Question

In Exercises, sketch the graph of the function. $$ j(x)=e^{-x+2} $$

Short answer

The graph of the function \(j(x) = e^{-x+2}\) will look like an 'n' shape whose bottom is shifted two units to the right from the origin, it starts at y = \(e^2\), and it asymptotically approaches y = 0 as x approaches either infinity or negative infinity.

Step-by-step solution

Identify key features of the function

First, it's important to recognize that this function is an exponential function, but it's been transformed. Specifically, it's been reflected across the y-axis (due to the negative sign before 'x') and shifted two units to the right (due to the '+2'). This indicates an initial point at \(x = 2\).

Compute the y-intercept

The y-intercept can be found by setting \(x = 0\) in the function: \(j(0) = e^{-0+2} = e^2\). So the y-intercept is at the point \(0, e^2\).

Sketch the function

Start by marking the y-intercept on the graph. The graph decreases as x moves away from 2 in either direction due to the negative exponent, which indicates a reflection over the y-axis. So it asymptotically approaches y = 0 in either direction. Thus, the overall shape will look like an 'n' shape shifted to the right, with the graph decreasing as it moves away from x = 2.

Key concepts

Graph Transformations

Exponential functions can undergo several transformations, altering their shape and position on a graph. The function we are examining is \( j(x) = e^{-x+2} \). This function has two key transformations: a reflection across the y-axis and a horizontal shift.

• The negative sign in front of \(x\) indicates a reflection across the y-axis. This means that the direction of the graph will be reversed. Normally, exponential functions increase rapidly; after reflection, they decrease.
• The "+2" inside the exponent causes a horizontal shift of two units to the right. This shifts the entire graph from its typical starting position.

These transformations affect how the graph relates to the original exponential function \( e^x \). Understanding these changes helps to accurately sketch and interpret the function's behavior across the graph.

Y-Intercept

The y-intercept is the point where the graph intersects the y-axis. For the function \( j(x) = e^{-x+2} \), this occurs when \( x = 0 \). Calculating the y-intercept involves setting \( x \) to 0 in the function, which results in: \[ j(0) = e^{-0+2} = e^2 \] This computation reveals that the y-intercept is at the point \((0, e^2)\). This particular point is vital when sketching the graph as it aids in identifying the graph's initial crossing point on the y-axis. Remember, the y-intercept gives a clear indication of how high the graph starts on the y-axis and serves as a reference point for further transformations.

Asymptotic Behavior

Asymptotic behavior describes how a graph behaves as it moves toward a specific line, known as an asymptote. For most exponential functions, there is a horizontal asymptote that the graph approaches but never quite reaches. In the function \( j(x) = e^{-x+2} \), the graph exhibits this behavior as it approaches the line \( y = 0 \) on both ends.

• The negative exponent indicates that the graph will decrease as \( x \) increases or decreases. This is typical of exponential decay.
• As \( x \) becomes very large in the positive or negative direction, the graph gets closer and closer to the x-axis.

This behavior helps predict and understand what happens at the graph's extremes. While it will never actually touch the x-axis, it gets infinitely close, reflecting one of the key characteristics of exponential functions.