer by using a graphing utility and find the intersection point of the two equations. This will provide the solution to the system as \(x\) and \(y\) coordinates, accurate to two decimal places.

Step 1: Graph the First Equation

The first equation is \(y=e^{x}\). To graph this equation, plot \(x\) values from an appropriate range and find the corresponding \(y\) values by applying the exponential function. This will create an upward sloping curve starting from the point (0,1) when using a graphing utility.

Step 2: Graph the Second Equation

The second equation is \(x - y + 1 = 0\). This can be rearranged to \(y = x + 1\) which is a linear function. Plot \(x\) values from the same range as the first equation and find the corresponding \(y\) values by applying the function. This will create a diagonal line starting from the point (0,1).

Step 3: Find the intersection

The solution to the system of equations is where the graphs intersect. Use the graphing utility to find the intersection accurate to two decimal places.

Key Concepts

Graphical Solution

A graphical solution involves solving systems of equations by plotting them on a graph to determine their intersection points. This method is particularly useful because it provides a visual representation of where two equations overlap or meet. In our exercise, we are utilizing a graphing utility to represent both the exponential and linear functions visually.

By graphing, we can identify the intersection, which indicates the values of \(x\) and \(y\) that satisfy both equations simultaneously. It's similar to solving the equations algebraically, but instead, we use visual insight. Additionally, the graphical method can handle any number of intersection points and provides a quick way to estimate solutions.

Exponential Function

An exponential function is a type of mathematical function in the form \(y = a \cdot b^x\), where \(a\) and \(b\) are constants, and \(b\) is the base of the exponential function. In our specific case, the base is \(e\) (approximately 2.718), and the equation is \(y = e^x\).

Exponential functions are characterized by their rapid growth or decay. For the graph of \(y = e^x\), the curve starts slowly for negative values of \(x\) and increases sharply as \(x\) becomes positive. It is critical to understand that these functions typically portray growth processes, such as population or investments over time, where rates of change multiply.

• The curve of \(y = e^x\) is always above the x-axis due to being positive for all \(x\).
• The value of \(y\) approaches infinity as \(x\) increases.
• At \(x = 0\), the function has a value \(y = 1\).

Linear Function

Linear functions are among the simplest forms of functions and are represented by a line on a graph. In our example, the linear function is expressed as \(y = x + 1\). It has a constant rate of change, which means that for every unit increase in \(x\), \(y\) increases by the same factor.

The graph of a linear function is a straight line with a slope and a y-intercept. In the case of \(y = x + 1\), the slope is 1, and the y-intercept is at the point (0, 1). Understanding linear functions is crucial because they serve as the foundation for more complex mathematical concepts.

• Linear functions can be written in the form \(y = mx + b\).
• The line of \(y = x + 1\) has a slope (m) of 1, showing that the line rises one unit up for every unit moved along the x-axis.
• The y-intercept is the value of \(y\) when \(x = 0\).

Intersection Point

The intersection point of two graphs is where the solutions of the equations that form these graphs are the same. In other words, it is the point where both equations have identical \(x\) and \(y\) values.

Finding the intersection of \(y = e^x\) and \(y = x + 1\) means we are looking for a point on the graph where both curves cross. This indicates that at that specific \(x\) and \(y\), both equations are true. For graphical solutions, this is often found using a graphing utility, which provides an accurate representation and identifies intersection points to specified decimal places, like two decimal places in this exercise.

• The intersection point gives the solution to the system of equations.
• Using a graphing utility can quickly find this point by zooming in on the graphs.
• The exact coordinates of the intersection correspond to the exact values needed for the solution.