f(x) = (2x + 3)/(x - 1). Show the vertical asymptote, horizontal asymptote, x-intercept, y-intercept, and demonstrate the function's behavior near the asymptotes.

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Let's analyze the rational function f of x equals two x plus three divided by x minus one. This function has a vertical asymptote at x equals 1, where the denominator equals zero. It also has a horizontal asymptote at y equals 2, which is the ratio of the leading coefficients. The x-intercept occurs at negative three-halves, where the numerator equals zero. The y-intercept is at negative 3, which we find by evaluating the function at x equals zero. Now let's examine the behavior of our function near its asymptotes. Near the vertical asymptote at x equals 1, we observe that as x approaches 1 from the left, the function values decrease without bound, approaching negative infinity. This happens because the denominator approaches zero from the negative side. Conversely, as x approaches 1 from the right, the function values increase without bound, approaching positive infinity. For the horizontal asymptote at y equals 2, as x approaches positive infinity, the function approaches 2 from above. And as x approaches negative infinity, the function approaches 2 from below. This behavior is typical of rational functions with degree of numerator equal to degree of denominator. Let's analyze the asymptotes algebraically. For the vertical asymptote, we examine where the denominator equals zero. Setting x minus 1 equal to zero gives us x equals 1. Since the numerator at this point equals 5, which is not zero, we confirm that x equals 1 is indeed a vertical asymptote. For the horizontal asymptote, we analyze the behavior as x approaches infinity. Taking the limit of our function as x approaches infinity, we can divide both numerator and denominator by x. This gives us the limit of 2 plus 3 over x, divided by 1 minus 1 over x. As x approaches infinity, these fraction terms approach zero, leaving us with 2 divided by 1, which equals 2. This confirms that y equals 2 is our horizontal asymptote. Let's examine the intercepts and domain of our rational function. To find the x-intercept, we set f of x equal to zero. This means the numerator must equal zero, giving us 2x plus 3 equals 0, which solves to x equals negative three-halves. For the y-intercept, we evaluate f of 0, which gives us 3 divided by negative 1, or negative 3. Regarding the domain and range, since the function is undefined at x equals 1, the domain is all real numbers except 1. The range excludes y equals 2, which is the horizontal asymptote that the function approaches but never reaches. As we trace along the function, notice how it changes sign when crossing the x-axis, and how it approaches but never touches the horizontal asymptote at y equals 2. To summarize our analysis of the rational function f of x equals two x plus three divided by x minus one: First, the function has a vertical asymptote at x equals 1, where the denominator equals zero. Second, it has a horizontal asymptote at y equals 2, which the function approaches but never reaches as x approaches infinity. Third, the function crosses the x-axis at negative three-halves and the y-axis at negative 3. Fourth, near the vertical asymptote, the function approaches negative infinity from the left and positive infinity from the right. Finally, the domain consists of all real numbers except 1, and the range includes all real numbers except 2. Understanding these key features helps us fully characterize the behavior of rational functions.