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For questions sketch a graph of the function and state its: a) domain and range; b) coordinates of any \(x\) -intercept(s) and \(y\) -intercept; \(c\) ) and the equation of any asymptote(s). $$p(x)=e^{x^{2}}-e$$

Short Answer

Expert verified
Domain: "); Range: "); x-intercepts: (-1,0) and (1,0); y-intercept: (0,1-e); Asymptotes: None.

Step by step solution

01

Analyze the Domain

The domain of a function is all the possible input values (usually x-values) for which the function is defined. Since the function is \( p(x) = e^{x^2} - e \), and exponential functions are defined for all real numbers, the domain is \( (-\infty, \infty) \).
02

Determine the Range

To find the range, observe the behavior of \( e^{x^2} \). This is always positive since the exponential function grows rapidly as \( x \) increases in magnitude, whether positive or negative. The smallest value of \( e^{x^2} \) occurs at \( x = 0 \), which is 1. Thus, \( p(x) = e^{x^2} - e \) takes on values from \( 1 - e \) to infinity. So, the range is \( [1-e, \infty) \).
03

Find x-Intercepts

The x-intercepts occur where \( p(x) = 0 \). Set the function equal to zero: \( e^{x^2} - e = 0 \). Solve by adding \( e \) to both sides and then taking the natural logarithm: \( e^{x^2} = e \), resulting in \( x^2 = 1 \). Thus, the x-intercepts are \( x = -1 \) and \( x = 1 \).
04

Find y-Intercept

The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the function: \( p(0) = e^{0^2} - e = 1 - e \). Therefore, the y-intercept is at \( (0, 1-e) \).
05

Determine Asymptotes

Exponential functions are typically characterized by asymptotes along the horizontal axis, but in this case \( e^{x^2} \) minus a constant does not lead to horizontal or vertical asymptotes. As \( x \to \pm\infty \), \( p(x) \to \infty \), indicating there are no asymptotes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain and Range
Understanding the domain and range is a fundamental aspect of graphing functions. In general, the **domain** of a function includes all possible input values, often represented as the x-values of the graph. For the function \( p(x) = e^{x^2} - e \), there are no restrictions on the x-values because the function includes the exponential term \( e^{x^2} \), which is defined for all real numbers. Thus, the domain for this function is all real numbers, expressed in mathematical notation as \( (-\infty, \infty) \).

The **range** of a function, on the other hand, denotes all possible output values or y-values. For our function, the term \( e^{x^2} \) plays a crucial role in determining the range. Since \( e^{x^2} \) is always positive and increases without bound as \( x \) increases, the smallest output of \( p(x) \) occurs when \( x = 0 \), where \( e^{x^2} = 1 \). Substituting this into the function gives us \( p(0) = 1 - e \), establishing the minimum value of the range. Therefore, the range of \( p(x) \) is \( [1-e, \infty) \), indicating that while the function can achieve values starting from \( 1-e \), it can increase to infinity.
Intercepts
Intercepts are key points where the graph of a function crosses the axes. Understanding intercepts helps in sketching the general shape of the graph. There are two types of intercepts: **x-intercepts** and **y-intercepts**.

**X-intercepts** occur at points where the graph crosses the x-axis, meaning the function's value is zero at these points. To find the x-intercepts of the function \( p(x) = e^{x^2} - e \), we set \( p(x) = 0 \). Solving this equation leads to \( e^{x^2} = e \), resulting in \( x^2 = 1 \). Therefore, the x-intercepts are at \( x = -1 \) and \( x = 1 \), giving us the points \((-1, 0)\) and \((1, 0)\).

**Y-intercepts** occur at the point where the graph crosses the y-axis. This happens when \( x = 0 \). Substituting into \( p(x) = e^{x^2} - e \) gives \( p(0) = 1 - e \). Thus, the y-intercept is at the point \((0, 1-e)\). Intercepts provide significant landmarks for sketching the curve of functions.
Asymptotes
Asymptotes are lines that the graph of a function approaches but never touches. They can be either horizontal, vertical, or even slant asymptotes, each serving as a guiding line for the behavior of functions, particularly as values approach infinity.

In the case of our function \( p(x) = e^{x^2} - e \), we look to identify any potential asymptotes. **Vertical asymptotes** aren't present here because exponential functions like \( e^{x^2} \) are defined everywhere—there's no division by zero or logarithms of negative numbers to create breaks. Similarly, **horizontal asymptotes** are typically tied to rational functions or diminishing terms at infinity, neither of which apply to this scenario. As \( x \to \pm \infty \), \( e^{x^2} \) grows without bound, pulling \( p(x) \) to infinity, meaning there isn't a horizontal asymptote.

The absence of asymptotes in \( p(x) \) suggests a curve that continues to rise as it stretches out indefinitely in either direction, a characteristic of many exponential-related functions.

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