Determine The Properties Or Characteristics Of The Function Given By:$\[ F(x) = \frac{2}{x-3} + 1 \\]

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Determine the Properties or Characteristics of the Function Given by f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1

In mathematics, functions are a crucial concept that helps us describe the relationship between variables. A function is a relation between a set of inputs (called the domain) and a set of possible outputs (called the range). In this article, we will determine the properties or characteristics of the function given by f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1. This function is a rational function, which means it is the ratio of two polynomials.

To determine the domain and range of a function, we need to consider the values of x for which the function is defined and the possible values of y that the function can take.

  • Domain: The domain of a function is the set of all possible input values (x) for which the function is defined. In the case of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, the denominator of the fraction cannot be zero, because division by zero is undefined. Therefore, we need to find the values of x for which the denominator is not equal to zero.

    x−3≠0{ x - 3 \neq 0 }

    Solving for x, we get:

    x≠3{ x \neq 3 }

    Therefore, the domain of the function is all real numbers except 3.

  • Range: The range of a function is the set of all possible output values (y) that the function can take. To determine the range of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, we need to consider the behavior of the function as x approaches positive and negative infinity.

    As x approaches positive infinity, the value of 2x−3\frac{2}{x-3} approaches zero, and the value of f(x)f(x) approaches 1.

    As x approaches negative infinity, the value of 2x−3\frac{2}{x-3} approaches zero, and the value of f(x)f(x) approaches 1.

    Therefore, the range of the function is all real numbers except 1.

A vertical asymptote is a vertical line that the graph of a function approaches but never touches. In the case of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, the vertical asymptote is the line x = 3.

To determine the vertical asymptote, we need to find the values of x for which the denominator of the fraction is equal to zero.

x−3=0{ x - 3 = 0 }

Solving for x, we get:

x=3{ x = 3 }

Therefore, the vertical asymptote is the line x = 3.

A horizontal asymptote is a horizontal line that the graph of a function approaches but never touches. In the case of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, the horizontal asymptote is the line y = 1.

To determine the horizontal asymptote, we need to consider the behavior of the function as x approaches positive and negative infinity.

As x approaches positive infinity, the value of 2x−3\frac{2}{x-3} approaches zero, and the value of f(x)f(x) approaches 1.

As x approaches negative infinity, the value of 2x−3\frac{2}{x-3} approaches zero, and the value of f(x)f(x) approaches 1.

Therefore, the horizontal asymptote is the line y = 1.

The x-intercept of a function is the value of x for which the function is equal to zero. The y-intercept of a function is the value of y for which the function is equal to zero.

To determine the x-intercept of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, we need to set the function equal to zero and solve for x.

2x−3+1=0{ \frac{2}{x-3} + 1 = 0 }

Subtracting 1 from both sides, we get:

2x−3=−1{ \frac{2}{x-3} = -1 }

Multiplying both sides by x−3x-3, we get:

2=−x+3{ 2 = -x + 3 }

Subtracting 3 from both sides, we get:

−x=−1{ -x = -1 }

Dividing both sides by -1, we get:

x=1{ x = 1 }

Therefore, the x-intercept of the function is the point (1, 0).

To determine the y-intercept of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, we need to set the function equal to zero and solve for y.

2x−3+1=0{ \frac{2}{x-3} + 1 = 0 }

Subtracting 1 from both sides, we get:

2x−3=−1{ \frac{2}{x-3} = -1 }

Multiplying both sides by x−3x-3, we get:

2=−x+3{ 2 = -x + 3 }

Subtracting 3 from both sides, we get:

−x=−1{ -x = -1 }

Dividing both sides by -1, we get:

x=1{ x = 1 }

Therefore, the y-intercept of the function is the point (0, 1).

In this article, we determined the properties or characteristics of the function given by f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1. We found that the domain of the function is all real numbers except 3, and the range of the function is all real numbers except 1. We also found that the vertical asymptote of the function is the line x = 3, and the horizontal asymptote of the function is the line y = 1. Finally, we found that the x-intercept of the function is the point (1, 0), and the y-intercept of the function is the point (0, 1).

  • [1] Calculus, James Stewart, 7th edition.
  • [2] Algebra and Trigonometry, Michael Sullivan, 7th edition.
  • [3] Precalculus, Michael Sullivan, 7th edition.
    Q&A: Determining the Properties or Characteristics of the Function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1

In our previous article, we determined the properties or characteristics of the function given by f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1. In this article, we will answer some frequently asked questions related to this function.

Q: What is the domain of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The domain of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 is all real numbers except 3. This is because the denominator of the fraction cannot be zero, and x = 3 makes the denominator equal to zero.

Q: What is the range of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The range of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 is all real numbers except 1. This is because the function approaches 1 as x approaches positive and negative infinity.

Q: What is the vertical asymptote of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The vertical asymptote of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 is the line x = 3. This is because the denominator of the fraction is equal to zero when x = 3, making the function undefined at this point.

Q: What is the horizontal asymptote of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The horizontal asymptote of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 is the line y = 1. This is because the function approaches 1 as x approaches positive and negative infinity.

Q: What is the x-intercept of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The x-intercept of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 is the point (1, 0). This is because the function is equal to zero when x = 1.

Q: What is the y-intercept of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The y-intercept of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 is the point (0, 1). This is because the function is equal to 1 when x = 0.

Q: How do I graph the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: To graph the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1, you can use a graphing calculator or a computer program. You can also use a table of values to plot the function.

Q: What are some real-world applications of the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1?

A: The function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1 has many real-world applications, including:

  • Modeling population growth
  • Modeling chemical reactions
  • Modeling electrical circuits
  • Modeling financial markets

In this article, we answered some frequently asked questions related to the function f(x)=2x−3+1f(x) = \frac{2}{x-3} + 1. We hope that this article has been helpful in understanding the properties and characteristics of this function.

  • [1] Calculus, James Stewart, 7th edition.
  • [2] Algebra and Trigonometry, Michael Sullivan, 7th edition.
  • [3] Precalculus, Michael Sullivan, 7th edition.