Math Problem Statement

Given the rational function f(x) = (x - 3)/(x - 2), determine the x-intercept, y-intercept, zeros of the function, vertical asymptote, and horizontal asymptote.

Solution

The image you've uploaded contains a table with rational functions and associated details such as xx-intercept, yy-intercept, zeros of the function, vertical asymptote, and horizontal asymptote. Let’s go over the first function provided in the table:

Function 1: f(x)=x3x2f(x) = \frac{x - 3}{x - 2}

We need to calculate the following:

  1. xx-intercept: The xx-intercept is found by setting f(x)=0f(x) = 0 (i.e., solving for xx when f(x)=0f(x) = 0).

    x3x2=0\frac{x - 3}{x - 2} = 0 This occurs when the numerator is zero, so: x3=0x=3x - 3 = 0 \quad \Rightarrow \quad x = 3

  2. yy-intercept: The yy-intercept is found by setting x=0x = 0 and evaluating f(0)f(0).

    f(0)=0302=32=32f(0) = \frac{0 - 3}{0 - 2} = \frac{-3}{-2} = \frac{3}{2}

  3. Zeros of the function: The zeros of the function are the values of xx where f(x)=0f(x) = 0, which we already calculated for the xx-intercept: x=3x = 3

  4. Vertical asymptote: A vertical asymptote occurs where the denominator is zero, so: x2=0x=2x - 2 = 0 \quad \Rightarrow \quad x = 2 Therefore, there is a vertical asymptote at x=2x = 2.

  5. Horizontal asymptote: To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Since the degrees are the same (both are degree 1), the horizontal asymptote is found by dividing the leading coefficients: y=11=1y = \frac{1}{1} = 1 So, the horizontal asymptote is y=1y = 1.

Summary for f(x)=x3x2f(x) = \frac{x - 3}{x - 2}:

  • xx-intercept: x=3x = 3
  • yy-intercept: y=32y = \frac{3}{2}
  • Zeros of the function: x=3x = 3
  • Vertical asymptote: x=2x = 2
  • Horizontal asymptote: y=1y = 1

Would you like details for the second function, or more clarification?

Here are 5 related questions to deepen your understanding:

  1. How do you find the oblique asymptote for rational functions if the degree of the numerator is greater than the denominator?
  2. How does the behavior of a rational function change near its vertical asymptotes?
  3. What happens to the graph of a rational function when there is a hole in it, and how do you identify it?
  4. Can a rational function have more than one horizontal asymptote?
  5. How do you find slant (or oblique) asymptotes in rational functions?

Tip: The horizontal asymptote of a rational function describes the end behavior of the function as xx \to \infty or xx \to -\infty.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Intercepts
Asymptotes

Formulas

x-intercept: Set f(x) = 0 and solve for x
y-intercept: Evaluate f(0)
Zeros of the function: Values of x where f(x) = 0
Vertical asymptote: Set the denominator equal to 0 and solve for x
Horizontal asymptote: Compare the degrees of the numerator and denominator

Theorems

Properties of Rational Functions

Suitable Grade Level

Grades 9-12